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## S Chand Class 12 ICSE Maths Solutions Chapter 24 The Plane Ex 24(f)

Question 1.

Find the equation of the plane which

(i) passes through P(3, -2, 4) and is perpendicular to a line whose direction ratios are 2, 2, -3;

(ii) passes through P(2, -3, 5) and has the line joining A(1, -3, -5) and B(2, 2, 3) as a normal;

(iii) bisects the line joining (5, -2, 6) and (7, 2, 0) at right angles;

(iv) passes through P(1, -2, -4) and is parallel to the plane 7 x – 4 y + 6 z + 2 = 0;

(v) passes through the points (-8, 6, 0),(0, 12, 0), and (-10, 0, -9);

(vi) passes through the points (6, 2, 3),(3, 3, -2),(2, -2, -1);

(vii) passes through the y-axis and the point (4, 2, -3).

Answer:

(i) Given direction ratios of normal to plane are ∴ < 2, 2, 3 >.

Thus, eqn. of plane through the point (3, -2, 4) and having direction ratios of normal to plane are

∴ < 2, 2, -3 > is given by 2(x – 3) + 2(y + 2) – 3(z – 4) = 0

⇒ 2 x + 2 y – 3 z + 10 = 0 be the required eqn. of plane.

(ii) ∴ directon ratios of normal to required plane are

∴< 2 – 1, 2 + 3, 3 + 5 >

i.e. ∴ < 1, 5, 8 > Thus required eqn. of plane through the point P(2, -3, 5) is given by

1(x – 2) + 5(y + 3) + 8(z – 5) = 0

⇒ x + 5 y + 8 z – 27 = 0

(iii) D ratios of line AB are

< 7 – 5, 2 + 2, 0 – 6 > i.e.

< 2, 4, -6 > Thus A B be normal to required plane. Also the required plane passes through the mid point of AB

i.e. \(\frac{5+7}{2}\), \(\frac{-2+2}{2}\), \(\frac{6+0}{2}\) i.e. (6, 0, 3).

∴ eqn. of plane through the point (6, 0, 3) is given by

1(x – 6) + 2(y – 0) – 3(z – 3) = 0

⇒ x + 2 y – 3 z + 3 = 0

(iv) eqn. of plane parallel to given plane

7 x – 4 y + 6 z + 2 = 0 be given by 7 x – 4 y + 6 z + k = 0

Since eqn. (1) passes through the point P(1,-2,-4).

∴ 7 × 1 – 4 × (-2) + 6(-4) + k = 0

⇒ 7 + 8 – 24 + k = 0 ⇒ k = 9

putting the value of k in eqn. (1) ; we have

7 x – 4 y + 6 z + 9 = 0 be the reqd. eqn. of plane.

(v) Let the eqn. of plane through the point (-8, 6, 0) be given by

a(x + 8) + b(y – 6) + c(z – 0) = 0

Now eqn. (1) passes through the point (0, 12, 0).

∴ 8 a + 6 b + 0 c = 0

⇒ 4 a + 3 b + 0 c = 0

Also, plane (1) passes through the point (-10, 0, -9)

∴ a(-10 + 8) + b(0 – 6) + c(-9 – 0) = 0

⇒ -2 a – 6 b – 9 c = 0

on solving eqn. (2) and eqn. (3); we have

\(\frac{a}{-27-0}\) = \(\frac{b}{0+36}\)

= \(\frac{c}{-24+6}\)

⇒ \(\frac{a}{-27}\)

= \(\frac{b}{36}\) = \(\frac{c}{-18}\)

⇒ \(\frac{a}{3}\) = \(\frac{b}{-4}\)

= \(\frac{c}{2}\) = k (say) where k ≠ 0

∴ a = 3 k ; b = -4 k and c = 2 k

putting the values of a, b and c in eqn. (1); we have

3 k(x + 8) – 4 k(y – 6) + 2 k(z) = 0

⇒ 3 x – 4 y + 2 z + 48 = 0

(vi) Let the eqn. of plane through the point (6, 2, 3) is given by

a(x – 6) + b(y – 2) + c(z – 3) = 0

where ∴ < a, b, c > are the direction rates of normal to plane (1). eqn. (1) passes through the point (3, 3, -2).

a(3 – 6) + b(3 – 2) + c(-2 – 3) = 0

⇒ -3 a + b – 5 c = 0

Also, plane (1) passes through the point (2, -2, -1).

a(2 – 6) + b(-2 – 2) + c(-1 – 3) = 0

a + b + c = 0

⇒ -4 a – 4 b – 4 c = 0

By cross-multiplication method; we have

\(\frac{a}{1+5}\) = \(\frac{b}{-5+3}\)

= \(\frac{c}{-3-1}\)

⇒ \(\frac{a}{6}\) = \(\frac{b}{-2}\)

= \(\frac{c}{-4}\)

⇒ \(\frac{a}{3}\) = \(\frac{b}{-1}\) = \(\frac{c}{-2}\) = k (say); k ≠ 0

∴ a = 3 k ; b = -k ; c = -2 k

putting the values of a, b and c in eqn. (1); we have

3 k(x – 6) – k(y – 2) – 2 k(z – 3) = 0

⇒ 3 x – y – 2 z – 10 = 0 be the required eqn. of plane.

(vii) since y-axis be the line of intersection of x o y plane (i.e. z = 0 ) and y o z plane ( i.e. x = 0 ) is given by

z + k x = 0

Since plane (1) passes through the point (4, 2, -3)

∴ -3 + 4 k = 0

⇒ k = \(\frac{3}{4}\)

∴ from (1); z + \(\frac{3 x}{4}\) = 0

⇒ 3 x + 4 z = 0 be the reqd. plane.

Question 2.

Find the equation of the plane

(i) parallel to the plane 4 x – 4 y + 7 z – 3 = 0 and distant 4 units from the point (4, 1, -2);

(ii) which passes through the point (3, -2, 4) and is perpendicular to each of the planes 7 x – 3 y + z – 5 = 0 and 4 x – y – z + 9 = 0.

(iii) perpendicular to each of the planes 3 x – y + z = 0 and x + 5 y + 3 z = 0 and is at a distance of \(\sqrt{6}\) from the origin ;

(iv) through (2, 2, 2) and (0, -2, 0) and perpendicular to the plane x – 2 y + 3 z – 7 = 0.

Answer:

(i) eqn. of given plane be

4 x – 4 y + 7 z – 3 = 0 …………………….. (1)

Thus eqn. of plane parallel to plane (1) be given by

4 x – 4 y + 7 z + k = 0 …………………….. (2)

also it is given that ⊥ distance of point (4, 1, -2) from given plane (2) = 4 units

\(\frac{|4 \times 4-4 \times 1+7 \times(-2)+k|}{\sqrt{4^2+(-4)^2+7^2}}\) = 4

⇒ \(\frac{|16-4-14+k|}{\sqrt{16+16+49}}\) = 4

⇒ \(\frac{|k-2|}{9}\) = 4

⇒ |k – 2| = 36 ⇒ k – 2 = ± 36

⇒ k = ± 36 + 2

⇒ k = 38, – 34

putting the values of k in eqn. (2); we have

4 x – 4 y + 7 z + 38 = 0 and 4 x – 4 y + 7 z – 34 = 0 be the required eqns. of planes.

(ii) The equations of given planes are ;

7 x – 3 y + z – 5 = 0 …………………….. (1)

4 x – y – z + 9 = 0 …………………….. (2)

Thus the eqn. of plane through the point (3, -2, 4) is given by

a(x – 3) + b(y + 2) + c(z – 4) = 0 …………………….. (3)

where < a, b, c > are the direction ratios of normal to plane (3).

Since the required plane (3) is ⊥ to plane (1) and (2).

∴ 7 a – 3 b + c = 0 …………………….. (4)

4 a – b – c = 0 …………………….. (5)

on solving eqn. (4) and (5) simultaneously using cross multiplication method, we have

\(\frac{a}{3+1}\) = \(\frac{b}{4+7}\)

= \(\frac{c}{-7+12}\)

i.e. \(\frac{a}{4}\) = \(\frac{b}{11}\) = \(\frac{c}{5}\) = k (say)

⇒ a = 4 k ; b = 11 k and c = 5 k

putting the values of a, b, c in eqn. (3); we have

4 k(x – 3) + 11 k(y + 2) + 5 k(z – 4) = 0

⇒ 4 x + 11 y + 5 z – 10 = 0 be the reqd. plane.

(iii) Let the eqn. of required plane be

a x + b y + c z + d = 0 …………………….. (1)

where < a, b, c > be the direction ratios of normal to plane (1).

The eqns. of given planes are

3 x – y + z = 0 …………………….. (2)

x + 5 y + 3 z = 0 …………………….. (3)

and

hiven planes.

Since the plane (1) is ⊥ to both given planes.

∴ 3 a – b + c = 0 …………………….. (4)

a + 5 b + 3 c = 0 …………………….. (5)

on solving (4) and (5) simultaneously

∴ \(\frac{a}{-8}\) = \(\frac{b}{1-9}\) = \(\frac{c}{15+1}\)

⇒ \(\frac{a}{1}\) = \(\frac{b}{1}\)

= \(\frac{c}{-2}\) = k (say) ; where k ≠ 0

∴ a = k ; b = k ; c = -2 k

putting the values of a, b, c in eqn. (1); we have

x + y – 2 z + \(\frac{d}{k}\) = 0 …………………….. (6)

⇒ x + y – 2 z + d = 0

Also it is given that ⊥ distance from (0, 0, 0) to given plane (6) = \(\sqrt{6}\)

\(\frac{\left|0+0-2 \times 0+d^{\prime}\right|}{\sqrt{1^2+1^2+(-2)^2}}\)

= \(\sqrt{6}\)

⇒ d = ± 6 .

∴ from eqn. (6); we have x + y – 2 z ± 6 = 0 be the reqd. equations of planes.

(iv) Let the eqn. of plane through the point (2, 2, 2) is given by

a(x – 2) + b(y – 2) + c(z – 2) = 0 …………………….. (1)

plane (1) passes through the point (0, -2, 0).

∴ a(0 – 2) + b(-2 – 2) + c(0 – 2) = 0

⇒ a + 2 b + c = 0 …………………….. (2)

⇒ -2 a – 4 b – 2 c = 0

Since eqn. (1) is ⊥ to plane

x – 2 y + 3 z – 7 = 0

a – 2 b + 3 c = 0 …………………….. (3)

\(\frac{a}{6+2}\) = \(\frac{b}{1-3}\)

= \(\frac{c}{-2-2}\)

i.e. \(\frac{a}{8}\) = \(\frac{b}{-2}\) = \(\frac{c}{-4}\)

i.e. \(\frac{a}{4}\) = \(\frac{b}{-1}\) = \(\frac{c}{-2}\) = k

∴ a = 4 k ; b = -k ; c = -2 k ; k ≠ 0

putting the values of a, b and c in eqn. (1); we have

4 k(x – 2) – k(y – 2) – 2 k(z – 2) = 0

⇒ 4 x – y – 2 z – 2 = 0 be the required eqn. of plane.

Question 3.

Find the equation of the plane which contains the line of intersection of the planes x + 2 y + 3 z – 4 = 0 and 2 x + y – z + 5 = 0 and is perpendicular to the plane 5 x + 3 y – 6 z + 8 = 0.

Answer:

Given eqns. of planes are ; and

x + 2 y + 3 z – 4 = 0 …………………….. (1)

2 x + y – z + 5 = 0 …………………….. (2)

Thus the eqn. of any plane through the line of intersection of given planes be

(x + 2 y + 3 z – 4) + k(2 x + y – z + 5) = 0

(1 + 2 k) x + (2 + k) y + (3 – k) z – 4 + 5 k = 0 …………………….. (3)

Now plane (3) is ⊥ to given plane

5 x + 3 y – 6 z + 8 = 0

∴ (1 + 2 k) 5 + ( 2 + k) 3 + (3 – k)(-6) = 0 …………………….. (4)

⇒ 5 + 10 k + 6 + 3 k – 18 + 6 k = 0

⇒ 19 k – 7 = 0

⇒ k = \(\frac{7}{19}\)

putting the value of k = \(\frac{7}{19}\) in eqn. (3) ; we get

(1 + \(\frac{14}{19}\)) x + (2 + \(\frac{7}{19}\)) y + (3 – \(\frac{7}{19}\)) z – 4 + \(\frac{35}{19}\) = 0

⇒ 33 x + 45 y + 50 z – 41 = 0 be the reqd. plane.

Question 4.

Find the equation of the plane through the intersection of the planes x + y + z = 1 and 2 x + 3 y – z + 4 = 0 and parallel to the x-axis.

Answer:

The eqn. of any palen through the line if intersection of two given planes x + y + z – 1 = 0 and 2 x + 3 y – z + 4 = 0 is given by

(x + y + z – 1) + k(2 x + 3 y – z + 4) = 0

⇒ (1 + 2 k) x + (1 + 3 k) y + (1 – k) z – 1 + 4 k = 0 …………………….. (1)

Now plane (1) is parallel to x-axis whose direction ratios are < 1, 0, 0 >

∴ Normal to plane (1) is ⊥ to x-axis.

Thus, (1 + 2 k) 1 + (1 + 3 k) 0 + (1 – k) 0 = 0

⇒ 2 k = -1

⇒ k = \(\frac{-1}{2}\)

∴ from (1); we have

\(\frac{-1}{2}\) y + \(\frac{3}{2}\) z – 1 – 2 = 0

⇒ -y + 3 z – 6 = 0

⇒ y – 3 z + 6 = 0 be the required eqn. of plane.

Question 5.

(i) Find the equation of the plane through (2, 3, -4) and (1, -1, 3) and parallel to the x-axis.

(ii) Find the equation of the plane passing through the points (2, 3, 1) and (4, -5, 3) and parallel to the x-axis.

Answer:

(i) The eqn. of any plane through the point (2, 3, -4) is given by

a(x – 2) + b(y – 3) + c(z + 4) = 0 …………………….. (1)

where < a, b, c > are the direction ratios of normal to plane (1).

Also paine (1) passes through the point (1, -1, 3).

∴ a(1 – 2) + b(-1 – 3) + c(3 + 4) = 0

⇒ -a – 4 b + 7 c = 0 ⇒ a + 4 b – 7 c = 0 …………………….. (2)

Also plane (1) is parallel to x-axis.

∴ Normal to plane (1) is ⊥ to x-axis.

∴ a + 0 b + 0 c = 0 …………………….. (3)

on solving eqn. (2) and eqn. (3) simultaneously by cross-multiplication method, we have

\(\frac{a}{0-0}\) = \(\frac{b}{-7-0}\) = \(\frac{c}{0-4}\)

⇒ \(\frac{a}{0}\) = \(\frac{b}{-7}\) = \(\frac{c}{-4}\) = k (say)

∴ a = 0 ; b = -7 k ; c = -4 k ; k ≠ 0

∴ from (1); we have

0(x – 2) – 7 k (y – 3) – 4 k(z + 4) = 0

⇒ -7 y + 21 – 4 z – 16 = 0

⇒ -7 y – 4 z + 5 = 0

⇒ 7 y + 4 z – 5 = 0 which is the required plane.

(ii) eqn. of any plane through the point (2, 3, 1) is given by

a(x – 2) + b(y – 3) + c(z – 1) = 0 …………………….. (1)

where < a, b, c > be the d ratios of normal to plane (1).

Now plane (1) passes through the point (4, -5, 3).

a(4 – 2) + b(-5 – 3) + c(3 – 1) = 0

⇒ 2 a – 8 b + 2 c = 0

⇒ a – 4 b + c = 0 …………………….. (2)

Since plane (1) is parallel to x-axis.

∴ Normal to plane (1) is ⊥ to x-axis whose direction ratios are < 1, 0, 0 > i.e. a + 0 b + 0 c = 0 …………………….. (3)

on solving eqn. (2) and eqn. (3) simultaneously

\(\frac{a}{0}\) = \(\frac{b}{1-0}\) = \(\frac{c}{0+4}\) = k (say); k ≠ 0

⇒ a = 0 k = 0 ; b = k ; c = 4 k

putting the values of a, b and c in eqn. (1); we get

0(x – 2) + k(y – 3) + 4 k(z – 1) = 0

⇒ y + 4 z – 7 = 0

which is the required eqn. of plane.

Question 6.

Find the equation of a plane which is perpendicular to the plane 2 x – 3 y + 6 z + 8 = 0 and passes through the intersection of the planes x + 2 y + 3 z – 4 = 0 and 2 x – y – z + 5 = 0.

Answer:

eqns. of given planes are ;

x + 2 y + 3 z – 4 = 0 …………………….. (1)

2 x – y – z + 5 = 0 …………………….. (2)

and

Thus the eqn. of any plane through the line of intersection of given planes is given by

x + 2 y + 3 z – 4 + k(2 x – y – z + 5) = 0

⇒ (1 + 2 k) x + (2 – k) y + (3 – k) z – 4 + 5 k = 0 …………………….. (1)

Now plane (1) is normal to plane 2 x – 3 y + 6 z + 8 = 0

∴ 2(1 + 2 k) – 3(2 – k) + 6(3 – k) = 0

⇒ 2 + 4 k – 6 + 3 k + 18 – 6 k = 0

⇒ k + 14 = 0

⇒ k = -14

putting the value of k in eqn. (1); we have

-27 x + 16 y + 17 z – 74

⇒ 27 x – 16 y – 17 z + 74 = 0

⇒ 27, which is the reqd. plane.

Question 7.

(i) Find the equation of the plane passing through A(-1, 1, 1) and B(1, 1, 1) and perpendicular to the plane x – 2 y + 2 z = 3.

(ii) Also, find the distance of the point A from the plane x – 2 y + 2 z = 3.

Answer:

(i) Any plane through the point A(-1, 1, 1) be given by

a(x + 1) + b(y – 1) + c(z – 1) = 0 …………………….. (1)

where < a, b, c > are the direction ratios of normal to plane (1).

The point B(1, -1, 1) lies an eqn. (1); we have

a(1 + 1) + b(-1 – 1) + c(1 – 1) = 0

⇒ 2 a – 2 b + 0 c = 0

⇒ a – b + 0 c = 0 …………………….. (2)

Now plane (1) ⊥ to given plane

x – 2 y + 2 z = 3

a – 2 b + 2 c = 0 …………………….. (3)

∴ on solving eqn. (2) and eqn. (3) simultaneously by cross multiplication method, we have

\(\frac{a}{-2-0}\) = \(\frac{b}{0-2}\) = \(\frac{c}{-2+1}\)

i.e. \(\frac{a}{-2}\) = \(\frac{b}{-2}\) = \(\frac{c}{-1}\)

i.e. \(\frac{a}{2}\) = \(\frac{b}{2}\) = \(\frac{c}{1}\) = k (say) ; k ≠ 0

∴ a = 2 k ; b = 2 k and c = k

putting the value of a, b and c in eqn. (1); we have

2 k(x + 1) + 2 k(y – 1) + k(z – 1) = 0

⇒ 2 x + 2 y + z – 1 = 0 be the reqd. plane

(ii) Required distance of A(-1, 1, 1) from x – 2 y + 2 z – 3 = 0

= \(\frac{|-1-2 \times 1+2 \times 1-3|}{\sqrt{1^2+(-2)^2+2^2}}\)

= \(\frac{|-1-2+2-3|}{3}\) = \(\frac{4}{3}\) units

Question 8.

A plane meets the plane x = 0, where x = 0, 2 y – 3 z = 5, and the plane z = 0 where z = 0, 7 x + 4 y = 10. Find the equation to the plane.

Answer:

The eqn. of any plane through the line of intersection of planes x = 0 and 2 y – 3 z – 5 = 0 be given by

2 y – 3 z – 5 + k x = 0 …………………….. (1)

Now plane (1) meets the planes z = 0 and 7 x + 4 y = 10

∴ from (1); 2 y – 5 + k (\(\frac{10-4 y}{7}\)) = 0

i.e. 14 y – 35 + 10 k – 4 k y = 0

i.e. (14 – 4 k) y + 10 k – 35 = 0

∴ 14 – 4 k = 0 and 10 k – 35 = 0

i.e. k = \(\frac{7}{2}\) and k = \(\frac{7}{2}\)

∴ from (1); 2 y – 3 z – 5 + \(\frac{7}{2}\) x = 0

⇒ 4 y – 6 z – 10 + 7 x = 0

which is the required eqn. of plane.

Question 9.

Prove that the plane 2 x + y – 3 z + 5 = 0, 5 x – 7 y + 2 z + 3 = 0, 5 and x + 10 y – 11 z + 12 = 0 have a line in common.

Answer:

The eqn. of plane through the line of intersection of first two planes is given by

2 x + y – 3 z + 5 + k(5 x – 7 y + 2 z + 3) = 0

⇒ (2 + 5 k) x + (1 – 7 k) y + (-3 + 2 k) z + 5 + 3 k = 0 …………………….. (1)

Now plane (1) is identical to given plane

if

x + 10 y – 11 z + 12 = 0

\(\frac{2-15 k}{1}\) = \(\frac{1-7 k}{10}\)

= \(\frac{-3-12 k}{-11}\) = \(\frac{5+3 k}{12}\) …………………….. (2)

From first two fractions ; 20 + 50 k = 1 – 7 k

⇒ 57 k = -19 ⇒ k = \(\frac{-1}{3}\)

putting k = \(\frac{-1}{3}\) in last two fractions ; we have

\(\frac{-3-\frac{2}{3}}{-11}\) = \(\frac{5-1}{12}\)

⇒ \(\frac{1}{3}\) = \(\frac{1}{3}\) which is true.

Thus the given three planes have same line of intersection.

Question 10.

Find the equation of the plane passing through the intersection of the plane.

\(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) = 1 and \(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\)) + 4 = 0 and parallel to x-axis.

Answer:

The eqn. of plane passing through the line of intersection of given planes

\(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) = 1 and \(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\)) + 4 = 0

be given by \(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) – 1] + λ[\(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\)) + 4] = 0

⇒ \(\vec{r}\) [(1 + 2λ) \(\hat{i}\) + (1 + 3λ) \(\hat{j}\) + (1 – λ) \(\hat{k}\)] + 4λ – 1 = 0 …………………….. (1)

Since eqn. (1) is parallel to x-axis.

∴ Normal to plane is ⊥ to x-axis.

∴(1 + 2 λ) 1 + (1 + 3λ) 0 + (1 – λ) 0 = 0

⇒ 1 + 2λ = 0

⇒ λ = \(\frac{-1}{2}\)

∴ from (1) ; we have

\(\vec{r}\) [\(\frac{-\hat{j}}{2}\) + \(\frac{3}{2}\) \(\hat{k}\)] – 3 = 0

Examples:

Question 1.

Show that the equation b y + c z + d = 0 represents a plane parallel to the axis OX. Find the equation to a plane through the points (2, 3, 1), (4, -5, 3) and parallel to OX.

Answer:

The eqn. of given plane be b y + c z + d = 0

∴ direction no’s of normal to plane (1) are < 0, b, c >

Now plane (1) parallel to x-axis whose direction ratios are < 1, 0, 0>

if normal to plane (1) is ⊥ to x-axis i.e. if 0 1 + b 0 + c 0 = 0 if 0 = 0, which is true.

Thus given plane (1) is parallel to x-axis.

Eqn. of any plane through the point (2, 3, 1) is given by

a(x – 2) + b(y – 3) + c(z – 1) = 0 …………………….. (1)

where < a, b, c > be the d ratios of normal to plane (1).

Now plane (1) passes through the point (4, -5, 3).

a(4 – 2) + b(-5 – 3) + c(3 – 1) = 0

⇒ 2 a – 8 b + 2 c = 0

⇒ a – 4 b + c = 0 …………………….. (2)

Since plane (1) is parallel to x-axis.

∴ Normal to plane (1) is ⊥ to x-axis whose direction ratios are < 1, 0, 0> i.e.

a + 0 b + 0 c = 0 …………………….. (3)

on solving eqn. (2) and eqn. (3) simultaneously

\(\frac{a}{0}\) = \(\frac{b}{1-0}\) = \(\frac{c}{0+4}\) = k (say) ; k ≠ 0

⇒ a = 0 k = 0 ; b = k ; c = 4 k

putting the values of a, b and c in eqn. (1); we get

0(x – 2) + k(y – 3) + 4 k(z – 1) = 0

⇒ y + 4 z – 7 = 0 which is the required eqn. of plane.

Question 2.

Find the equations of the planes parallel to the plane 3 x – 6 y + 2 z = 12 and 6 units away from it.

Answer:

The eqn. of plane || to plane 3 x – 6 y + 2 z – 12 = 0 is given by

3 x – 6 y + 2 z + k = 0 …………………….. (1)

Let P(x_{1}, y_{2}, z_{1}) be any point on given plane

3 x – 6 y + 2 z = 12 …………………….. (2)

3 x_{1} – 6 y_{1} + 2 z_{1} = 12 …………………….. (3)

i.e. also it is given that distance between planes (1) and (2) = 6

⇒ \(\frac{|3 x_1-6 y_1+2 z_1+k|}{\sqrt{3^2+(-6)^2+2^2}}\) = 6

⇒ \(\frac{|12+k|}{7}\) = 6

⇒ 12 + k = ± 42

⇒ k = ± 42 – 12

∴ k = 30, – 54 [using (3)]

∴ from (1); 3 x – 6 y + 2 z + 30 = 0 and 3 x – 6 y + 2 z – 54 = 0 be the required eqns. of planes.

Question 3.

Find the equation of the plane passing through the point (2, 3, 4) and making equal intercepts on the axis.

Answer:

The eqn. of any plane making equal intercepts on coordiantes axes be given by

\(\frac{x}{a}\) + \(\frac{y}{a}\) + \(\frac{z}{a}\) = 1 …………………….. (1)

where a be the length of intercept made by plane on coordiante axes.

Since plane (1) passes through the point (2, 3, 4).

2 + 3 + 4 = a

⇒ a = 9

∴ from (1); x + y + z = 9 be the required eqn. of plane.

Question 4.

Show that the four points (0, 4, 3), (-1, -5, -3), (-2, -2, 1) and (1, 1, -1) are coplanar and find the equation of the common plane.

Answer:

The eqn. of any plane through the point (0,4,3) is given by

a(x – 0) + b(y – 4) + c(z – 3) = 0 …………………….. (1)

where < a, b, c > are the direction ratios of normal to plane.

The point (-1, -5, -3) lies on eqn. (1).

∴ a(-1) + b(-5 – 4) + c(-3 – 3) = 0

⇒ a + 9 b + 6 c = 0 …………………….. (2)

The plane (1) passes through the point (-2, -2, 1).

-2 a + b(-2 – 4) + c(1 – 3) = 0

a + 3 b + c = 0

⇒ -2 a – 6 b – 2 c = 0 …………………….. (3)

on solving eqn. (2) and eqn. (3) simultaneously by cross multiplication method, we have

\(\frac{a}{9-18}\) = \(\frac{b}{6-1}\) = \(\frac{c}{3-9}\)

⇒ \(\frac{a}{-9}\) = \(\frac{b}{5}\) = \(\frac{c}{-6}\) = k (say)

∴ a = -9 k ; b = 5 k ; c = -6 k ; k ≠ 0

putting the value of a, b and c in eqn. (1); we have

-9 k x + 5 k(y – 4) – 6 k(z – 3) = 0

9 x – 5 y + 6 z + 2 = 0

⇒ -9 x + 5 y – 6 z – 2 = 0 …………………….. (4)

Also the point (1, 1, -1) lies on plane (4).

if 9 – 5 – 6 + 2 = 0

⇒ 0 = 0, which is true.

Hence the given four points are coplanar.

Question 5.

Find the equation of the plane which is parallel to x-axis and passes through the points (2, 3, 1) and (4, -5, 3).

Answer:

Eqn. of any plane through the point (2,3,1) is given by

a(x – 2) + b(y – 3) + c(z – 1) = 0 …………………….. (1)

where < a, b, c > be the d ratios of normal to plane (1).

Now plane (1) passes through the point (4, -5, 3).

a(4 – 2) + b(-5 – 3) + c(3 – 1) = 0

⇒ 2 a – 8 b + 2 c = 0

⇒ a – 4 b + c = 0 …………………….. (2)

Since plane (1) is parallel to x-axis.

∴ Normal to plane (1) is ⊥ to x-axis whose direction ratios are < 1, 0, 0 >

i.e. a + 0 b + 0 c = 0 …………………….. (3)

on solving eqn. (2) and eqn. (3) simultaneously

\(\frac{a}{0}\) = \(\frac{b}{1-0}\) = \(\frac{c}{0+4}\) = k (say); k ≠ 0

⇒ a = 0 k = 0 ; b = k ; c = 4 k

putting the values of a, b and c in eqn. (1); we get

0(x – 2) + k(y – 3) + 4 k(z – 1) = 0

⇒ y + 4 z – 7 = 0

which is the required eqn. of plane.

Question 6.

Find the equation of the plane passing through to the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2 x + 6 y + 6 z = 9.

Answer:

The eqn. of any plane through the point (2, 2, 1) be given by

a(x – 2) + b(y – 2) + c(z – 1) = 0 …………………….. (1)

where < a, b, c > are the direction ratios of normal to plane (1).

Now plane (1) passes through the point (9, 3, 6).

∴ a(9 – 2) + b(3 – 2) + c(6 – 1) = 0

i.e. 7 a + b + 5 c = 0 …………………….. (2)

Also plane (1) is ⊥ to given plane

2 x + 6 y + 6 z – 9 = 0

2 a + 6 b + 6 c = 0 …………………….. (3)

on solving eqn. (2) and (3) simultaneously

By cross-multiplication method, we have

\(\frac{a}{6-30}\) = \(\frac{b}{10-42}\) = \(\frac{c}{42-2}\)

i.e. \(\frac{a}{-24}\) = \(\frac{b}{-32}\) = \(\frac{c}{40}\)

i.e. \(\frac{a}{3}\) = \(\frac{b}{4}\) = \(\frac{c}{-5}\) = k (say); where k ≠ 0

∴ a = 3 k ; b = 4 k and c = -5 k

∴ from (1); we have

3 k(x – 2) + 4 k(y – 2) – 5 k(z – 1) = 0

⇒ 3 x + 4 y – 5 z – 9 = 0 be the required eqn. of plane.

Question 7.

A plane is passing through the point (2, -3, 1) and perpendicular to the straight line joining the points (3, 4, -1) and (2, -1, 5). Find the equation of the plane.

Answer:

Here D ratios of normal to plane are < 2 – 3, -1 – 4, 5 + 1 > i.e. < -1, -5, 6 > The eqn. of any plane through the point (2, -3, 1) be given by

-1(x – 2) – 5(y + 3) + 6(z – 1) = 0

⇒ – x – 5 y + 6 z – 19 = 0

⇒ x + 5 y – 6 z + 19 = 0

Question 8.

Find the equation of the plane passing through (1, 2, 3) and perpendicular to the straight line \(\frac{x}{-2}\) = \(\frac{y}{4}\) = \(\frac{z}{3}\)

Answer:

The eqn. of given line be \(\frac{x}{-2}\) = \(\frac{y}{4}\) = \(\frac{z}{3}\)

∴ direction ratios of line (1) are < -2, 4, 3 > …………………….. (1)

Since line (1) is ⊥ to required plane.

∴ a ratios of normal to reqd. plane are < -2, 4, 3 >

Hence the eqn. of plane through the point (1, 2, 3) be given by

-2(x – 1) + 4(y – 2) + 3(z – 3) = 0

2 x – 4 y – 3 z + 15 = 0

⇒ -2 x + 4 y + 3 z – 15 = 0

Question 9.

Find the cosine of the angle between the planes x + 2 y – 2 z + 6 = 0 and 2 x + 2 y + z + 6 = 0.

Answer:

Given eqns. of planes are

x + 2 y – 2 z = -6 …………………….. (1)

2 x + 2 y + z = -6 …………………….. (2)

and

2 x + 2 y + z = -6

direction ratios of normal to plane (1) are < 1, 2, -2 > and d ratios of normal to plane (2) are < 2, 2, 1 >

Let θ be the angle between given planes

∴ cosθ = \(\frac{(1 \times 2+2 \times 2-2 \times 1)}{\sqrt{1^2+2^2+(-2)^2} \sqrt{2^2+2^2+1^2}}\) = \(\frac{4}{3 \times 3}\) = \(\frac{4}{9}\)

∴ θ = cos^{-1} \(\frac{4}{9}\)

Question 10.

Find the angle between the line

\(\frac{x-6}{3}\) = \(\frac{y-7}{2}\) = \(\frac{z-7}{-2}\) and the plane x + y + 2 z = 0 .

Answer:

eqn. of given line be \(\frac{x-6}{3}\) = \(\frac{y-7}{2}\) = \(\frac{z-7}{-2}\)

∴ direction ratios of given line are < 3, 2, -2 >

and eqn. of given plane be x + y + 2 z = 0

∴ direction ratios of normal to given plane are < 1, 1, 2 >

Let θ be angle between line and plane then 90° – θ be the angle between normal to plane and given line.

∴ cos(90° – θ ) = \(\frac{3 \times 1+2 \times 1-2 \times 2}{\sqrt{3^2+2^2+(-2)^2} \sqrt{1^2+1^2+2^2}}\)

⇒ sinθ = \(\frac{1}{\sqrt{17} \sqrt{6}}\) = \(\frac{1}{\sqrt{102}}\)

∴ θ = sin^{-1} (\(\frac{1}{\sqrt{102}}\))

Question 11.

Find the equation of a plane through the point (-1, -1, 2) and perpendicular to the planes 3 x + 2 y – 3 z = 1 and 5 x – 4 y + z = 5.

Answer:

The eqn. of plane through a the point (-1 ,-1, 2) be given by

a(x + 1) + b(y + 1) + c(z – 2) = 0 …………………….. (1)

where < a, b, c > are the direction ratios of normal to plane (1).

Since plane (1) is ⊥ to planes 3 x + 2 y – 3 z = 1 and 5 x – 4 y + z = 5

∴ 3 a + 2 b – 3 c = 0 …………………….. (2)

5 a – 4 b + c = 0

and

5 a – 4 b + c = 0 …………………….. (3)

on solving eqn. (2) and (3) simultaneously by cross multiplication method, we have

\(\frac{a}{2-12}\) = \(\frac{b}{-15-3}\) = \(\frac{c}{-12-10}\)

i.e. \(\frac{a}{-10}\) = \(\frac{b}{-18}\) = \(\frac{c}{-22}\)

i.e. \(\frac{a}{5}\) = \(\frac{b}{9}\) = \(\frac{c}{11}\) = k (say) where k ≠ 0

∴ a = 5 k ; b = 9 k and c = 11 k

putting the vaues of a, b and c in eqn. (1); we have

5 k(x + 1) + 9 k(y + 1) + 11 k(z – 2) = 0

⇒ 5 x + 9 y + 11 z – 8 = 0 be the reqd. plane.

Question 12.

Find the equations to the planes passing through the points (0, 4, – 3) and (6, -4, 3) if the sum of their intercepts on the three axes is zero.

Answer:

Let the eqn. of plane (using intercept form) be given by

\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1 …………………….. (1)

It passes through the point (0, 4, – 3) and (6, -4, 3).

∴ \(\frac{0}{a}\) + \(\frac{4}{b}\) – \(\frac{3}{c}\) = 1

⇒ \(\frac{4}{b}\) – \(\frac{3}{c}\) = 1 …………………….. (2)

\(\frac{6}{a}\) – \(\frac{4}{b}\) + \(\frac{3}{c}\) = 1 …………………….. (3)

a + b + c = 0

also

a + b + c = 0 …………………….. (4)

on adding (2) and (3); we get

\(\frac{6}{a}\) = 2 ⇒ a = 3

∴ from (4); b + c = -3 ⇒ c = -3 – b …………………….. (5)

∴ from (2); 4 c – 3 b = b c ⇒ 4(-3 – b) – 3 b = b(-3 – b)

⇒ 2 – 7 b = -3 b – b^{2}

⇒ b^{2} – 4 b – 12 = 0

⇒ (b + 2)(b – 6) = 0

⇒ b = -2, 6

when b = -2 ∴ from (5); c = -3 + 2 = -1

when b = 6 ∴ from (5); c = -3 – 6 = -9

putting the values of a, b and c in eqn. (1); we have

\(\frac{x}{3}\) + \(\frac{y}{-2}\) + \(\frac{z}{-1}\) = 1 and

\(\frac{x}{3}\) + \(\frac{y}{6}\) + \(\frac{z}{-9}\) = 1 be the required eqn’s of planes.

Question 13.

Find the equation of the plane through the point (1, 2, 3) and perpendicular to the planes

x + y + 2 z = 3 and 3 x + 2 y + z = 4

Answer:

The eqn. of any plane through the point (1, 2, 3) be given by

a(x – 1) + b(y – 2) + c(z – 3) = 0 …………………….. (1)

where < a, b, c > are the direction ratios of normal to plane (1).

Now plane (2) is ⊥ to given planes x + y + 2 z = 3 and 3 x + 2 y + z = 4

a + b + 2 c = 0 …………………….. (2)

3 a + 2 b + c = 0 …………………….. (3)

and on solving eqn. (2) and eqn. (3) simultaneusly by cross multiplication method, we have

\(\frac{a}{1-4}\) = \(\frac{b}{6-1}\) = \(\frac{c}{2-3}\)

i.e. \(\frac{a}{-3}\) = \(\frac{b}{5}\) = \(\frac{c}{-1}\) = k (say) where k ≠ 0

∴ a = -3 k ; b = 5 k ; c = -k

putting the values of a, b and c in eqn. (1); we get

-3 k(x – 1) + 5 k(y – 2) – k(z – 3) = 0

⇒ -3 x + 5 y – z – 4 = 0

⇒ 3 x – 5 y + z + 4 = 0 which is the required plane.

Question 14.

Find the coordinates of the point where the line joining the points (1, -2, 3) and (2, -1, 5) cuts the plane x – 2 y + 3 z = 19. Hence, find the distance of this point from the point (5, 4, 1).

Answer:

The eqn. of line through the points (1, -2, 3) and (2, -1, 5) is given by

\(\frac{x-1}{2-1}\) = \(\frac{y+2}{-1+2}\) = \(\frac{z-3}{5-3}\) …………………….. (1)

i.e. \(\frac{x-1}{1}\) = \(\frac{y+2}{1}\) = \(\frac{z-3}{2}\) = t (say)

Any point on line (1) be given by (t + 1, t – 2, 2 t + 3). It is given that this point lies on given plane

x – 2 y + 3 z = 19

⇒ t + 1 – z(t – 2) + 3(2 t + 3) = 19

⇒ 5 t = 5 ⇒ t = 1

∴ required point of intersection be (1 + 1, 1 – 2, 2 + 3) i.e. (2, -1, 5)

Thus required distance of point P(2, -1, 5) from given point (5, 4, 1)

= \(\sqrt{(5-2)^2+(4+1)^2+(1-5)^2}\)

= \(\sqrt{9+25+16}\)

= \(\sqrt{50}\) = 5 \(\sqrt{2}\) units

Question 15.

Find the equation of the plane which contains the line \(\frac{x-1}{2}\) = \(\frac{y+1}{-1}\) = \(\frac{z-3}{4}\) and is perpendicular to the plane x + 2 y + z = 12.

Answer:

Given eqn. of line be \(\frac{x-1}{2}\) = \(\frac{y+1}{-1}\) = \(\frac{z-3}{4}\) …………………….. (1)

The eqn. of any plane through the line (1) be given by

a(x – 1) + b(y + 1) + c(z – 3) = 0 …………………….. (2)

where

2 a – b + 4 c = 0 …………………….. (3)

plane (1) is ⊥ to given plane x + 2 y + z – 12 = 0

∴ a + 2 b + c = 0 …………………….. (4)

on solving eqn. (3) and enq. (4) simultaneously by cross-multiplication method, we have

\(\frac{a}{-1-8}\) = \(\frac{b}{4-2}\) = \(\frac{c}{4+1}\)

i.e. \(\frac{a}{-9}\) = \(\frac{b}{2}\) = \(\frac{c}{5}\) = k (say)

∴ a = -9 k ; b = 2 k ; c = 5 k ; where k ≠ 0

putting the value of a, b, c in eqn. (2); we have

-9 k(x – 1) + 2 k(y + 1) + 5 k(z – 3) = 0

⇒ -9 x + 2 y + 5 z – 4 = 0

9 x – 2 y – 5 z + 4 = 0 which is the reqd. plane of plane.

Question 16.

Find the shortest distance between the lines whose vector equations are

\(\vec{r}\) = (4 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) + λ(\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) and \(\vec{r}\) = (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + μ(3 \(\hat{i}\) + 2 \(\hat{j}\) – 4 \(\hat{k}\))

Answer:

Given eqns. of lines are ;

\(\vec{r}\) = (4 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) + λ(\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) and \(\vec{r}\)

= (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + λ(3 \(\hat{i}\) + 2 \(\hat{j}\) – 4 \(\hat{k}\))

on comparing with \(\vec{r}\)

= \(\vec{a}_1\) + λ \(\overrightarrow{b_1}\) and \(\vec{r}\)

= \(\vec{a}_2\) + λ \(\overrightarrow{b_2}\) we have

\(\overrightarrow{a_1}\) = 4 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\) ;

\(\overrightarrow{a_2}\) = 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)

\(\overrightarrow{b_1}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\) ;

\(\overrightarrow{b_2}\)

= 3 \(\hat{i}\) + 2 \(\hat{j}\) – 4 \(\hat{k}\)

∴ \(\overrightarrow{b_1}\) × \(\vec{b}_2\)

= \(|\begin{array}{rrr}

\hat{i} & \hat{j} & \hat{k}

1 & 2 & -3

3 & 2 & -4

\end{array}|\)

= \(\hat{i}\)(-8 + 6) – \(\hat{j}\)(-4 + 9) + \(\hat{k}\)(2 – 6)

= -2 \(\hat{i}\) – 5 \(\hat{j}\) – 4 \(\hat{k}\)

∴ \(|\vec{b}_1 \times \vec{b}_2|\)

= \(\sqrt{(-2)^2+(-5)^2+(-4)^2}\) = \(\sqrt{4+25+16}\) = \(\sqrt{45}\)

and \(\vec{a}_2\) – \(\vec{a}_1\)

= (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) – (4 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))

= -2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)

Thus (\(\vec{a}_2\) – \(\vec{a}_1\)) \(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\)

= (-2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) (-2 \(\hat{i}\) – 5 \(\hat{j}\) – 4 \(\hat{k}\))

= (-2)(-2) + 2(-5) – 3(-4) = 4 – 10 + 12 = 6

Thus, required S.D between given lines = \(\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1}) \cdot(\overrightarrow{b_1} \times \vec{b}_2)|}{|\overrightarrow{b_1} \times \overrightarrow{b_2}|}\)

= \(\frac{6}{\sqrt{45}}\) = \(\frac{6}{3 \sqrt{5}}\)

= \(\frac{2}{\sqrt{5}}\) units.

Question 17.

Find the equation of the plane passing through the line of intersection of the planes x + 2 y + 3 z – 4 = 0 and 3 z – y = 0 and perpendicular to the plane 3 x + 4 y – 2 z + 6 = 0.

Answer:

The given planes are

x + 2 y + 3 z – 4 = 0 …………………….. (1)

3 z – y = 0 …………………….. (2)

3 x + 4 y – 2 z + 6 = 0 …………………….. (3)

and

The eqn. of any plane through the intersection of plane (1) and plane (2) be given by

x + 2 y + 3 z – 4 + λ(3 z – y) = 0

⇒ x + (2 – λ) y + (3 + 3λ) z – 4 = 0 …………………….. (4)

Also d ratios of normal to plane (4) are < 1, 2, – λ, 3 + 3 λ >

D ratios of normal to plane (3) are given by < 3, 4, -2 >

Since pane (4) is ⊥ plane (3).

∴ 3.1 + 4(2 – λ) – 2(3 + 3λ) = 0

⇒ 3 + 8 – 4 λ – 6 – 6 λ = 0

⇒ 5 – 10λ = 0

⇒ λ = \(\frac{1}{2}\)

putting the value of λ in eqn. (4); we have

x + 2 y + 3 z – 4 + \(\frac{1}{2}\)(3 z – y) = 0

⇒ 2 x + 3 y + 9 z – 8 = 0 be the required plane.

Question 18.

Find the vector equation of the line passing through the point (-1, 2, 1) and parallel to the line \(\vec{r}\) = 2 \(\hat{i}\) + 3 \( \hat{j}\) – \(\hat{k}\) + λ(\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)). Also, find the distance between them.

Answer:

Given vector eqn. of line be \(\vec{r}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\) + λ(\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)) …………………….. (1)

So line (1) has direction ratios < 1, -2, 1 >

Since the required line is || to line (1)

∴ d ratios of required line are < 1, -2, 1 >

Hence required vector eqn. of line passing through the point where P.V

\(\overrightarrow{a_2}\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\) and || to vector \(\vec{b}\) = \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\) be given by

\(\vec{r}\) = \(\overrightarrow{a_2}\) + λ \(\vec{b}\)

= (\(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)) + λ(\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\))

…………………….. (2)

where λ be the parameter

Here \(\vec{a}_1\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\) ;

\(\vec{a}_2\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)

and \(\vec{b}\) = \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)

Thus required distance between || lines =

S.D between lines = \(\frac{|\vec{b} \times(\overrightarrow{a_2}-\overrightarrow{a_1})|}{|\vec{b}|}\) …………………….. (3)

Now, \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\) = -3 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)

∴ \(\overrightarrow{b_1}\) × \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)

= \(|\hat{i} \hat{j} \hat{k}

1 -2 1

-3 -1 2

|\)

= \(\hat{i}\)(-4 + 1) – \(\hat{j}\)(2 + 3) + \(\hat{k}\)(-1 – 6)

= -1 \(\hat{i}\) – 5 \(\hat{j}\) – 7 \(\hat{k}\)

∴ from (3); required distance = \(\frac{|-3 \hat{i}-5 \hat{j}-7 \hat{k}|}{\sqrt{1^2+(-2)^2+1^2}}\) = \(\frac{\sqrt{(-3)^2+(-5)^2+(-7)^2}}{\sqrt{6}}\) = \(\sqrt{\frac{83}{6}}\) units

Question 19.

Find the equation of the plane passing through the points A(2, 1, -3), B(-3, -2, 1) and C(2, 4, -1).

Answer:

The eqn. of any plane through the point (2, 1, -3) be given by

a(x – 2) + b(y – 1) + c(z + 3) = 0 …………………….. (1)

where < a, b, c > are the direction ratios of normal to plane (1).

The point B(-3, -2, 1) lies on eqn. (1); we have

a(-3 – 2) + b(-2 – 1) + c(1 + 3) = 0

⇒ -5 a – 3 b + 4 c = 0 …………………….. (2)

Also the plane (1) passes through the point C(2, 4, -1).

i.e. a(2 – 2) + b(4 – 1) + c(-1 + 3) = 0

i.e. 0 a + 3 b + 2 c = 0 …………………….. (3)

on solving eqn. (2) and eqn. (3) simultaneously using cross multiplication method, we have

\(\frac{a}{-6-12}\) = \(\frac{b}{0+10}\)

= \(\frac{c}{-15-0}\)

i.e. \(\frac{a}{-18}\) = \(\frac{b}{10}\) = \(\frac{c}{-15}\) = k (say);

where k ≠ 0

∴ a = -18 k ; b = 10 k ; c = -15 k

putting the values of a, b and c in eqn. (1) we have

-18 k(x – 2) + 10 k(y – 1) – 15 k(z + 3) = 0

⇒ -18 x + 10 y – 15 z – 19 = 0

18 x – 10 y + 15 z + 19 = 0 which is the required eqn. of plane.

Question 20.

Find the shortest distance between the line

\(\frac{x-8}{3}\) = \(\frac{y+9}{-16}\) = \(\frac{z-10}{7}\) and \(\frac{x-15}{3}\)

= \(\frac{y-29}{8}\) = \(\frac{5-z}{5}\)

Answer:

Equations of given lines are

and

\(\frac{x-8}{3}\) = \(\frac{y+9}{-16}\) = \(\frac{z-10}{7}\) …………………………… (1)

\(\frac{x-15}{3}\) = \(\frac{y-29}{8}\) = \(\frac{z-5}{-5}\) ……………………………. (2)

Let < l, m, n > be the d cosines of line of S.D PQ

Since line P Q is \perp to both given lines (1) and (2).

3 l – 16 m + 7 n = 0 …………………………… (3)

3 l + 8 m – 5 n = 0 …………………………… (4)

on solving eqn. (3) and (4) simultaneously using cross-multiplication methods, we have

\(\frac{l}{80-56}\) = \(\frac{m}{21+15}\)

= \(\frac{n}{24+48}\)

i.e. \(\frac{l}{24}\) = \(\frac{m}{36}\) = \(\frac{n}{72}\)

i.e. \(\frac{l}{2}\) = \(\frac{m}{3}\) = \(\frac{n}{6}\) = k say

∴ l = 2 k ; m = 3 k and n = 6 k; where k ≠ 0

Since l^{2} + m^{2} + n^{2} = 1

⇒ 4 k^{2} + 9 k^{2} + 36 k^{2} = 1

⇒ 49 k^{2} = 1

⇒ k = ± \(\frac{1}{7}\)

∴ l = ± \(\frac{2}{7}\) ; m = ± \(\frac{3}{7}\) and n = ± \(\frac{6}{7}\)

Thus D cosines of line of S.D are < \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{6}{7}\) >

∴ SD = |PQ| = projection of line of AB and PQ

= |l(x_{2} – x_{1}) + m(y_{2} – y_{1}) + n(z_{2} – z_{1})|

= |\(\frac{2}{7}\)(15 – 8) + \(\frac{3}{7}\) (29 + 9) + \(\frac{6}{7}\)(5 – 10)|

= \(\frac{2}{7}\) × 7 + \(\frac{3}{7}\) × 38 + \(\frac{6}{7}\) × (-5)

= \(\frac{14+114-30}{7}\) = 14 units

Alter : In vector form;

\(\vec{a}_1\) = 8 \(\hat{i}\) – 9 \(\hat{j}\) + 10 \(\hat{k}\) ;

\(\vec{a}_2\) = 15 \(\hat{i}\) + 29 \(\hat{j}\) + 5 \(\hat{k}\)

\(\vec{b}_1\) = 3 \(\hat{i}\) – 16 \(\hat{j}\) + 7 \(\hat{k}\) ;

\(\vec{b}_2\) = 3 \(\hat{i}\) + 8 \(\hat{j}\) – 5 \(\hat{k}\)

∴ \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\) = 7 \(\hat{i}\) + 38 \(\hat{j}\) – 5 \(\hat{k}\)

\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\)

= \(|\hat{i} \hat{j} \hat{k}

3 -16 7

3 8 -5

|\)

= \(\hat{i}\) (80 – 56) – \(\hat{j}\)(-15 – 21) + \(\hat{k}\)(24 + 48)

= 24 \(\hat{i}\) + 36 \(\hat{j}\) + 72 \(\hat{k}\)

∴ |\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\)|

= 12(2 \(\hat{i}\) + 3 \(\hat{j}\) + 6 \(\hat{k}\)

= 12 \(\sqrt{2^2+3^2+6^2}\) = 12 × 7 = 84

Thus, (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) (\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\))

= (7 \(\hat{i}\) + 38 \(\hat{j}\) – 5 \(\hat{k}\)) (24 \(\hat{i}\) + 36 \(\hat{j}\) + 72 \(\hat{k}\))

= 7 × 24 + 38 × 36 – 5 × 72

= 12[14 + 114 – 30] = 12 × 98

∴ required S.D between lines

= \(\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1}) (\overrightarrow{b_1} \times \overrightarrow{b_2})|}{|(\overrightarrow{b_1} \times \overrightarrow{b_2})|}\)

= \(\frac{12 \times 98}{84}\)

= 14 units

Question 21.

Find the equation of the plane passing through the line of intersection of the planes x + 2 y + 3 z – 5 = 0 and 3 x – 2 y – z + 1 = 0 and cutting off equal intercepts on the x and z axes.

Answer:

The eqns. of given planes are ;

x + 2 y + 3 z – 5 = 0 …………………………… (1)

3 x – 2 y – z + 1 = 0 …………………………… (2)

and

Thus the eqn. of any plane through the line of intersection of two given planes be given by

(x + 2 y + 3 z – 5) + λ (3 x – 2 y – z + 1) = 0

⇒ (1+3λ) x + (2 – 2λ) y + (3 – λ) z – 5 + λ = 0

⇒ \(\frac{x}{\frac{5-\lambda}{1+3 \lambda}}\) + \(\frac{y}{\frac{5-\lambda}{2-2 \lambda}}\) + \(\frac{z}{\frac{5-\lambda}{3-\lambda}}\) = 1 …………………………… (3)

Here intercepts made by plane (3) on x and z-axis are ; \(\frac{5-\lambda}{1+3 \lambda}\) and \(\frac{5-\lambda}{3-\lambda}\)

According to given condition, we have

\(\frac{5-\lambda}{1+3 \lambda}\) = \(\frac{5-\lambda}{3-\lambda}\)

⇒ (5 – λ)[3 – λ – 1 – 3λ] = 0

⇒ (5 – λ)(2 – 4 λ) = 0

⇒ λ = 5, \(\frac{1}{2}\)

but 5 ≠ λ

∴ λ = \(\frac{1}{2}\)

putting the value of λ in eqn. (3); we have

\(\frac{5}{x}\) + y + \(\frac{5}{2}\) z – \(\frac{9}{2}\) = 0

⇒ 5 x + 2 y + 5 z – 9 = 0

be the required plane.

Question 22.

Find the equation of a line passing through the point (-1, 3, -2) and perpendicular to the lines: \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) and \(\frac{x+2}{-3}\) = \(\frac{y-1}{2}\) = \(\frac{z+1}{5}\)

Answer:

The eqns. of given lines are ;

\(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) …………………………… (1)

and

\(\frac{x+2}{-3}\) = \(\frac{y-1}{2}\) = \(\frac{z+1}{5}\) …………………………… (2)

∴ D ratios of given lines (1) and (2) are < 1, 2, 3 > and < -3, 2, 5 >

Let the direction ratios of required line are < a, b, c >

Since the required line is ⊥ to line (1) and (2)

∴ a + 2 b + 3 c = 0 …………………………… (3)

-3 a + 2 b + 5 c = 0 …………………………… (4)

on solving eqn. (3) and eqn. (4) simultaneously

\(\frac{a}{10-6}\) = \(\frac{b}{-9-5}\) = \(\frac{c}{2+6}\)

i.e. \(\frac{a}{4}\) = \(\frac{b}{-14}\) = \(\frac{c}{8}\)

i.e. \(\frac{a}{2}\) = \(\frac{b}{-7}\) = \(\frac{c}{4}\)

Thus the required eqn. of line throught he point (-1, 3, -2) having d ratios < 2, -7, 4 > be given by \(\frac{x+1}{2}\) = \(\frac{y-3}{-7}\) = \(\frac{z+2}{4}\)

Question 23.

Find the equations of planes parallel to the plane 2 x – 4 y + 4 z = 7 and which are at a distance of five units from the point (3, -1, 2).

Answer:

eqn. of given plane be 2 x – 4 y + 4 z – 7 = 0 …………………………… (1)

∴ eqn. of any plane parallel to plane (1) be given by

2 x – 4 y + 4 z + k = 0 …………………………… (2)

Also given ⊥ distance from P(3, -1, 2) to plane (2) = 5 units

\(\frac{|2 \times 3-4 \times(-1)+4 \times 2+k|}{\sqrt{2^2+(-4)^2+4^2}}\) = 5

⇒ \(\frac{|18+k|}{6}\) = 5

⇒ |18 + k| = 30

⇒ 18 + k = ± 30

∴ k = ± 30 – 18 = 12, -48

∴ from (2); we have

2 x – 4 y + 4 z + 12 = 0 and 2 x – 4 y + 4 z – 48 = 0

which are the required eqns. of planes.

Question 24.

Find the equation of a line passing through the points P(-1, 3, 2) and Q(-4, 2, -2). Also, Q the point R(5, 5, λ) is collinear w,ith the points. P and Q, then find the values of λ

Answer:

Now, direction ratios of the line joining the points P(-1, 3, 2) and Q(-4, 2, -2) are

< -4 + 1, 2 – 3, -2 – 2 >

i.e. < -3, -1, -4 >

i.e. < 3, 1, 4 >

Thus the required eqn. of line through the point (-1, 3, 2) and is having direction ratios < 3, 1, 4 > be given by

\(\frac{x+1}{3}\) = \(\frac{y-3}{1}\) = \(\frac{z-2}{4}\) …………………………… (1)

Since the point R(5, 5, λ) is collinear with P and Q.

∴ P, Q and R lies on same straight line.

∴ R(5, 5, λ) lies on (1); we get

\(\frac{5+1}{3}\) = \(\frac{5-3}{1}\) = \(\frac{\lambda-2}{4}\)

⇒ 2 = 2 = \(\frac{\lambda-2}{4}\)

λ – 2 = 8

⇒ λ = 10

Question 25.

Find the equation of the plane passing through the points (2, -3, 1) and (-1, 1, -7) and perpendicular to the plane x – 2 y + 5 z + 1 = 0.

Answer:

The eqn. of any plane through the point (2, -3, 1) be given by

a(x – 2) + b(y + 3) + c(z – 1) = 0 …………………………… (1)

where < a, b, c > are the direction ratios of normal to plane (1).

Now plane (1) passes through the point (-1, 1, -7).

a(-1 – 2) + b(1 + 3) + c(-7 – 1) = 0

-3 a + 4 b – 8 c = 0 …………………………… (2)

i.e. x – 2 y + 5 z + 1 = 0

a – 2 b + 5 c = 0 …………………………… (3)

on solving eqn. (2) and eqn. (3) simultaneously by cross-multiplication method, we have

\(\frac{a}{20-16}\) = \(\frac{b}{-8+15}\)

= \(\frac{c}{6-4}\)

i.e. \(\frac{a}{4}\) = \(\frac{b}{7}\)

= \(\frac{c}{2}\) = k (say ); where k ≠ 0

∴ a = 4 k ; b = 7 k ; c = 2 k

putting the value of a, b and c in eqn. (1); we have

4 k(x – 2) + 7 k(y + 3) + 2 k(z – 1) = 0

⇒ 4 x + 7 y + 2 z + 11 = 0 be the reqd. eqn. of plane.

Question 26.

Find the equation of the plane passing through the intersection of the planes:

x + y + 1 = 0 and 2 x – 3 y + 5 z – 2 = 0 and the point (-1, 2, 1).

Answer:

The eqn. of any plane through the line of intersection of given planes be given by

x + y + z – 1 + k(2 x – 3 y + 5 z – 2) = 0 …………………………… (1)

Now plane (1) passes through the point (-1, 2, 1).

∴ -1 + 2 + 1 – 1 + k(-2 – 6 + 5 – 2) = 0

⇒ 1 – 5 k = 0

⇒ k = 1 / 5

putting the value of k in eqn. (1); we have

x + y + z – 1 + \(\frac{1}{5}\)(2 x – 3 y + 5 z – 2) = 0

⇒ 7 x + 2 y + 10 z – 7 = 0

which is the reqd. eqn. of plane.

Question 27.

Find the shortest distance between the lines

\(\vec{r}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\) + λ(2

\(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) and \(\vec{r}\) = 2 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) + µ(4 \(\hat{i}\) + 6 \(\hat{j}\) + 8 \(\hat{k}\))

Answer:

Given lines are ;

and

\(\vec{r}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) …………………………… (1)

\(\vec{r}\) = 2 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) + µ(4 \(\hat{i}\) + 6 \(\hat{j}\) + 8 \(\hat{k}\))

⇒ \(\vec{r}\) = 2 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) …………………………… (2)

where λ = 2µ

on comparing eqn. (1) and eqn. (2) with

\(\vec{r}\) = \(\overrightarrow{a_1}\) + λ\(\vec{b}\) and \(\vec{r}_2\) = \(\overrightarrow{a_2}\) + λ \(\vec{b}\)

∴ \(\vec{a}_1\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\) ;

\(\vec{a}_2\) = 2 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\)

and \(\vec{b}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)

⇒ \(\vec{a}_2\) – \(\vec{a}_1\) = \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)

∴ \(\vec{b}\) × (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\))

= \(|\hat{i} \hat{j} \hat{k}

2 3 4

1 2 2

|\)

= \(\hat{i}\)(6 – 8) – \(\hat{j}\)(4 – 4) + \(\hat{k}\)(4 – 3)

= -2 \(\hat{i}\) + 0 \(\hat{j}\) + \(\hat{k}\)

Thus |\(\vec{b}\) × (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\))|

= \(\sqrt{(-2)^2+0^2+1^2}\) = \(\sqrt{5}\) and

\(\vec{b}\) = \(\sqrt{2^2+3^2+4^2}\) = \(\sqrt{29}\)

∴ required S.D between parallel lines = \(\frac{|\vec{b} \times (\overrightarrow{a_2}-\overrightarrow{a_1})|}{|\vec{b}|}\)

= \(\sqrt{\frac{5}{29}}\) units

Question 28.

Find the image of the point (2, -1, 5) in the line \(\frac{x-11}{10}\) = \(\frac{y+2}{-4}\) = \(\frac{z+8}{-11}\). Also, find the length of the perpendicular from the point (2, -1, 5) to the line.

Answer:

Let P(2, -1, 5) be the given point and eqn. of given line AB be

\(\frac{x-11}{10}\) = \(\frac{y+2}{-4}\) = \(\frac{z+8}{-11}\) = t (say)

So any point on given line AB be M(10 t + 11, – 4 t – 2, -11 t – 8)

Let this point M} be the foot of ⊥ drawn from P on AB.

Produce PM to P s.t PM = MP.

Then P be the image of P in AB and let (α, β, γ) be its coordinates.

∴ d ratios of line PM are

< 10 t + 11 – 2, -4 t – 2 + 1, -11 t – 8 – 5 >

i.e. < 10 t + 9, -4 t – 1, -11 t – 13 >

and d ratios of given line AB are < 10, -4, -11 >

Since line PM is ⊥ to line AB.

∴ 10(10 t + 9) – 4(-4 t – 1) – 11(-11 t – 13) = 0

⇒ 100 t + 90 + 16 t + 4 + 121 t + 143 = 0

⇒ 237 t + 237 = 0

⇒ t = -1

∴ coordinates of M are (-10 + 11, 4 – 2, 11 – 8) i.e. (1, 2 , 3)

∴ coordinates of M are (-10 + 11, 4 – 2, 11 – 8) i.e. (1, 2, 3)

Since M be the mid point of PP.

∴ \(\frac{α+2}{2}\) = 1

⇒ α + 2 = 2

⇒ α = 0

\(\frac{β-1}{2}\) = 2

⇒ β – 1 = 4

\(\frac{γ+5}{2}\) = 3

⇒ γ + 5 = 6

⇒ β = 5

γ = 1

Thus, the required image of P be P(0, 5, 1).

Question 29.

Find the cartesian equation of the plane passing through the line of intersection of the planes:

\(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) + 5 = 0 and

\(\vec{r}\)(\(\hat{i}\) – 5 \(\hat{j}\) + 7 \(\hat{k}\)) + 2 = 0 and intersecting y-axis at (0, \(\mathbf{3}\), \(\mathbf{0}\)).

Answer:

Cartesian eqns. of given planes are

\(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) + 5 = 0

⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) + 5 = 0

⇒ 2 x + 3 y – 4 z + 5 = 0 …………………………… (1)

and \(\vec{r}\)(\(\hat{i}\) – 5 \(\hat{j}\) + 7 \(\hat{k}\)) = -2

⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (\(\hat{i}\) – 5 \(\hat{j}\) + 7 \(\hat{k}\)) + 2 = 0

⇒ x – 5 y + 7 z + 2 = 0 …………………………… (2)

Thus, the eqn. of any plane through the line of intersection of eqn. (1) and (2) be given by

(2 x + 3 y – 4 z + 5) + k(x – 5 y + 7 z + 2) = 0 …………………………… (3)

Since plane (3) passes through the point (0, 3, 0).

∴(0 + 9 – 0 + 5) + k(0 – 15 + 0 + 2) = 0

⇒ 14 – 13 k = 0

⇒ k = \(\frac{14}{13}\)

putting the value of k in eqn. (3); we have

(2 x + 3 y – 4 z + 5) + \(\frac{14}{13}\)(x – 5 y + 7 z + 2) = 0

⇒ 40 x – 31 y + 46 z + 93 = 0

which is the required eqn. of plane.

Question 30.

A line making angle 45° and 60° with the positive directions of the axes of x and y makes with the positive direction of z-axis, an angle of

(a) 60°

(b) 120°

(c) 60° and 120°

(d) None of these

Answer:

Let θ be the acute angle made by line with z-axis.

Then direction cosines of required line are < cos \(\frac{\pi}{4}\), cos \(\frac{\pi}{3}\), cosθ >

i.e. < \(\frac{1}{\sqrt{2}}\), \(\frac{1}{2}\), cosθ >

Here l = \(\frac{1}{\sqrt{2}}\), m = \(\frac{1}{2}\) ; n = cosθ

Now l^{2} + m^{2} + n^{2} = 1

⇒ (\(\frac{1}{\sqrt{2}}\))^{2} + (\(\frac{1}{2}\))^{2} + cos^{2}θ = 1

⇒ \(\frac{1}{2}\) + \(\frac{1}{4}\) + cos^{2}θ = 1

⇒ cos^{2}θ = \(\frac{1}{4}\)

⇒ cosθ = \(\frac{1}{2}\)

⇒ θ = \(\frac{\pi}{3}\)

[ ∵ cosθ > 0 since θ be the acute angle]

Question 31.

A line makes equal angles with the coordinate axes. Its direction cosines are

(a) < f{0 , 0 , 0} >

(b) < ± 1, ± 1, ± 1 >

(c) < ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\) >

(d) < ± \(\frac{1}{3}\), ± \(\frac{1}{3}\), ± \(\frac{1}{3}\) >

Answer:

Let α be the equal angle made by line with the coordinate axes. Then direction cosines of line are < cosα, cosα, cosα > i.e. cos^{2} α + cos^{2} α + cos^{2}α = 1

⇒ cosα = ± \(\frac{1}{\sqrt{3}}\)

∴ required direction cosines of line are < ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\) >

∴ Ans. (c)

Question 32.

The direction cosines of the line \(\frac{x+2}{2}\) = \(\frac{2 y-5}{3}\), z = -1 are

(a) (\(\frac{4}{5}\), \(\frac{3}{5}\), 0)

(b) (\(\frac{3}{5}\), \(\frac{4}{5}\), \(\frac{1}{5}\))

(c) (\(-\frac{3}{5}\), \(\frac{4}{5}\), 0)

(d) (\(\frac{4}{5}\), – \(\frac{1}{5}\), \(\frac{1}{5}\))

Answer:

Given eqn. of line be \(\frac{x+2}{2}\) = \(\frac{2 y-5}{3}\), z = -1

i.e. \(\frac{x+2}{2}\) = \(\frac{y-5 / 2}{3 / 2}\) = \(\frac{z+1}{0}\)

i.e. \(\frac{x+2}{4}\) = \(\frac{y-5 / 2}{3}\) = \(\frac{z+1}{0}\)

The direction ratios of given line are proportional to < 4, 3, 0 >.

∴ Direction cosines of given line are ;

< \(\frac{4}{\sqrt{16+9+0}}\), \(\frac{3}{\sqrt{16+9+0}}\), \(\frac{0}{\sqrt{16+9+0}}\)

i.e. < \(\frac{4}{5}\), \(\frac{3}{5}\), 0 >

∴ Ans. (a)

Question 33.

Find the length of the perpendicular to the line \(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z-2}{3}\) from the point

(1, 6, 3).

(a) \(\sqrt{13}\)

(b) \(\sqrt{10}\)

(c) 3

(d) 2

Answer:

eqn. of given line be

\(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z-2}{3}\) = t (say) ……………………… (1)

Any point on given line be Q(t, 2 t + 1, 3 t + 2) and coordinates of given point P are (1, 6, 3)

∴ D’ratios of line PQ are < t – 1, 2 t + 1 – 6, 3 t + 2 – 3 >

i.e. < t – 1, 2 t – 5, 3 t – 1 >

Since PQ is ⊥ to given line (1).

∴(t – 1) 1 + (2 t – 5) 2 + (2 t – 1) 3 = 0

⇒ 14 t – 14 = 0 ⇒ t = 1

Thus coordinates of Q are (1, 3, 5)

∴|P Q| = \(\sqrt{(1-1)^2+(6-3)^2+(3-5)^2}\)

= \(\sqrt{0+9+4}\) = \(\sqrt{13}\)

∴ Ans. (a)

Question 34.

(i) The angle between the lines

\(\frac{x-5}{-3}\) = \(\frac{y+3}{-4}\) = \(\frac{z-7}{0}\)

\(\frac{x}{1}\) = \(\frac{y-1}{-2}\) = \(\frac{z-6}{2}\) is

(a) \(\frac{\pi}{3}\)

(b) tan^{-1} (\(\frac{1}{5}\))

(c) cos^{-1} (\(\frac{1}{3}\))

(d) \(\frac{\pi}{2}\)

(ii) The angle between the lines \(\vec{r}\)

= (2 \(\hat{i}\) – 3 \(\hat{j}\) + \(\hat{k}\)) + λ(\(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\)) and \(\vec{r}\)

= (\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) + µ(\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) is

(a) cos^{-1} (\(\frac{9}{\sqrt{91}}\))

(b) cos ^{-1} (\(\frac{7}{\sqrt{84}}\))

(c) \(\frac{\pi}{3}\)

(d) \(\frac{\pi}{2}\)

Answer:

(i) D’ ratio’s of given lines are < -3, -4, 0 > and < 1, -2, 2 >

Let θ be the angle between given lines.

Then cos θ = \(\frac{-3 \times 1-4 \times(-2)+0 \times 2}{\sqrt{9+16+0} \sqrt{1+4+4}}\)

= \(\frac{5}{5 \times 3}\) = \(\frac{1}{3}\)

⇒ θ = cos^{-1} (\(\frac{1}{3}\))

∴ Ans. (c)

(ii) Comparing given lines with

\(\vec{r}\) = \(\overrightarrow{a_1}\) + λ \(\overrightarrow{b_1}\) and \(\vec{r}\)

= \(\overrightarrow{a_2}\) + λ \(\overrightarrow{b_2}\)

∴ \(\overrightarrow{b_1}\) = \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\);

\(\overrightarrow{b_2}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)

⇒ \(\overrightarrow{b_1}\) \(\overrightarrow{b_2}\)

= 1 × 1 + 4 × 2 + 3 × (-3) = 0

Let θ be the angle between given lines.

Then cos θ

= \(\frac{\overrightarrow{b_1}\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\)

= 0 ⇒ θ = \(\frac{\pi}{2}\)

∴ Ans. (d)

Question 35.

The value of λ for which the lines are \(\frac{1-x}{3}\) = \(\frac{y-2}{2 \lambda}\)

= \(\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}\) = \(\frac{y-1}{1}\)

= \(\frac{6-z}{7}\) are perpendicular to each other is

(a) -1

(b) -2

(c) 1

(d) 2

Answer:

Given eqns. of lines are

\(\frac{1-x}{3}\) = \(\frac{y-2}{2 \lambda}\) = \(\frac{z-3}{2}\)

Here a_{1} = -3 ; b_{1} = 2λ ; c_{1} = 2

and \(\frac{x-1}{3 \lambda}\) = \(\frac{y-1}{1}\) = \(\frac{6-z}{7}\)

i.e. a_{2} = 3λ ; b_{2} = 1 ; c_{2} = -7

Since given linnes are perpendicular to each other.

∴ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}

= 0 – 3(3λ) + 2 λ(1) + 2(-7) = 0

⇒ -7λ – 14 = 0

⇒ λ = -2

Question 36.

A straight line joining the points (1, 1, 1) and (0, 0, 0) intersects the plane 2 x + 2 y + z = 10 at

(a) (1, 2, 5)

(b) (2, 2, 2)

(c) (2, 1, 5)

(d) (1, 1, 6)

Answer:

eqn. of any line joining the points (1, 1, 1) and (0, 0, 0) be given by

\(\frac{x-1}{1}\) = \(\frac{y-1}{1}\) = \(\frac{z-1}{1}\) = t (say)

So any point on line (1) be given by

P(t + 1, t + 1, t + 1)

Now given line intersects given plane

2 x + 2 y + z = 10

⇒ 2(t + 1) + 2(t + 1) + t + 1 = 10

⇒ 5 t + 5 = 10

⇒ t = 1

Thus the required point of intersection of given line and plane be (2, 2, 2).

Question 37.

Lines \(\frac{x-2}{1}\) = \(\frac{y-3}{1}\) = \(\frac{z-4}{-k}\) and \(\frac{x-1}{k}\)

= \(\frac{y-4}{2}\) = \(\frac{z-5}{1}\) are coplanar if

(a) k = 2

(b) k = 0

(c) k = 3

(d) k = -1

Answer:

We know that, the lines \(\frac{x-x_1}{a_1}\) = \(\frac{y-y_1}{b_1}\) = \(\frac{z-z_1}{c_1}\) and

\(\frac{x-x_2}{a_2}\) = \(\frac{y-y_2}{b_2}\) = \(\frac{z-z_2}{c_2}\)

are coplanar if

\(|\begin{array}{ccc}

x_2-x_1 y_2-y_1 z_2-z_1

a_1 b_1 c_1

a_2 b_2 c_2

\end{array}|\) = 0

For given lines ;

x_{1} = 2 ; y_{1} = 3 ; z_{1} = 4

x_{2} = 1 ; y_{2} = 4 ; z_{2} = 5

a_{1} = 1 ; b_{1} = 1 ; c_{1} = -k

a_{2} = k ; b_{2} = 2 ; c_{2} = 1

Thus,

\(|\begin{array}{ccc}

1-2 4-3 5-4

1 1 -k

k 2 1

\end{array}|\) = 0

⇒ \( |\begin{array}{rrr}

-1 1 1

1 1 -k

k 2 1

\end{array}|\) = 0

Expanding along R_{3}

-1(1 + 2 k) – 1(1 + k^{2} ) + 1(2 – k) = 0

⇒ -1 – 2 k – 1 – k^{2} + 2 – k = 0

⇒ -3 k – k^{2} = 0

⇒ -k(3 + k) = 0

⇒ k = 0,-3

Question 38.

Angle between the planes x + y + 2 z = 6 and 2 x – y + z = 9 is

(a) \(\frac{\pi}{4}\)

(b) \(\frac{\pi}{6}\)

(c) \(\frac{\pi}{3}\)

(d) \(\frac{\pi}{2}\)

Answer:

D’ratios of normal to given planes are < 1, 1, 2 > and < 2, -1, 1 > Let θ be the angle between given planes.

Then cos θ = \(\frac{1 \times 2-1 \times 1+2 \times 1}{\sqrt{4+1+1} \sqrt{1+1+4}}\)

= \(\frac{3}{\sqrt{6} \sqrt{6}}\)

= \(\frac{3}{6}\)

= \(\frac{1}{2}\)

⇒ θ = \(\frac{\pi}{3}\)

Question 39.

If the planes \(\vec{r}\) (2 \(\hat{i}\) – λ\(\hat{j}\) + 3 \(\hat{k}\)) = 0 and

\(\vec{r}\) (λ\(\hat{i}\) + 5 \(\hat{j}\) – \(\hat{k}\)) = 5 are perpendicular to each other then the value of λ^{2} + λ is

(a) 0

(b) -2

(c) -1

(d) 2

Answer:

Cartesian eqns. of given planes are ;

2 x – λ y + 3 z = 0

and λx + 5 y – z = 5

On comparing with a_{1} x + b_{1}y + c_{1}z = d_{1} and a_{2} x + b_{2}y + c_{2}z

= d_{2}

∴ a_{1} = 2 ; b_{1} = -λ ; c_{1} = 3

a_{2} = λ ; b_{2} = 5 ; c_{2} = -1

planes (1) and (2) are ⊥ to each other.

∴ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0

2 × λ – λ × 5 + 3 × (-1) = 0

⇒ -3λ – 3 = 0

⇒ λ = -1

∴ λ^{2} + λ = (-1)^{2} – 1 = 0

Question 40.

If α, β and γ are the direction cosine of a line in space, then the value of sin^{2}α + sin^{2}β + sin^{2}γ.

(a) 0

(b) 1

(c) -1

(d) 2

Answer:

Let \(\vec{a}\) be the vector that makes α, β, γ with OX, OY and OZ respectively.

Let < l, m, n> be the direction cosines of \(\vec{a}\)

∴ l = cos α ; m = cos β ; n = cos γ

Also l^{2} + m^{2} + n^{2} = 1

cos^{2} α + cos^{2} β + cos^{2}γ = 1

⇒ (1 – sin^{2}α) + (1 – sin^{2}β) + (1 – sin^{2}γ) = 1

⇒ sin^{2}α + sin^{2}β + sin^{2}γ = 3 – 1 = 2

Question 41.

The length of the perpendicular drawn from (1, 2, 3) to the line \(\frac{x-6}{3}\) = \(\frac{y-7}{2}\) = \(\frac{z-7}{-2}\) is

(a) 4

(b) 5

(c) 6

(d) 7

Answer:

Let L be the foot of ⊥ drawn from P(1, 2, 3) on given line. any point on given line be,

\(\frac{x-6}{3}\) = \(\frac{y-7}{2}\) = \(\frac{z-7}{-2}\) = t (say)

i.e. x = 3 t + 6 ; y = 2 t + 7 ; z = -2 t + 7

Thus the coordinates of L are (3 t + 6, 2 t + 7, -2 t + 7)

∴ D’ ratios of line PL are proportional to < 3 t + 6 – 1, 2 t + 7 – 2, -2 t + 7 – 3 >

i.e. < 3 t + 5, 2 t + 5, -2 t + 4 >

also D’ ratios of given line be proportional to < 3, 2, -2 >.

Since P L is ⊥ to given line.

∴ (3 t + 5) 3 + (2 t + 5) 2 + (-2 t + 4)(-2) = 0

⇒ 9 t + 15 + 4 t + 10 + 4 t – 8 = 0

⇒ 17 t + 17 = 0

⇒ t = -1

∴ Coordinates of point L are (3, 5, 9).

∴ required ⊥ distance = |PL|

= \(\sqrt{(3-1)^2+(5-2)^2+(9-3)^2}\)

= \(\sqrt{4+9+36}\) = 7

Question 42.

The lines \(\frac{x-2}{1}\) = \(\frac{y-3}{1}\) = \(\frac{4-z}{k}\) and \(\frac{x-1}{k}\) = \(\frac{y-4}{2}\) = \(\frac{z-5}{-2}\)

are mutually perpendicular, if the value of k is

(a) \(-\frac{2}{3}\)

(b) \(\frac{2}{3}\)

(c) -2

(d) 2

Answer:

eqns. of given lines are ; \(\frac{x-2}{1}\) = \(\frac{y-3}{1}\) = \(\frac{4-z}{k}\)

and \(\frac{x-1}{k}\) = \(\frac{y-4}{2}\) = \(\frac{z-5}{-2}\)

Since lines (1) and (2) are mutually ⊥ to each other

∴ 1(k) + 1(2) + (-k)(-2) = 0

⇒ k + 2 + 2 k = 0

[∵ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0]

⇒ 3 k = -2

⇒ k = \(-\frac{2}{3}\)

Question 43.

Similar question. The two lines x = a y + b, z = c y + d; and x = a y + b, z = c y + d are perpendicular to each other, if

(a) \(\frac{a}{a^{\prime}}\) + \(\frac{c}{c^{\prime}}\) = 1

(b) \(\frac{a}{a^{\prime}}\) + \(\frac{c}{c^{\prime}}\) = -1

(c) a a + c c = 1

(d) a a + c c = -1

Answer:

eqns. of given lines in cartesian form, can be written as ;

\(\frac{x-b}{a}\) = \(\frac{y}{1}\) = \(\frac{z-d}{c}\)

and \(\frac{x-b^{\prime}}{a^{\prime}}\) = \(\frac{y}{1}\)

= \(\frac{z-d^{\prime}}{c^{\prime}}\)

Direction ratios of given lines (1) and (2) are < a, 1, c > and < a, 1, c >

Now lines (1) and (2) are \perp to each other

if a × a + 1 × 1 + c × c = 0

if a a + 1 + c c = 0

Question 44.

The distance of the origin from the plane -2 x + 6 y – 3 z = -7 is

(a) 1 unit

(b) \(\sqrt{2}\) units

(c) 2 \(\sqrt{2}\) units

(d) 3 units

Answer:

∴ required ⊥ distance of O(0, 0, 0) from given plane = \(\frac{|-2 \times 0+6 \times 0-3 \times 0+7|}{\sqrt{4+36+9}}\)

= \(\frac{7}{7}\) = 1

Question 45.

The two planes x – 2 y + 4 z = 10 and 18 x + 17 y – k z = 50 are perpendicular, if k is equal to

(a) -4

(b) 4

(c) 2

(d) -2

Answer:

eqns. of given planes are ;

x – 2 y + 4 z = 10

18 x + 17 y – k z = 50

On comparing eqns. (1) and (2) with

a_{1} x + b_{1}y + c_{1}z = d_{1}

and a_{2}x + b_{2}y + c_{2}z = d_{2}

Here a_{1} = 1 ; b_{1} = -2 ; c_{1} = 4 ;

a_{2} = 18 ; b_{2} = 17 ; c_{2} = -k

We know that planes (1) and (2) are \perp to each other.

Then a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0

⇒ 1 × 18 – 2 × 17 + 4 × (-k) = 0

⇒ 18 – 34 – 4 k = 0

⇒ 4 k = -16

⇒ k = -4

Question 46.

The line \(\frac{x-2}{3}\) = \(\frac{y-3}{4}\) = \(\frac{z-4}{5}\) is parallel to the plane

(a) 2 x + 3 y + 4 z = 0

(b) 3 x + 4 y – 5 z = 7

(c) 2 x + y – 2 z = 0

(d) x – y + z = 2

Answer:

eqn. of given line be,

\(\frac{x-2}{3}\) = \(\frac{y-3}{4}\) = \(\frac{z-4}{5}\)

∴ D’ratios of given line (1) are < 3, 4, 5 > Let us take the option (b) ; the eqn. of given plane be

3 x + 4 y – 5 z = 7

∴ D’ratios of normal to plane (2) are < 3, 4, -5 >

Now line (1) is parallel to plane (2) iff normal to plane (2) is ⊥ to line (1)

Here, 3 × 3 + 4 × 4 + 5(-5) = 9 + 16 – 25 = 0

Question 47.

What is the distance (in units) between the two planes 3 x + 5 y + 7 z = 3 and 9 x + 15 y + 21 z = 9 ?

(a) 0

(b) 3

(c) \(\frac{6}{\sqrt{83}}\)

(d) 6

Answer:

eqns. of given planes are

3 x + 5 y + 7 z = 3

and 9 x + 15 y + 21 z = 9

Here \(\frac{3}{9}\) = \(\frac{5}{15}\) = \(\frac{7}{21}\) = \(\frac{1}{3}\)

Thus given planes (1) and (2) are coincident.

Let P(x, y, z) be any point on plane (1).

∴ distance between given planes = ⊥ distance of P(x, y, z) from plane (2)

= \(\frac{|3(3 x+5 y+7 z)-9|}{\sqrt{9^2+15^2+21^2}}\)

= \(\frac{|3 \times 3-9|}{\sqrt{81+225+441}}\) = 0

Question 48.

The equation of the line in vector form passing through the point (-1, 3, 5) and parallel to the line \(\frac{x-3}{2}\) = \(\frac{y-4}{3}\), z = 2 is

(a) \(\vec{r}\) = (\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\))

(b) \(\vec{r}\) = (\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)) + λ(-2 \(\hat{i}\) + 3 \(\hat{j}\))

(c) \(\vec{r}\) = (2 \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\)) + λ(\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\))

(d) \(\vec{r}\) = (2 \(\hat{i}\) + 3 \(\hat{j}\)) + λ(\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\))

Answer:

eqn. of given line can be written as ;

\(\frac{x-3}{2}\) = \(\frac{y-4}{3}\) = \(\frac{z-2}{0}\)

∴ direction ratios of required line which is parallel to line (1) are < 2, 3, 0 >.

Hence vector eqn. of line through the point (-1, 3, 5) whose position vector \(\vec{a}\)

= \(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\) and having direction ratios < 2, 3, 0 >

i.e. || to vector

\(\vec{b}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) be given by \(\vec{r}\)

= \(\vec{a}\) + λ\(\vec{b}\)

where λ be the parameter \(\vec{r}\) = (\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\))

Question 49.

(i) The sine of the angle between the straight line \(\frac{x-2}{3}\) = \(\frac{y-3}{4}\) = \(\frac{z-4}{5}\) and the plane 2 x – 2 y + z = 5 is

(a) \(\frac{10}{6 \sqrt{5}}\)

(b) \(\frac{4}{5 \sqrt{2}}\)

(c) \(\frac{2 \sqrt{3}}{5}\)

(d) \(\frac{\sqrt{2}}{10}\)

Answer:

Equation of given line be \(\frac{x-2}{3}\) = \(\frac{y-3}{4}\) = \(\frac{z-4}{5}\)

Thus the line passing through the point (2, 3, 4) and having direction ratios < 3, 4, 5 >.

Thus vector equation of line passing through the point whose P.V be 2

\(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) and parallel to \(\vec{b}\)

= 3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) is given by \(\vec{r}\)

= 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) + λ(3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\)) and

given equation of plane be 2 x – 2 y + z = 5

D’ ratios of normal to plane are < 2, -2, 1 >

∴ \(\vec{n}\) = 2 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)

Let θ be the angle between given plane and given line.

Then sinθ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)

= \(\frac{(3 \hat{i}+4 \hat{j}+5 \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{3^2+4^2+5^2} \sqrt{2^2+(-2)^2+1^2}}\)

= \(\frac{3(2)+4(-2)+5(1)}{\sqrt{9+16+25} \sqrt{4+4+1}}\)

= \(\frac{3}{5 \sqrt{2} \times 3}\) = \(\frac{1}{5 \sqrt{2}}\)

= \(\frac{\sqrt{2}}{10}\)

(ii) Similar questions. The plane 2 x – 3 y + 6 z – 11 = 0 makes an angle sin^{-1} (α) with x-axis. The value of α is equal to

(a) \(\frac{\sqrt{3}}{2}\)

(b) \(\frac{\sqrt{2}}{3}\)

(c) \(\frac{2}{7}\)

(d) \(\frac{3}{7}\)

Answer:

The given plane be 2 x – 3 y + 6 z – 11 = 0

∴ \(\vec{n}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + 6 \(\hat{k}\)

equation of given line be x-axis and direction ratios of x-axis are

< 1,0,0 > i.e., \(\vec{b}\) = \(\hat{i}\) + 0 \(\hat{j}\) + 0 \(\hat{k}\)

Since θ be the angle between given plane and given line.

∴ sinθ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)

= \(\frac{(2 \hat{i}-3 \hat{j}+6 \hat{k}) \cdot(\hat{i}+0 \hat{j}+0 \hat{k})}{\sqrt{2^2+(-3)^2+6^2} \sqrt{1^2+0^2+0^2}}\)

= \(\frac{2(1)-3(0)+6(0)}{7 \times 1}\) = \(\frac{2}{7}\)

⇒ θ = sin^{-1} \(\frac{2}{7}\)

Also given plane makes an angle sin^{-1} α with given line.

⇒ α = \(\frac{2}{7}\)

Question 50.

(i) The point of intersection of the straight line \(\frac{x-2}{2}\) = \(\frac{y-1}{-3}\) = \(\frac{z+2}{1}\) with the plane x + 3 y – z + 1 = 0 is

(a) (4, -2, -1)

(b) (-5, 1, -1)

(c) (2, 0, 3)

(d) (5, -1, 3)

(ii) Similar questions. The value of λ for which the straight line \(\frac{x-\lambda}{3}\) = \(\frac{y-1}{2+\lambda}\)

= \(\frac{z-3}{-1}\) may line on the plane x – 2 y = 0 is

(a) 2

(b) 0

(c) -1 / 2

(d) no such value of λ exists.

Answer:

(i) eqn. of given line be, \(\frac{x-2}{2}\) = \(\frac{y-1}{-3}\) = \(\frac{z+2}{1}\) = t (say)

and eqn. of given plane be, x + 3 y – z + 1 = 0

Any point on line (1) be P(2 t + 2, -3 t + 1, t – 2)

For point of intersection of line (1) and plane (2) then any point P lies on plane (2).

2 t + 2 + 3(-3 t + 1) – (t – 2) + 1 = 0

⇒ 2 t + 2 – 9 t + 3 – t + 2 + 1 = 0

⇒ -8 t + 8 = 0 ⇒ t = 1

Hence, the coordinates of required point of intersection of line (1) and plane (2) are (4, -2, -1)

(ii) eqn. of given line be \(\frac{x-\lambda}{3}\) = \(\frac{y-1}{2+\lambda}\) = \(\frac{z-3}{-1}\)

and eqn. of given plane be x – 2 y = 0

D’ ratios of normal to plane (2) are < 1, -2, 0 >

and D’ratios of line (1) are < 3, 2 + λ, -1 >

Now line (1) lies on plane (2) if normal to plane (2) is ⊥ to line (1).

∴ 1 × 3 – 2(2 + λ) + 0 × (-1) = 0

⇒ 3 – 4 – 2λ = 0

⇒ -1 – 2λ = 0

⇒ λ = \(-\frac{1}{2}\)

Question 51.

The distance of the point (1, 0, 2) from the point of intersection of the line \(\frac{x-2}{3}\) = \(\frac{y+1}{4}\) = \(\frac{z-2}{12}\) and the plane x – y + z = 16, is

(a) 3 \(\sqrt{21}\)

(b) 13

(c) 2 \(\sqrt{14}\)

(d) 8

Answer:

eqn. of given line be, \(\frac{x-2}{3}\) = \(\frac{y+1}{4}\) = \(\frac{z-2}{12}\) = t (say)

and eqn. of given plane be, x – y + z – 16 = 0

Any point on line (1) be P(3 t + 2, 4 t – 1, 12 t + 2)

Now line (1) intersects plane (2). Then point P lies on plane (2).

∴ 3 t + 2 – (4 t – 1) + 12 t + 2 = 16 ⇒ 11 t = 11 ⇒ t = 1

Thus coordinates of point of intersection of line (1) and plane (2) are (3 + 2, 4 – 1, 12 + 2) i.e. (5, 3, 14)

∴ Required distance of Q(1, 0, 2) from P(5, 3, 14) = |P Q|

= \(\sqrt{(5-1)^2+(3-0)^2+(14-2)^2}\)

= \(\sqrt{16+9+144}\) = 13

Question 52.

Let P(-7, 1, -5) be a point on a plane and let O be the origin. If O P is a normal to the plane, then the equation of the plane is

(a) 7 x-y+5 z+75=0

(b) 7 x-y+5 z+80=0

(c) 7 x+y+5 z+80=0

(d) 7 x-y-5 z-75=0

Answer:

Let the eqn. of plane through P(-7, 1, -5) be a(x + 7) + b(y – 1) + c(z + 5) = 0 where < a, b, c > are the D’ratios of normal to plane (1).

given O P be also the normal to plane (1) and its direction ratios are <-7,1,-5 >

Thus eqn. (1) becomes ;

-7(x + 7) + 1(y – 1) – 5(z + 5) = 0

⇒ – 7 x + y – 5 z – 75 = 0

⇒ 7 x – y + 5 z + 75 = 0

Question 53.

If the straight lines \(\frac{x-2}{1}\) = \(\frac{y-3}{1}\) = \(\frac{z-4}{0}\) and \(\frac{x-1}{k}\) = \(\frac{y-4}{2}\) = \(\frac{z-5}{1}\) are coplanar, then the value of k is

(a) -3

(b) 0

(c) 1

(d) 6

Answer:

We know that, \(\frac{x-x_1}{a_1}\) = \(\frac{y-y_1}{b_1}\) = \(\frac{z-z_1}{c_1}\) and \(\frac{x-x_2}{a_2}\) = \(\frac{y-y_2}{b_2}\) = \(\frac{z-z_2}{c_2}\) are coplanar

if \(|\begin{array}{ccc}x_2-x_1 y_2-y_1 z_2-z_1 a_1 b_1 c_1 a_2 b_2 c_2\end{array}|\) = 0

For given lines ;

x_{1} = 2 ; y_{1} = 3 ; z_{1} = 4 ; x_{2} = 1 ; y_{2} = 4 ; z_{2} = 5

a_{1} = 1 = b_{1} ; c_{1} = 0 ; a_{2} = k ; b_{2} = 2 ; c_{2} = 1

Here, \(|\begin{array}{ccc}-1 1 1 1 1 0 k 2 1\end{array}|\) = 0;

Expanding along R 1

-1(1 – 0) – 1(1 – 0) + 1(2 – k) = 0

⇒ -1 – 1 + 2 – k = 0

⇒ k = 0

Question 54.

The distance of the point (2, 1, 0) from the plane 2 x + y + 2 z + 5 = 0 is

(a) 10

(b) \(\frac{10}{3}\)

(c) \(\frac{10}{9}\)

(d) 5

Answer:

Required ⊥ distance of P(2, 1, 0) from given plane

= \(\frac{|2 \times 2+1 \times 1+2 \times 0+5|}{\sqrt{2^2+1^2+2^2}}\)

= \(\frac{10}{3}\)

Question 55.

If the distance of point 2 \(\hat{i}\) + 3 \(\hat{j}\) + λ \(\hat{k}\) from the plane \(\vec{r}\) (3 \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\)) = 13 is 5 units, then

(a) 6,- \(\frac{17}{3}\)

(b) -6, \(\frac{17}{3}\)

(c) -6, \(-\frac{17}{3}\)

(d) -6, \(\frac{17}{3}\)

Answer:

Cartesian eqn. of given plane be,

3 x + 2 y + 6 z = 13

given ⊥ distance of point (2, 3λ) from plane (1) = 5

⇒ \(\frac{|3 \times 2+2 \times 3+6 \times \lambda-13|}{\sqrt{3^2+2^2+6^2}}\) = 5

⇒ \(\frac{|12+6 \lambda-13|}{7}\) = 5

⇒ |6λ + 12 – 13| = 35

⇒ (6λ – 1)= ± 35

⇒ 6 λ = ± 35 + 1

⇒ λ = 6, – \(\frac{17}{3}\)

Question 56.

If the foot of the perpendicular from the origin to a plane is (1, 2, 3), then equation of the plane is

(a) 2 x – y + z = 3

(b) x + y + z = 6

(c) x – y – z = -4

(d) x + 2 y + 3 z = 14

Answer:

The eqn. of required plane through P(1, 2, 3) be given by a(x – 1) + b(y – 2) + c(z – 3) = 0 where < a, b, c > be the direction ratios of normal to plane (1).

Now OP is normal to plane (1).

∴ its D’ratios are < 1 – 0, 2 – 0, 3 – 0 > i.e. < 1, 2, 3 >

Therefore eqn. (1) becomes ;

1(x – 1) + 2(y – 2) + 3(z – 3) = 0

⇒ x + 2 y + 3 z = 14

Question 57.

(i) If a line makes angles \(\frac{\pi}{2}\), \(\frac{3 \pi}{4}\) and \(\frac{\pi}{4}\) with x, y, z axes respectively, then its direction cosines are?

(ii) If a line has direction ratio 2, -1, 2, then what are the-direction cosives ?

(iii) Write the direction cosines of the line joining the points (1, 0, 0) and (0, 1, 1).

(iv) What are the direction cosines of a line which makes equal angles with the coordinate axes.

(v) Find the direction cosives of the vector joining the points A(1, 2, -3) and B(-1, -2, 1) directed from B to A.

Answer:

(i) We know that, direction cosines of a line are the cosines of the angles made by line with positive direction of coordinate axes.

Since the line makes an angle of \(\frac{\pi}{2}\), \(\frac{3 \pi}{4}\) and \(\frac{\pi}{4}\) with coordinate axes.

Then direction cosines of line are < cos \(\frac{\pi}{2}\), cos \(\frac{3 \pi}{4}\), cos \(\frac{\pi}{4}\) >

i.e. < 0, \(\frac{-1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\) >

[∵ cos \(\frac{3 \pi}{4}\) = cos (π –\(\frac{\pi}{4}\)) = -cos \(\frac{\pi}{4}\) = \(-\frac{1}{\sqrt{2}}\)

(ii) Given direction ratios of line are < 2, -1, -2 >

∴ Its direction cosines are

< \(\frac{2}{\sqrt{2^2+(-1)^2+2^2}}\), \(\frac{-1}{\sqrt{2^2+(-1)^2+2^2}}\), \(\frac{2}{\sqrt{2^2+(-1)^2+2^2}}\) >

i.e. < \(\frac{2}{3}\), \(-\frac{1}{3}\), \(\frac{2}{3}\) >

(iii) D’ratios of line joining the points (1, 0, 0) and (0, 1, 1) are < 0 – 1, 1 – 0, 1 – 0 >

i.e. < -1, 1, 1 >

∴ D’ cosines of line are < \(\frac{-1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\) >

(iv) Let α be the equal angle made by line with the coordinate axes. Then direction cosines of line are < cos α, cos α, cos α >.

i.e. cos^{2} α + cos^{2} α + cos^{2} α = 1

⇒ cos α = ± \(\frac{1}{\sqrt{3}}\)

∴ required direction cosines of line are < ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\) >

(v) D’ratios of \(\overrightarrow{\mathrm{BA}}\) are < 1-(-1), 2-(-2),-3 – 1 >

i.e. < 2, 4, -4 >

∴ D cosines of \(\overrightarrow{B_A}\) are < \(\frac{2}{\sqrt{4+16+16}}\), \(\frac{4}{\sqrt{4+16+16}}\), \(\frac{-4}{\sqrt{4+16+16}}\) >

< \(\frac{2}{6}\), \(\frac{4}{6}\), \(\frac{-4}{6}\) >

i.e. < \(\frac{1}{3}\), \(\frac{2}{3}\), \(\frac{-2}{3}\) >

Question 58.

Write the distance of the point (2, 3, 4) from the x-axis.

Answer:.

Required distance of P(2, 3, 4) from x-axis = distance of P(2, 3, 4) from (2, 0, 0)

= (2 – 2)^{2} + (3 – 0)^{2} + (4 – 0)^{2}

= \(\sqrt{9+16}\) = 5 units

Question 59.

The equations of a line are 5 x – 3 = 15 y + 7 = 3 – 10 z. Write the distance cosines of the line.

Answer:

Given line can be written as ;

5(x – \(\frac{3}{5}\)) = 15(y + \(\frac{7}{15}\)) = -10(z – \(\frac{3}{10}\))

⇒ \(\frac{x-\frac{3}{5}}{6}\) = \(\frac{y+\frac{7}{15}}{2}\) = \(\frac{z-\frac{3}{10}}{-3}\)

∴ D cosines of given line are

< \(\frac{6}{\sqrt{6^2+2^2+(-3)^2}}\), \(\frac{2}{\sqrt{6^2+2^2+(-3)^2}}\), \(\frac{-3}{\sqrt{6^2+2^2+(-3)^2}}\) >

i.e. < \(\frac{6}{7}\), \(\frac{2}{7}\), \(\frac{-3}{7}\) >

Question 60.

Find the direction cosines of the line \(\frac{4-x}{2}\) = \(\frac{y}{6}\) = \(\frac{1-z}{3}\)

Answer:

eqn. of given line can be written as ; \(\frac{x-4}{-2}\) = \(\frac{y}{6}\) = \(\frac{z-1}{-3}\)

∴ direction cosines of given line (1) be

< \(\frac{-2}{\sqrt{4+36+9}}\), \(\frac{6}{\sqrt{4+36+9}}\), \(\frac{-3}{\sqrt{4+36+9}}\) >

i.e. < \(\frac{-2}{7}\), \(\frac{6}{7}\), \(\frac{-3}{7}\) >

Question 61.

The equation of a line are given by \(\frac{3-x}{-3}\) = \(\frac{y+2}{-2}\) = \(\frac{z+2}{6}\). Write the direction cosines of a line parallel to the above line.

Answer:

eqn. of given line can be written as ;

\(\frac{x-3}{3}\) = \(\frac{y+2}{-2}\) = \(\frac{z+2}{6}\)

∴ D’ratios of line parallel to line (1) are < 3, -2, 6 >

∴ D’ratios of line parallel to given line (1) are ;

< \(\frac{3}{\sqrt{9+4+36}}\), \(\frac{-2}{\sqrt{9+4+36}}\), \(\frac{6}{\sqrt{9+4+36}}\) >

i.e. < \(\frac{3}{7}\), \(-\frac{2}{7}\), \(\frac{8}{7}\) >

Question 62.

Write the equation of a line parallel to the line \(\frac{x-2}{-3}\) = \(\frac{y+3}{2}\) = \(\frac{z+5}{6}\) and passing through the point (1, 2, 3).

Answer:

eqn. of given line be, \(\frac{x-2}{-3}\) = \(\frac{y+3}{2}\) = \(\frac{z+5}{6}\)

∴ D’ ratios of line || to line (1) are < -3, 2, 6 >

Hence eqn. of line passing through the point (1, 2, 3)

and having direction ratios are < -3, 2, 6 > be \(\frac{x-1}{-3}\) = \(\frac{y-2}{2}\) = \(\frac{z-3}{6}\)

Question 63.

Find the vector equation of a line which passes through the points (3, 4, -7) and (1, -1, 6).

Answer:

We know that, vector equation of line passing through the points with position vectors \(\vec{a}\) and \(\vec{b}\) is \(\vec{r}\) = \(\vec{a}\) + λ(\(\vec{b}\) – \(\vec{a}\))

Here \(\vec{a}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) – 7 \(\hat{k}\) ;

\(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) + 6 \(\hat{k}\)

∴ Required vector eqn. of line be

\(\vec{r}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) – 7 \(\hat{k}\) + λ\(\hat{i}\) – \(\hat{j}\) + 6 \(\hat{k}\) – 3 \(\hat{i}\) – 4 \(\hat{j}\) + 7 \(\hat{k}\)

⇒ \(\vec{r}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) – 7 \(\hat{k}\) + λ(-2 \(\hat{i}\) – 5 \(\hat{j}\) + 13 \(\hat{k}\))

Question 64.

Write the vector equation of the line given by \(\frac{x-5}{3}\) = \(\frac{y+4}{7}\) = \(\frac{z-6}{2}\)

Answer:

Given equation of line in cartesian form be given by \(\frac{x-5}{3}\) = \(\frac{y+4}{7}\) = \(\frac{z-6}{2}\)

So the given line pass through the point (5, -4, 6) whose P.V be \(\vec{a}\) = 5 \(\hat{i}\) – 4 \(\hat{j}\) + 6 \(\hat{k}\) and having direction ratios < 3, 7, 2 > i.e. || to vector \(\vec{b}\) = 3 \(\hat{i}\) + 7 \(\hat{j}\) + 2 \(\hat{k}\)

Thus the required vector equation of line be \(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\) i.e. \(\vec{r}\) = (5 \(\hat{i}\) – 4 \(\hat{j}\) + 6 \(\hat{k}\)) + λ(3 \(\hat{i}\) + 7 \(\hat{j}\) + 2 \(\hat{k}\)) where λ be any scalar.

Question 65.

Find the vector equation of the line which passes through the point (3, 4, 5) and is parallel to the vector 2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)

Answer:

We know that, vector eqn. of line passing through the point with P.V \(\vec{a}\) and || to \(\vec{b}\) be \(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\)

Here, \(\vec{a}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) and \(\vec{b}\) = 2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)

∴ required vector eqn. of line be \(\vec{r}\) = (3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\)) +

λ(2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\))

Question 66.

Find a Cartesian form of the equation of a line which passes through a point with position vector 2 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\) and makes angles 60° , 120° and 45° with x, y, z-axis respectively.

Answer:

Since the required line pass through the point with P.V. 2 \(\hat{i}[latex] – 3 [latex]\hat{j}\) + 4 \(\hat{k}\)

∴ line must pass through the point (2, -3, 4).

D’ cosines of required line are < cos 60°, cos 120°, cos 45° >

i.e. < \(\frac{1}{2}\), \(-\frac{1}{2}\), \(\frac{1}{\sqrt{2}}\) >.

∴ D’ ratios of required line are < 1, -1, \(\sqrt{2}\) >

Hence eqn. of required line through (2, -3, 4) and having direction ratios < 1, -1, \(\sqrt{2}\) > be \(\frac{x-2}{1}\) = \(\frac{y+3}{-1}\) = \(\frac{z-4}{\sqrt{2}}\)

Question 67.

Find the acute angle between the lines \(\frac{x-4}{3}\) = \(\frac{y+3}{4}\) = \(\frac{z+1}{5}\) and \(\frac{x-1}{4}\) = \(\frac{y+1}{-3}\) = \(\frac{z+10}{5}\)

Answer:

We know that, The acute angle between the given lines \(\frac{x-x_1}{a_1}\) = \(\frac{y-y_1}{b_1}\) = \(\frac{z-z_1}{c_1}\)

and \(\frac{x-x_2}{a_2}\) = \(\frac{y-y_2}{b_2}\) = \(\frac{z-z_1}{c_2}\) be

cosθ = \(\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)

Here a_{1} = 3 ; b_{1} = 3 ; c_{1} = 5 and a_{2} = 4 ; b_{2} = -3 ; c_{2} = 5

∴ cosθ = \(\frac{|3 \times 4+4 \times(-3)+5 \times 5|}{\sqrt{9+16+25} \sqrt{9+16+25}}\)

= \(\frac{25}{\sqrt{50} \sqrt{50}}\) = \(\frac{1}{2}\)

⇒ θ = \(\frac{\pi}{3}\)

Question 68.

Find the Cartesian equation of the plane \(\vec{r}\) ( \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) = 1.

Answer:

Let P(x, y, z) be any point on given plane so its P. V be \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)

Thus eqn. of plane in cartesian form becomes ;

(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\))(\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) = 1

⇒ x + y – z = 1

Question 69.

Find the equation of the plane which passes through the points (2, 0, 0)(0, 3, 0) and (0, 0, 4).

Answer:

Since the required plane passes through the points A(2, 0, 0), B(0, 3, 0) and C(0, 0, 4). Thus the plane made intercepts on x-axis, y-axis and z-axis are 2, 3 , and 4.

We know that equation of plane having intercepts a, b, c on coordinate axes is given by

\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1

Here a = 2 ; b = 3 and c = 4

Thus equation (1) reduces to \(\frac{x}{2}\) + \(\frac{y}{3}\) + \(\frac{z}{4}\) = 1.

Question 70.

Find the sum of the intercepts made by the plane 2 x + y – z = 5 on the coordinate axes.

Answer:

Given eqn. of plane be 2 x + y – z = 5

⇒ \(\frac{x}{\frac{5}{2}}\) + \(\frac{y}{5}\) + \(\frac{z}{-5}\) = 1

We know that, lengths of intercepts made by plane \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1

on coordinate axes are a, b and c respectively.

∴ a = \(\frac{5}{2}\) ; b = 5 and c = -5

∴ Required sum = \(\frac{5}{2}\) + 5 – 5 = \(\frac{5}{2}\)

Question 71.

Find the acute angle between the planes \(\vec{r}\) (\(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)) = 1 and \(\vec{r}\) (3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\)) = 0.

Answer:

Given eqn. of planes are \(\vec{r}\) (\(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)) = 1

and \(\vec{r}\)(3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\)) = 0

We know that, the angle between the planes \(\vec{r}\) \(\overrightarrow{n_1}\) = d_{1} and \(\vec{r}\) \(\overrightarrow{n_2}\) = d_{2}be given by

cosθ = \(\frac{|\overrightarrow{n_1} \cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}\)

Here \(\overrightarrow{n_1}\) = \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\);

\(\overrightarrow{n_2}\) = 3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\)

∴ \(\overrightarrow{n_1}\) \(\overrightarrow{n_2}\) = (\(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)) (3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\))

= 1(3) – 2(-6) – 2(2) = 3 + 12 – 4 = 11

and |\(\vec{n}_1\)| = \(\sqrt{1^2+(-2)^2+(-2)^2}\)

= 3 and |\(\overrightarrow{n_2}\)| = \(\sqrt{3^2+(-5)^2+2^2}\) = 7

∴ cosθ = \(\frac{11}{3 \times 7}\) = \(\frac{11}{21}\)

⇒ θ = cos^{-1} \(\frac{11}{21}\)

Question 72.

Find the angle between the line \(\vec{r}\) = (5 \(\hat{i}\) – \(\hat{j}\) – 4 \(\hat{k}\))

+ λ(2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) and the plane

\(\vec{r}\)(3 \(\hat{i}\) – 4 \(\hat{j}\) – \(\hat{k}\)) + 5 = 0 .

Answer:

Equation of given line be \(\vec{r}\) = (5 \(\hat{i}\) – \(\hat{j}\) – 4 \(\hat{k}\)) + λ(2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))

on comparing with \(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\)

Here \(\vec{b}\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)

and equation of given plane be \(\vec{r}\) (3 \(\hat{i}\) – 4 \(\hat{j}\) – \(\hat{k}\)) + 5 = 0

Here \(\vec{n}\) = 3 \(\hat{i}\) – 4 \(\hat{j}\)

\(-\hat{k}\)

Let θ be the angle between given line and given plane

Then

sinθ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)

= \(\frac{(2 \hat{i}-\hat{j}+\hat{k}) (3 \hat{i}-4 \hat{j}-\hat{k})}{\sqrt{2^2+(-1)^2+1^2} \sqrt{3^2+(-4)^2+(-1)^2}}\)

= \(\frac{2(3)-1(-4)+1(-1)}{\sqrt{4+1+1} \sqrt{9+16+1}}\)

= \(\frac{9}{\sqrt{6} \sqrt{26}}\)