Students often turn to ISC Class 12 Maths OP Malhotra Solutions Chapter 24 The Plane Ex 24(a) to clarify doubts and improve problem-solving skills.
S Chand Class 12 ICSE Maths Solutions Chapter 24 The Plane Ex 24(a)
Question 1.
Find the direction cosines of the normal to the plane
(i) 2x – 3y + 6z = 7
(ii) x + 2 y + 2 z – 1 = 0
Answer:
(i) Given eqn. of plane be 2 x – 3 y + 6z = 7
∴ direction ratios of normal to plane be <2, -3, 6>
Thus, direction cosines of normal to given plane be
< \(\frac{2}{\sqrt{2^2+(-3)^2+6^2}}\), \(\frac{-3}{\sqrt{2^2+(-3)^2+6^2}}\), \(\frac{6}{\sqrt{2^2+(-3)^2+7^2}}\) >
i.e.< \(\frac{2}{7}\), \(\frac{-3}{7}\), \(\frac{6}{7}\)>
(ii) Given eqn. of plane be
x + 2 y + 2 z – 1 = 0
∴ direction ratios of normal to plane be < 1, 2, 2 >
Thus, direction cosines of normal to plane are
< \(\frac{1}{\sqrt{1^2+2^2+2^2}}\), \(\frac{2}{\sqrt{1^2+2^2+2^2}}\), \(\frac{2}{\sqrt{1^2+2^2+2^2}}\) >
i.e.< \(\frac{1}{3}\), \(\frac{2}{3}\), \(\frac{2}{3}\)>
Question 2.
Show that the normals to the planes x – y + z = 1, 3 x + 2 y – z + 2 = 0 are inclined to each other at an angle of 90°.
Answer:
Given eqn. of plane are
x – y + 2 z = 1
and 3 x + 2 y – z + 2 = 0
∴ Direction numbers of normals to plane (1) and (2) are
< 1, -1, 1 > and < 3, 2, -1>
Here a1 a2 + b1 b2 + c1 c2
= 1(3) – 1(2) + 1(-1) = 0
Hence both normals are inclined to each other at an angle of 90°.
Question 3.
Find the intercepts of the plane 2 x – 3 y + 4 z = 12 on the coordinate axes.
Answer:
Given eqn. of plane be 2 x – 3 y + 4 z = 12
⇒ \(\frac{x}{6}\) – \(\frac{y}{4}\) + \(\frac{z}{3}\) = 1
Thus the intercepts made by plane (1) an coordinate axes are 6,-4 and 3 .
Question 4.
Write the equation of each of the following planes in the intercept form:
(i) 2 x – 3 y + 4 z = 12
(ii) 3 x + 2 y – 5 z = 15
(iii) x + 3 y + 4 z = 12
Answer:
(i) Given eqn. of plane be,
2 x – 3 y + 4 z = 12
On dividing throughout eqn. (1) by 12 , we have
\(\frac{2 x}{12}\) – \(\frac{3 y}{12}\) + \(\frac{4 z}{12}\) = 1
⇒ \(\frac{x}{6}\) – \(\frac{y}{4}\) + \(\frac{z}{3}\) = 1
which is the required intercept form of plane.
(ii) Given eqn. of plane be
3 x + 2 y – 5 z = 15
⇒ \(\frac{3 x}{15}\) + \(\frac{2 y}{15}\) – \(\frac{5 z}{15}\) = 1
⇒ \(\frac{x}{5}\) + \(\frac{y}{\frac{15}{2}}\) – \(\frac{z}{3}\) = 1
⇒ \(\frac{x}{5}\) + \(\frac{y}{\frac{15}{2}}\) + \(\frac{z}{-3}\) = 1
which is the required intercept from.
(iii) Given eqn. of plane be,
x + 3 y + 4 z = 12
⇒ \(\frac{x}{12}\) + \(\frac{y}{4}\) + \(\frac{z}{3}\) = 1
which is the required intercept form of plane
Question 5.
Find the equation of the plane
(i) which cuts the axes of x, y and z at (-2, 0, 0),(0, 3, 0),(0, 0, 5) respectively.
(ii) which cuts the axes of x and y at (3, 0, 0) and (0, -2, 0) and does not cut the z-axis.
(iii) which cuts the x-axis at (4, 0, 0) and does not cut the y-axis and the z-axis.
(iv) passing hrough the point (2, 3, 4) and making equal intercepts on the coordinates axes.
Answer:
(i) Let the eqn. of plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
eqn. (1) passes through (-2, 0, 0)
∴ \(\frac{-2}{a}\) = 1
⇒ a = -2
eqn. (1) passes through (0, 3, 0) and (0, 0, 5)
∴ \(\frac{3}{b}\) = 1
⇒ b = 3
and \(\frac{5}{c}\) = 1
⇒ c = 5
Hence, the required eqn. of plane be
\(\frac{x}{-2}\) + \(\frac{y}{3}\) + \(\frac{z}{5}\) = 1
(ii) eqn. of plane parallel to z-axis be
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
eqn. (1) passes through (3, 0, 0)
∴ \(\frac{3}{a}\) = 1
⇒ a = 3
Also, eqn. (1) passes through (0, -2, 0)
∴ \(-\frac{2}{b}\) = 1
⇒ b = -2
∴ eqn. (1) becomes,
\(\frac{x}{3}\) – \(\frac{y}{2}\) = 1
be the required eqn. of plane.
(iii) eqn. of plane does not cut y-axis and z-axis i.e. eqn. of plane parallel to y o z plane be
x = k
eqn. (1) given to be passes through (4, 0, 0)
∴ k = 4
Thus eqn. (1) becomes; x = 4 which gives the required eqn. of plane.
(iv) Let the eqn. of plane having equal intercepts a on coordinates axes be
\(\frac{x}{a}\) + \(\frac{y}{a}\) + \(\frac{z}{a}\) = 1
Also eqn. (1) passes through (2, 3, 4)
∴\(\frac{2}{a}\) + \(\frac{3}{a}\) + \(\frac{4}{a}\) = 1
⇒ a = 9
Thus eqn. (1) becomes:
x + y + z = 9
be the req. eqn. of plane
Question 6.
Find the equations of the two 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 be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
It is given that sum of their intercepts an axes be 0
∴ a + b + c = 0
eqn. (1) passes through the point (0, 4, -3).
∴ \(\frac{0}{a}\) + \(\frac{4}{b}\) – \(\frac{3}{c}\) = 1
⇒ 4 c – 3 b = b c
Also the point (6, -4, 3) lies on eqn. (1); we have
\(\frac{6}{a}\) – \(\frac{4}{b}\) + \(\frac{3}{c}\) = 1
using eqn. (3) in eqn. (4); we have
\(\frac{6}{a}\) – 1 = 1
⇒ \(\frac{6}{a}\) = 2
⇒ a = 3
∴ from (2); c = -3 – b
∴ from (3); 4(-3 – b) – 3 b = b(- 3 – b)
⇒ – 12 – 7 b = -3 b – b2
⇒ b2 – 4 b – 12 = 0
⇒ (b – 6)(b + 2) = 0
⇒ b = 6,-2
when b = 6
∴ c = -3 – b = -9
when b = -2
∴ c = -3 + 2 = -1
when a = 3 ; b = 6
and c = -9
∴ eqn. (1) becomes ; \(\frac{x}{3}\) + \(\frac{y}{6}\) + \(\frac{z}{-9}\) = 1
when a = 3 ; b = -2 and c = -1
∴ eqn. (1) becomes;
\(\frac{x}{3}\) + \(\frac{y}{-2}\) + \(\frac{z}{-1}\) = 1
be the reqd. eqns. of planes.
Question 7.
A plane meets the coordinate axes at A, B, C respectively such that the centroid of the triangle A B C is (1, -2, 3). Find the equation of the plane.
Answer:
Let the required eqn. of plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
Then the coordinates of A, B and C are A (a, 0, 0), B(0, b, 0) and C(0, 0, c).
∴ centroid of the ∆ ABC are (\(\frac{a}{3}\), \(\frac{b}{3}\), \(\frac{c}{3}\)) also given centroid of ∆ ABC are (1, -2, 3)
i.e. \(\frac{a}{3}\) = 1
⇒ a = 3 ;
\(\frac{b}{3}\) = -2
⇒ b = -6
and \(\frac{c}{3}\) = 3
⇒ c = 9
putting the values of a, b, c in eqn. (1); we get
\(\frac{x}{3}\) + \(\frac{y}{-6}\) + \(\frac{z}{9}\) = 1
⇒ 6 x – 3 y + 2 z = 18
be the required equation of plane.
Question 8.
(i) Find the equation of the plane the normal to which from the origin is of length of 5 and which makes angles, with the axes of x, y and z, equal to 120°, 45° and 120° respectively.
(ii) Find the equation of the plane the normal to which from the origin is of length 4 and which makes angles with the axes of x, y, z equal to 90°, 135°, 45° respectively.
(iii) Find the equation of the plane such that the foot of the normal to which from the origin is the point (2, 3, 1).
(iv) Find the equation of the plane such that the length of the perpendicular to which from the origin is equal to 2 units, and the angles α, β, γ that this perpendicular makes with the axes of x, y, z are connected by the relation
\(\frac{\cos \alpha}{-1}\) = \(\frac{\cos \beta}{4}\) = \(\frac{\cos \gamma}{4}\) .
Answer:
(i) Here p = length of the normal from origin = 5
∴D cosines of normal to plane are < cos 120°, cos 45°, cos 120° >
i.e. < \(-\frac{1}{2}\), \(\frac{1}{\sqrt{2}}\), \(-\frac{1}{2}\) >
i.e. l = \(-\frac{1}{2}\) ; m = \(\frac{1}{\sqrt{2}}\) ; n = \(-\frac{1}{2}\)
Thus required eqn. of plane be
b x + m y + n z = p
\(\frac{-x}{2}\) + \(\frac{y}{\sqrt{2}}\) y – \(\frac{1}{2}\) z = 5
⇒ x – \(\sqrt{2}\) y + z + 10 = 0
(ii) Here p = length of the normal from origin = 4
∴ D cosines of normal to plane are < cos 90°, cos 135°, cos 45° >
i.e. < 0, \(-\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\) >
i.e. l = 0 ; m = \(-\frac{1}{\sqrt{2}}\) ; n = \(\frac{1}{\sqrt{2}}\)
Hence the required eqn. of plane be
b e + m y + n z = p
⇒ 0 x – \(\frac{1}{\sqrt{2}}\) y + \(\frac{1}{\sqrt{2}}\) z = 4
⇒ -y + z = 4 \(\sqrt{2}\)
⇒ y – z + 4 \(\sqrt{2}\) = 0
be the required eqn. of plane
(iii) Let the eqn. of plane through the point (2, 3, 1) be given by
a(x – 2) + b(y – 3) + c(z – 1) = 0 .
Where < a, b, c > are the direction ratios of the normal to plane
∴ direction ratios of normal to plane ON are < 2, 3, 1 >
Thus eqn. (1) becomes;
2(x – 2) + 3(y – 3) + 1(z – 1) = 0
⇒ 2 x + 3 y + z – 14 = 0
be the reqd. eqn. of plane.
(iv) Given p = length of ⊥ to the reqd. plane from origin =2 units
Since α, β , and γ are the angles made by this perpendicular with coordinates areas
Then direction cosines of this normal are < cosα, cosβ , cosγ >
Given \(\frac{\cos \alpha}{-1}\) = \(\frac{\cos \beta}{4}\) = \(\frac{\cos \gamma}{8}\)
= \(\frac{\sqrt{\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma}}{\sqrt{1+16+64}}\)
= ± \(\frac{1}{9}\)
⇒ cos α = ± \(\frac{1}{9}\) ; cos β = ± \(\frac{4}{9}\)
and cos γ = ± \(\frac{8}{9}\)
Thus direction cosines of normal to plane are < ± \(\frac{1}{9}\), ± \(\frac{4}{9}\), ± \(\frac{8}{9}\) >
i.e. l = ± \(\frac{1}{9}\) ; m = ± \(\frac{4}{9}\) and n = ± \(\frac{8}{9}\)
Hence the reqd. eqn. of plane be
b e + m y + n z = p
\(\frac{-1}{9}\) x + \(\frac{4}{9}\) y + \(\frac{8}{9}\) z = 2
i.e x – 4 y – 8 z + 18 = 0
and \(\frac{x}{9}\) – \(\frac{4 y}{9}\) – \(\frac{8 z}{9}\) = 2
⇒ x – 4 y – 8 z – 18 = 0
Question 9.
Reduce each of the following equations to the normal form and then determine the direction cosines and the length of the normal from the origin :
(i) 2 x – 2 y + z – 12 = 0
(ii) 9 x + 6 y – 2 z + 7 = 0
Answer:
(i) Given eqn. of plane be
2 x – 2 y + z – 12 = 0
Dividing eqn. (1) throughout by \(\sqrt{2^2+(-2)^2+1^2}\) i.e. 3 , we get
\(\frac{2}{3}\) x – \(\frac{2}{3}\) y + \(\frac{z}{3}\) – \(\frac{12}{3}\) = 0
\(\frac{2}{3}\) x – \(\frac{2}{3}\) y + \(\frac{z}{3}\) = 4
which is the required normal form of plane.
On comparing with lx + m y + n z = p
Here l = \(\frac{2}{3}\) ; m = \(-\frac{2}{3}\)
and n = \(\frac{1}{3}\) and p = 4
Thus, the direction cosines of normal to plane are < \(\frac{2}{3}\), \(-\frac{2}{3}\), \(\frac{1}{3}\) >
and p = length of normal from origin = 4.
(ii) Given eqn. of plane be
9 x + 6 y – 2 z + 7 = 0
on dividing eqn. (1) throughout by
\(\sqrt{9^2+6^2+c-2^2}\)
i.e. \(\sqrt{81+36+4}\)
i.e. 11 ; we have
\(\frac{9}{11}\) x + \(\frac{6}{11}\) y – \(\frac{2}{11}\) z + \(\frac{7}{11}\) = 0
⇒ \(\frac{9 x}{11}\) + \(\frac{6 y}{11}\) – \(\frac{2 z}{11}\) = \(\frac{-7}{11}\)
⇒ \(\frac{-9}{11}\) x – \(\frac{6}{11}\) y + \(\frac{2}{11}\) z = \(\frac{7}{11}\)
which is the required normal form and on comparing with l x + m y + n z = p
Here, l = \(-\frac{9}{11}\) ; m = \(-\frac{6}{11}\)
and n = \(\frac{z}{11}\) and p = \(\frac{7}{11}\).
Thus reqd. direction cosines of normal to plane are < \(\frac{-9}{11}\), \(\frac{-6}{11}\), \(\frac{2}{11}\) >
and p = length of normal from origin = \(\frac{7}{11}\)
Question 10.
For each of the following planes, find the direction ratios and the direction cosines of the normal from the origin.
(i) 3 x + 2 y – 6 z + 22 = 0
(ii) 4 x – 3 y + 5 z – 30 = 0
(iii) 5 y – 12 z + 20 = 0
Answer:
(i) Given eqn. of plane be
3 x+2 y-6 z+2 z=0
∴ Direction ratios of normal to plane are < 3, 2, -6 >
⇒ 3 x + 2 y – 6 z = -22
⇒ -3x – 2 y + 6 z = 22
on dividing eqn. (1) by
(-2)2 + 62 i.e. 7 ;
we get
\(\frac{-3}{7}\) x – \(\frac{2}{7}\) y + \(\frac{6}{7}\) z = \(\frac{22}{7}\)
Thus the required direction cosines of normal to plane are < \(\frac{-3}{7}\), \(\frac{-2}{7}\), \(\frac{6}{7}\)
(ii) Given eqn. of plane be
4 x – 3 y + 5 z – 30 = 0
∴ direction ratios of normal to plane are < 4,-3,5 >
⇒ 4 x – 3 y + 5 z = 30
on diviaing throughout by \(\sqrt{16+9+25}\) i.e. 5 \(\sqrt{2}\); we get
\(\frac{4}{5 \sqrt{2}}\) x – \(\frac{3}{5 \sqrt{2}}\) y + \(\frac{1}{\sqrt{2}}\) z = \(\frac{30}{5 \sqrt{2}}\)
on comparing with b x + m y + n z = p
Here l = \(\frac{4}{5 \sqrt{2}}\) ; m = \(\frac{-3}{5 \sqrt{2}}\) and n = \(\frac{1}{\sqrt{2}}\) Hence the direction cosines of normal to plane are < \(\frac{4}{5 \sqrt{2}}\), \(\frac{-3}{5 \sqrt{2}}\), \(\frac{1}{\sqrt{2}}\) >.
(iii) Given eqn. of plane be 5 y – 12z = -20
∴ direction ratios of normal to plane are < 0, 5, -12 >
∴ -5 y + 12 z = 20
on dividing eqn. (1) throughout by
\(\sqrt{(-5)^2+12^2}\) = \(\sqrt{169}\) = 13 ; we get
Thus, the direction cosines of normal to plane are < 0, \(\frac{-5}{13}\), \(\frac{12}{13}\)
Question 11.
(i) Find the coordinates of the point, where the line \(\frac{x+1}{2}\) = \(\frac{y+2}{3}\) = \(\frac{z+3}{4}\) meets the plane x + y + 4 z = 6.
(ii) 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 = 1. Hence, find the distance of this point from the point (5, 4, 1).
(iii) Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the X Z plane: Also find the angle which the line makes with the X Z plane.
Answer:
(i) Given eqn. of line be
\(\frac{x+1}{2}\) = \(\frac{y+2}{3}\) = \(\frac{z+3}{4}\) = t (say)
So any point on line (1) be P(2 t – 1, 3 t – 2, 4 t – 3)
Given eqn. of plane be
x + y + 4 z = 6
For the point of intersection of line (1) and plane (2)
∴ point P lies on eqn. (2).
∴ 2 t – 1 + 3 t – 2 + 4(4 t – 3) = 6
⇒ 21 t = 21
⇒ t = 1
Hence the required point of intersection be P(2 – 1,3 – 2,4 – 3) i.e. P(1, 1, 1).
(ii) eqn. of line through the points (1, -2, 3) and (2, -1, 5) is given by
\(\frac{x-1}{2-1}\) = \(\frac{y+2}{-1+2}\) = \(\frac{z-3}{5-3}\)
i.e. \(\frac{x-1}{1}\) = \(\frac{y+2}{1}\) = \(\frac{z-3}{2}\)
∴ any point on line (1) be given by
\(\frac{z+1}{1}\) = \(\frac{y+2}{1}\) = \(\frac{z-3}{2}\) = t (say)
P (t + 1, t – 2, 2 t + 3)
Also given eqn. of plane be
x – 2 y + 3 z = 19
For point of intersection of line (1) and plane (2), the point P(t + 1, t – 2,2 t + 3) lies on plane (2).
∴ t + 1 – 2(t – 2) + 3(2 t + 3) = 19
⇒ 5 t + 14 = 19 ⇒ 5 t = 5 ⇒ t = 1
Thus the required point of intersection of given line (1) and (2) be
P(1 + 1, 1 – 2, 2 + 3) i.e. P(2, -1, 5)
∴ required distance of P(2, -1, 5) from given point (5, 4, 1)
= (5 – 2)2 + (4 + 1)2 + (1 – 5)2
= \(\sqrt{9+25+16}\) = \(\sqrt{50}\) = 5 \(\sqrt{2}\) units
(iii) eqn. of line passing through the points A(3, 4, 1) and B(5, 1, 6) is given by
\(\frac{x-3}{5-3} \) = \(\frac{y-4}{1-4}\) = \(\frac{z-1}{6-1}\)
i.e. \(\frac{x-3}{2}\) = \(\frac{y-4}{-3}\) = \(\frac{z-1}{5}\)
Any point on line (1) be
i.e \(\frac{x-3}{2}\) = \(\frac{y-4}{-3}\) = \(\frac{z-1}{5}\) = t (say)
given by P(2 t + 3, -3 t + 4,5 t + 1)
eqn. of XOZ plane be y=0
Since the line (1) crosses XOZ plane
∴ point P lies on eqn. (2)
∴-3 t + 4 = 0 ⇒ t = 4 / 3
∴ required point of intersection of line (1) and plane ( 2 ) be:
P (\(\frac{8}{3}\) + 3, -4 + 4, \(\frac{20}{3}\) + 1)
i.e. P (\(\frac{17}{3}\), 0, \(\frac{23}{3}\))
Let θ be the angle between plane (2) and line (1)
Then 90° – θ be the angle between line (1) and normal to plane (2).
Now, direction ratios of line (1) are < 2, -3, 5 >and direction ratios of normal to plane (2) are < 0, 1, 0 >.
∴ cos (90° – θ) = \(\frac{|2(0)-3(1)+5(10)|}{\sqrt{2^2+9+25} \sqrt{0^2+1^2+0}}\)
= \(\frac{3}{\sqrt{38}}\)
⇒ sinθ
= \(\frac{3}{\sqrt{38}}\)
Question 12.
Foot of the perpendicular from the origin to the plane is (2, 3, 4). Find the equation of the Plane.
Answer:
The required eqn. of plane through the point N(2, 3, 4) be given by
a(x – 2) + b(y – 3) + c(z – 4) = 0
Where < a, b, c >are the direction numbers of normal to piane.
Further ON be the normal to the plane (1) Also, direction ratios of ON are < 2, 3, 4 >
∴\(\frac{a}{2}\) = \(\frac{b}{3}\) = \(\frac{c}{4}\) = k (say)
i.e. a = 2 k ; b = 3 k and c = 4 k
∴ eqn. (1) becomes ;
2(x – 2) + 3(y – 3) + 4(z – 4) = 0
⇒ 2 x + 3 y + 4 z – 29 = 0
be the reqd. eqn. of plane.
Question 13.
Find the distance of the point (2, 3, 4) from the plane 3 x + 2 y + 2 z + 5 = 0, measured parallel to the line
\(\frac{x+3}{3}\) = \(\frac{y-2}{6}\) = \(\frac{z}{2}\) .
Answer:
eqn. of given line be
\(\frac{x+3}{3}\) = \(\frac{y-2}{6}\) = \(\frac{z}{2}\)
eqn.of any line through point A(2, 3, 4) and |l to line (1) be given by
\(\frac{x-2}{3}\) = \(\frac{y-3}{6}\) = \(\frac{z-4}{2}\)
So any point an line (2) be given by
\(\frac{x-2}{3}\) = \(\frac{y-3}{6}\) = \(\frac{z-4}{2}\) = t (say)
P(3 t + 2, 6 t + 3, 2 t + 4)
Since point P lies on given plane
3 x + 2 y + 2 z + 5 = 0
i.c. 3(3 t + 2) + 2(6 t + 3) + 2(2 t + 4) + 5 = 0
⇒ 25 t + 25 = 0 ⇒ t = -1
Hence the coordinates of P becomes
(-3 + 2, -6 + 3, -2 + 4) i.e. P(-1, -3, 2)
Thus required distance = |AP|
= \(\sqrt{(-1-2)^2+(-3-3)^2+(2-4)^2}\)
= \(\sqrt{9+36+4}\) = 7 units
Question 14.
Find the equation in Cartesian form of the plane passing through the point (3, – 3, 1) and normal to the line joining the points (3, 4, -1) and (2, -1, 5).
Answer:
The eqn. of plane through the point (3, -3, 1) be given by
a(x – 3) + b(y + 3) + c(z – 1) = 0
Where < a, b, c >are the direction ratios of normal to plane.
Also direction ratios of normal to plane are
< 2 – 3, -1 – 4, 5 + 1 > i.e. < -1, -5, 6 >
∴ \(\frac{a}{-1}\) = \(\frac{b}{-5}\) = \(\frac{c}{6}\) = k (say)
a = -k ; b = -5 k ; c = 6 k
putting all these values in eqn. (1); we have
⇒ -k(x – 3) – 5 k(y + 3) + 6 k(z – 1) = 0
⇒ -x – 5 y + 6 z – 18 = 0
⇒ x + 5 y – 6 z + 18 = 0
be the required eqn. of plane.
Question 15.
Find the image of the point
(i) (3, -2, 1) in the plane
3 x – y + 4 z = 2
(ii) (0,0,0) in the plane
3 x + 4 y – 6 z + 1 = 0
Answer:
(i) Let M be the foot of ⊥ drawn from P(3, -2, 1) on given plane
3 x – y + 4 z = 2
∴ eqn. of line PM i.e. line passing through the point P(3, -2, 1) and normal to plane (1) be
\(\frac{x-3}{3}\) = \(\frac{y+2}{-1}\) = \(\frac{z-1}{4}\) = t (say)
So any point on line (2) be
Q(3 t + 3, -t – 2, 4 t + 1)
NowQ be the image of P in plane (1) if mid point of PQ i.e. M lies on plane (1) Here coordinates of
M (\(\frac{3 t+3+3}{2}\), \(\frac{-t-2-2}{2}\), \(\frac{4 t+1+1}{2}\))
i.e. M (\(\frac{3 t+6}{2}\), \(\frac{-t-4}{2}\), \(\frac{4 t+2}{2}\))
Now M lies on plane (1)
∴ 3(\(\frac{3 t+6}{2}\)) – (\(\frac{-t-4}{2}\)) + 4(\(\frac{4 t+2}{2}\)) = 2
⇒ 9 t + 18 + t + 4 + 16 t + 8 = 4
⇒ 26 t + 26 = 0
⇒ t = -1
Hence the required coordinates of image of given point in given plane be
Q (3 – 3, + 1 – 2, -4 + 1)
i.e. Q(0, -1, -3)
(ii) Let M be the foot of ⊥ drawn from O(0, 0, 0) on given plane
3 x + 4 y – 6 z + 1 = 0
This eqn. of line OM i.e. line passing through the point O(0, 0, 0) and norinal to plane (1) be
\(\frac{x-0}{3}\) = \(\frac{y-0}{4}\) = \(\frac{z-0}{-6}\) = t (say)
So any point on line (2) be P(3 t, 4 t, – 6 t)
Now P be the image of point O in given plane
Then mid point of OP i.e. M lies on given plane (1)
Here coordinates of M(\(\frac{3 t}{2}\), \(\frac{4 t}{2}\), \(\frac{-6 t}{2}\)) and lies on plane (1)
3(\(\frac{3 t}{2}\)) + 4(\(\frac{4 t}{2}\)) – 6(\(-\frac{6 t}{2}\)) + 1 = 0
⇒ 9 t + 16 t + 36 t + 2 = 0
⇒ t = \(-\frac{2}{61}\)
Hence required coordinates of image of given point in plane (1) be
P( \(-\frac{6}{61}\), \(\frac{-8}{61}\), \(\frac{12}{61}\)).
Question 16.
Find the reflection of the point (1, 2, -1) in the plane 3 x – 5 y + 4 z = 5.
Answer:
Let M be the foot of ⊥ drawn from given point P(1, 2, -1) on given plane
3 x – 5 y + 4 z = 5
Thus eqn. of PM i.e. the line through the point P(1, 2, -1) and normal to plane (1) be given by
\(\frac{x-1}{3}\) = \(\frac{y-2}{-5}\) = \(\frac{z+1}{4}\) = t (say)
So any point on line (2) be
Q(3 t + 1, -5 t + 2, 4 t – 1)
Now point Q be the image of P in plane (1) if the mid point of line segment P Q
i.e. M (\(\frac{3 t+1+2}{2}\), \(\frac{-5 t+2+2}{2}\), \(\frac{4 t-1-1}{2}\))
lies on plane (1)
∴ 3(\(\frac{3 t+2}{2}\)) – 5(\(\frac{-5 t+4}{2}\)) + 4(\(\frac{4 t-2}{2}\)) = 5
⇒ 9 t + 6 + 25 t – 20 + 16 t – 8 = 10
⇒ 50 t = 32 ⇒ t = \(\frac{16}{25}\)
Hence the required coordinates of reflection of point P be Q(\(\frac{48}{25}\) + 1, \(\frac{-80}{25}\) + 2, \(\frac{64}{25}\) – 1) i.e. Q(\(\frac{73}{25}\), \(\frac{-6}{5}\), \(\frac{39}{25}\))
Question 17.
From the point P(1, 2, 4), a perpendicular is drawn on the plane 2 x + y – 2 z + 3 = 0. Find the equation, the length and the coordinates of the foot of the perpendicular.
Answer:
Let M be the foot of ⊥ drawn from given point P(1, 2, 4) on given plane
2 x + y – 2 z + 3 = 0
∴ eqn. of line PM i.e. the line through P(1, 2, 4) and normal to plane (1) be given by
\(\frac{x-1}{2}\) = \(\frac{y-2}{1}\) = \(\frac{z-4}{-2}\) = t (say)
So any point on line (2) be M(2 t + 1, t + 2, -2 t + 4) clearly M lies on plane (1).
∴ 2(2 t + 1) + t + 2 – 2(-2 t + 4) + 3 = 0
⇒ 4 t + 2 + t + 2 + 4 t – 8 + 3 = 0
⇒ 9 t – 1 = 0 ⇒ t = \(\frac{1}{9}\)
∴ Coordinates of foot of ⊥ M be given by
(\(\frac{2}{9}\) + 1, \(\frac{1}{9}\) + 2, \(\frac{-2}{9}\) + 4)
i.e. M(\(\frac{11}{9}\), \(\frac{19}{9}\), \(\frac{34}{9}\))
and
|P M| = \(\sqrt{(\frac{11}{9}-1)^2 + (\frac{19}{9}-2)^2 + (\frac{34}{9}-4)^2)}\)
= \(\sqrt{\frac{4}{81}+\frac{1}{81}+\frac{4}{81}}\) = \(\sqrt{\frac{1}{9}}\) = \(\frac{1}{3}\) units
Question 18.
Find the distance of the point (1, -2, 3) from the plane x – y + z = 5 measured along a line parallel to \(\frac{x}{2}\) = \(\frac{y}{3}\) = \(\frac{z}{-6}\).
Answer:
Given eqn. of line be
\(\frac{x}{2}\) = \(\frac{y}{3}\) = \(\frac{z}{-6}\)
Thus eqn. of line through the given point A(1, -2, 3) and parallel to line (1) be given by
\(\frac{x-1}{2}\) = \(\frac{y+2}{3}\) = \(\frac{z-3}{-6}\) = t (say)
So any point on line (2) be
P(2 t + 1, 3 t – 2, -6 t + 3)
Since the point P} lies on given plane
x – y + z = 5
2 t + 1 – (3 t – 2) + (-6 t + 3) = 5
⇒ -7 t = -1
⇒ t = 1 / 7
Thus required coordinates of point P are (\(\frac{2}{7}\) + 1, \(\frac{3}{7}\) – 2, \(\frac{-6}{7}\) + 3)
i.e. P (\(\frac{9}{7}\), \(\frac{-11}{7}\), \(\frac{15}{7}\))
Thus required distance = |A D|
= \(\sqrt{(\frac{9}{7}-1)^2 + (\frac{-11}{7}+2)^2 + (\frac{15}{7}-3)^2}\)
= \(\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}\)
= \(\sqrt{\frac{49}{49}}\) = 1 units.
Question 19.
Prove that a variable plane which moves so that the sum of reciprocals of its intercepts on the three coordinate axes is constant, passes through a fixed point.
Answer:
Let the eqn. of variable plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
Where a, b, c are the lengths of intercepts cut off by plane on coordinates ares.
also it is given that
\(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\) = k (constant)
⇒ \(\frac{1}{k a}\) + \(\frac{1}{k b}\) + \(\frac{1}{k c}\) = 1
⇒ \(\frac{1}{a}\) (\(\frac{1}{k}\)) + \(\frac{1}{b}\) (\(\frac{1}{k}\)) + \(\frac{1}{c}\) (\(\frac{1}{k}\)) =1
eqn. (2) shows that eqn. (1) passes through (\(\frac{1}{k}\), \(\frac{1}{k}\), \(\frac{1}{k}\)) which is a fixed point, since k be a constant.
Hence the variable plane passes through fixed point.
Question 20.
(i) A variable plane is at a constant distance p from the origin and meets the axes in A, B, C. Show that the locus of the centroid of triangle A B C is
x-2 + y-2 + z-2 = 9 p-2
(ii) A variable plane, which remains at a constant distance of 9 units from the origin, cuts the coordinates axes at the points A, B and C. Show that the locus of the centroid of ∆ MBC is
\(\frac{1}{x^2}\) + \(\frac{1}{y^2}\) + \(\frac{1}{z^2}\) = \(\frac{1}{9}\)
Answer:
(i) Let the eqn. of variable plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
p = ⊥ distance from (0, 0, 0) an plane (1)
⇒ p = \(\frac{\left|\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1\right|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\)
⇒ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\) = \(\frac{1}{p}\) ;
on squaring ; we have
\(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) + \(\frac{1}{c^2}\) = \(\frac{1}{p^2}\)
Since the plane (1) meets x-axis (y = 0 = z) at A
i.e: \(\frac{x}{a}\) = 1 ⇒ x = a
∴ coordinates of A are (a, 0,0)
Now plane (1) meets y-axis (x = 0 = z) at B
∴ from (1); \(\frac{y}{b}\) = 1 ⇒ y = b
Thus coordinates of point B are (0, b, 0) Hence (1) meets z-axis (x = 0 = y) at point C
∴ from (1); \(\frac{z}{c}\) = 1 ⇒ z = c
∴ coordinates of point C are (0,0, c) Thus centroid of ∆ABC be given by
(\(\frac{a+0+0}{3}\), \(\frac{0+b+0}{3}\), \(\frac{0+0+c}{3}\))
i.e. (\(\frac{a}{3}\), \(\frac{b}{3}\), \(\frac{c}{3}\))
Also, let the centroid of ∆ABC be G(α, β, γ)
∴ α = \(\frac{a}{3}\), β = \(\frac{b}{3}\) and γ = \(\frac{c}{3}\)
⇒ a = 3 α, b = 3 β and c = 3 γ
Putting the values of a, b and c in eqn. (1); we get
\(\frac{1}{9 α^2}\) + \(\frac{1}{9 β^2}\) + \(\frac{1}{9 γ^2}\) = \(\frac{1}{p^2}\)
Hence the locus of G(α, β, γ) be given by
[i.e. replacing α by x ; β by y and γ by z ]
∴ from (2); \(\frac{1}{9 x^2}\) + \(\frac{1}{9 b^2}\) + \(\frac{1}{9 z^2}\) = \(\frac{1}{p^2}\)
∴ x-2 + y-2 + z-2 = 9 p-2
(ii) Let the eqn. of variable plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
given 9 = ⊥ distance from (0, 0, 0) on plane (1)
⇒ 9 = \(\frac{|0+0+0-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\)
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\) = \(\frac{1}{9}\)
on squaring; we have
\(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) + \(\frac{1}{c^2}\) = \(\frac{1}{81}\)
eqn. (1) m eets x-axis at A i.e. y = 0 = z
∴ from (1); \(\frac{x}{a}\) = 1 ⇒ x = a
∴ coordinates of A} are (a, 0,0) eqn. (1) meets y-axis at B i.e. x = 0 = z
∴ from (1); \(\frac{y}{b}\) = 1 ⇒ y = b
∴ coordinates of B} are (0, b, 0)
eqn. (1) meets z-axis at c i.e. x = 0 = y
∴ from (1); \(\frac{z}{c}\) = 1 ⇒ z = c
∴ coordinates of C are (0,0, c)
∴ centroid of ∆ABC be (\(\frac{a}{3}\), \(\frac{b}{3}\), \(\frac{c}{3}\))
Also but centroid of ∆ABC be G(α, β, γ)
i.e. \(\frac{a}{3}\) = α ⇒ a = 3 α
and \(\frac{c}{3}\) = γ ⇒ c = 3 γ
and \(\frac{b}{3}\) = β ⇒ b = 3 β
∴ From (2); \(\frac{1}{α^2}\) + \(\frac{1}{β^2}\) + \(\frac{1}{γ^2}\) = \(\frac{9}{81}\) = \(\frac{1}{9}\)
Thus, the locus of G(α, β, γ) be given by
\(\frac{1}{x^2}\) + \(\frac{1}{y^2}\) + \(\frac{1}{z^2}\) = \(\frac{1}{9}\)
[Replacing α by x ; β by y and γ by z ].
Question 21.
Find the equation of the plane passing through the following points :
(i) A(2, 2, -1), B(3, 4, 2), Q(7, 0, 6)
(ii) A(2, 1, 0), B(3, -2, -2), C(3, 1, 7).
Answer:
(i) Let the eqn. of plane through the point A (2, 2, -1) be given by a(x – 2) + b(y – 2) + c(z + 1) = 0 ……….. (1)
Where < a, b, c > be the direction ratios of normal to plane (1)
plane (1) passes through the point B(3, 4, 2)
∴ a(3 – 2) + b(4 – 2) + c(2 + 1) = 0
i.e. a + 2 b + 3 c = 0
Also point C(7, 0, 6) lies on plane
a(7 – 2) + b(0 – 2) + c(6 + 1) = 0
5 a – 2 b + 7 c = 0
On solving eqn. (2) and (3) by using cross-multiplication methods we have
\(\frac{a}{14+6}\) = \(\frac{b}{15-7}\) = \(\frac{c}{-2-10}\)
⇒ \(\frac{a}{20}\) = \(\frac{b}{8}\) = \(\frac{c}{-12}\) = k (say) ; k < 0
⇒ a = 20 k ; b = 8 k ; c = -12 k
Putting the values of a, b & c in eqn. (1); we have
20 k(x – 1) + 8 k(y – 2) – 12 k(z + 1) = 0
⇒ 20 x + 8 y – 12 z – 48 = 0
⇒ 5 x + 2 y – 3 z – 12 = 0 .
(ii) eqn. of any plane through te point (2, 1, 0) is given by
a(x – 2) + b(y – 1) + c(z – 0) = 0
Since the plane passes through given points (3, -2, -2) and (3, 1, 7). Thus (3, -2, -2) and (3, 1, 7) lies on eqn. (1). a(3 – 2) + b(-2 – 1) + c(-2 – 0) = 0
⇒ a – 3 b – 2 c = 0
also, a(3 – 2) + b(1 – 1) + c(7 – 0) = 0
⇒ a + 0 b + 7 c = 0
On solving (2) and (3) ; we have
\(\frac{a}{-21-0}\) = \(\frac{b}{-2-7}\) = \(\frac{c}{0+3}\)
i.e. \(\frac{a}{-21}\) = \(\frac{b}{-9}\) = \(\frac{c}{3}\) = k (say)
i.e. a = -21 k ; b = -9 k ; c = 3 k
∴ From eqn. (1); we have
-21 k(x – 2) – 9 k(y – 1) + 3 k(z – 0) = 0
⇒ -3 k [7 x – 14 + 3 y – 3 – z] = 0
⇒ 7 x + 3 y – z = 17
be the required equation of plane.
Question 22.
Show that the following points are coplanar: (-6, 3, 2), (3, -2, 4), (5, 7, 3) and (-13, 17, -1)
Answer:
For proving any four points A, B, C D are coplanar if plane through any three points must pass through the remaining fourth point.
Let the eqn.of plane through the point A(-6, 3, 2) is given by
a(x + 6) + b(y – 3) + c(z – 2) = 0
Now plane (1) passes through the point B (3, -2, 4)
a(3 + 6) + b(-2 – 3) + c(4 – 2) = 0
∴ 9 a – 5 b + 2 c = 0
Again the point C (5, 7, 3) lies on plane (1).
∴ a(5 + 6) + b(7 – 3) + c(3 – 2) = 0
i.e. 11 a + 4 b + c = 0
On solving eqn. (2) and eqn. (3) using crossmultiplication method, we have
\(\frac{a}{-5-8}\)
= \(\frac{b}{22-9}\)
= \(\frac{c}{36+55}\)
i.e. \(\frac{a}{-13}\)
= \(\frac{b}{13}\)
= \(\frac{c}{91}\) = k (say) ; k < 0
⇒ a = -13 k ; b = 13 k ; c = 91
k putting the values of a, b c in eqn. (1); we have
-13 k(x + 6) + 13 k(y – 3) + 91 k(z – 2) = 0
⇒ -13 x + 13 y + 91 z – 299 = 0
⇒ x – y – 7 z + 23 = 0
Further point D lies on eqn …………… (4)
∴ D(-13, 17, -1) satisfies eqn …………… (4)
i.e. -13 – 17 + 7 + 23 = 0
⇒ 0 = 0, which is true
Hence all the given four points lies in a same plane
∴ all the given points are coplanar.
Question 23.
Show that the line joining the points (0, -1, 2) and (2, -1, -1) is coplanar with the line joining the points (1, 1, 1) and (0, 3, 3).
Answer:
Eqn. of plane through the point (0, -1, 2) is given by
a(x – 0) + b(y + 1) + c(z – 2) = 0
Now the point (2, -1, -1) lies on plane (1).
∴ a(2 – 0) + b(-1, + 1) + c(-1 – 2) = 0
⇒ 2 a + 0 b – 3 c = 0
Again the point (1, 1, 1) lies on plane (1).
∴ a + 2 b – c = 0
Solving eqn. (2) (3) by using crossmultipluation method, we have
\(\frac{a}{0+6}\)
= \(\frac{b}{-3+2}\) + \(\frac{c}{4-0}\)
\(\frac{a}{6}\)
= \(\frac{b}{-1}\) = \(\frac{c}{4}\) = k (say)
∴ a = 6 k ; b = -k ; c = 4 k where k < 0
Putting all these values in eqn. (1); we have
6 k x – k(y + 1) + 4 k(z – 2) = 0
⇒ 6 x – y + 4 z – 9 = 0
Now the remaining point (0, 3, 3) lies on plane (4) if (0, 3, 3) satisfies eqn. ………… (4).
if 6 × 0 – 3 + 12 – 9 = 0 if 0 = 0, which is true.
Hence all the four points (0, -1, 2), (2, -1, -1), (1, 1, 1) and (0, 3, 3).