ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry Ex 2.8

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ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry Ex 2.8

Very short answer type questions (1 to 3) :

Question 1.
In each of the following problems, find the perpendicular distance :
(i) from the origin to the plane 3x – 4y + 12z = 3.
(ii) from the point (- 6, 0, 0) to the plane 2x – 3y + 6z – 2 = 0. (NCERT)
(iii) from the point (3, – 2, 1) to the plane 2x – y + 2z + 3 = 0. (NCERT)
Solution:
(i) eqn. of given plane be,
3x – 4y + 12z – 3 = 0

∴ Required ⊥ distance from (0, 0, 0) to plane (1)
= \(\frac{|3 \times 0-4 \times 0+12 \times 0-3|}{\sqrt{3^2+(-4)^2+12^2}}\)
= \(\frac{3}{\sqrt{9+16+144}}\)
= \(\frac{13}{3}\) units

(ii) Required ⊥ distance from (- 6, 0, 0) to plane 2x – 3y + 6z – 2 = 0
= \(\frac{|2(-6)+0+0-2|}{\sqrt{4+9+36}}\)
= \(\frac{14}{7}\)
= 2 units

(iii) Here given point be (3, – 2, 1) and eqn. of given plane is 2x – y + 2z + 3 = 0
∴ Required ⊥ distance = \(\frac{|2(3)-(-2)+2(1)+3|}{\sqrt{4+1+4}}\)
= \(\frac{|6+2+2+3|}{3}\)
= \(\frac{13}{3}\) units

Question 1 (old).
(i) the origin to the plane 2x – 3y + 6z + 21 = 0.
Solution:
∴ required ⊥ distance from O (0, 0, 0) to given plane 2x – 3y + 6z + 21 = 0

= \(\frac{|2(0)-3(0)+6(0)+21|}{\sqrt{2^2+(-3)^2+6^2}}\)
= \(\frac{21}{\sqrt{49}}=\frac{21}{7}\)
= 3 units

Question 2.
(i) Find the distance of the point whose position vector is \(2 \hat{i}+\hat{j}-\hat{k}\) from the plane \(\vec{r} \cdot(\hat{i}-2 \hat{j}+4 \hat{k})\) = 9. (NCERT Exemplar)
(ii) Find the length of the perpendicular from origin to the plane \(\vec{r} \cdot(3 \hat{i}-4 \hat{j}-12 \hat{k})\) + 39 = 0. (ISC 2019)
(iii) Find the distance of the point (3, 4, 5) from the plane \(\vec{r} \cdot(2 \hat{i}-5 \hat{j}+3 \hat{k})\) = 13.
(iv) Write the distance from the point (2, 3, – 5) to the plane XY-plane.
Solution:
(i) We know that,
⊥ distance from point whose P.V is \(\vec{a}\) to the plane
\(\vec{r} \cdot \vec{n}\) = d
= \(\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|}\)
Here \(\vec{a}=2 \hat{i}+\hat{j}-\hat{k}\)
and \(\vec{n}=\hat{i}-2 \hat{j}+4 \hat{k}\) ;
d = 9
∴ required ⊥ distance = \(\frac{|(2 \hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}-2 \hat{j}+4 \hat{k})-9|}{\sqrt{1^2+(-2)^2+4^2}}\)
= \(\frac{|2(1)+1(-2)-1(4)-9|}{\sqrt{21}}\)
= \(\frac{13}{\sqrt{21}}\)

(ii) Given vector eqn. of plane be,
\(\vec{r} \cdot(3 \hat{i}-4 \hat{j}-12 \hat{k})\) + 39 = 0 ……………………(1)
Let P (x, y, z) be any plane (1) then PV. of point P be
\(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
∴ from (1) ;
\(x \hat{i}+y \hat{j}+z \hat{k}\)
\((3 \hat{i}-4 \hat{j}-12 \hat{k})\) + 39 = 0
⇒ 3x – 4y – 12z + 39 = 0 …………………..(2)
∴ required length of ⊥ from (0, 0, 0) to plane (2)
= \(\frac{|3 \times 0-4 \times 0-12 \times 0+39|}{\sqrt{3^2+(-4)^2+(-12)^2}}\)
= \(\frac{39}{\sqrt{9+16+144}}\)
= \(\frac{39}{\sqrt{169}}=\frac{39}{13}\)
= 3

(iii) Given eqn. of plane in cartesian form be given by
2x – 5y + 3z – 13 = 0 …………….(1)
∴ required ⊥ distance from (3, 4, 5) to plane (1)
= \(\frac{|2(3)-5(4)+3(5)-13|}{\sqrt{2^2+(-5)^2+3^2}}\)
= \(\frac{|6-20+15-13|}{\sqrt{4+25+9}}\)
= \(\frac{12}{\sqrt{38}}\)

(iv) Equation of xy plane be z = 0.
⊥ distance of point (2, 3, 5) from planez = 0
= \(\frac{|5-0|}{\sqrt{1^2}}\) = 0

Question 2 (old).
(ii) Find the distance of the plane \(\vec{r} \cdot(2 \hat{i}-6 \hat{j}+3 \hat{k})\) = 4 from origin.
Solution:
Given plane in cartesian form is given by
2x – 6y + 3z – 4 = 0 ………………..(1)
required ± distance from O (0, 0, 0) to plane (1)
= \(\frac{|2(0)-6(0)+3(0)-4|}{\sqrt{2^2+(-6)^2+3^2}}\)
= \(\frac{4}{7}\) units.

Question 3.
Show that the points with position vectors \(\hat{i}-\hat{j}+3 \hat{k}\) and 3 \((\hat{i}+\hat{j}+\hat{k})\) are equidistant from the plane \(\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})\) + 9 = 0. (NCERT Exemplar)
Solution:
We know that, ⊥ distance of point having P.V \(\vec{a}=\hat{i}-\hat{j}+3 \hat{k}\)
from the plane \(\vec{r} \cdot \vec{n}\) = d is given by
\(\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|}\)
⊥ distance of point \((\hat{i}-\hat{j}+3 \hat{k})\) from plane
\(\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})\) + 9 = 0
= \(\frac{|(\hat{i}-\hat{j}+3 \hat{k}) \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})-(-9)|}{\sqrt{5^2+2^2+(-7)^2}}\)
= \(\frac{|5-2-21+9|}{\sqrt{25+4+49}}\)
= \(\frac{9}{\sqrt{78}}\)
and ⊥ distance of point \((3 \hat{i}+3 \hat{j}+3 \hat{k})\) from given plane
= \(\frac{|(3 \hat{i}+3 \hat{j}+3 \hat{k}) \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})-(-9)|}{\sqrt{5^2+2^2+(-7)^2}}\)
= \(\frac{|3(5)+3(2)+3(-7)+9|}{\sqrt{78}}\)
= \(\frac{9}{\sqrt{78}}\)
Hence the point \(\hat{i}-\hat{j}+3 \hat{k}\) and \(3 \hat{i}+3 \hat{j}+3 \hat{k}\) are equidistant from the given plane.

Question 4.
(i) If the distance from the point (k, 3, – 5) to the plane x + 2y – 2z = 9 is 3 units, then find the value(s) of k.
(ii) If the distance from the point (- 6, 0, 0) to the plane 2x – 3y + z – k = 0 is \(\sqrt{14}\) units, find the value(s) of k.
Solution:
(i) Equation of given plane be
x + 2y – 2z = 9 ………………(1)
⊥ distance from (k, 3, – 5) to plane (1)
= \(\frac{|k+2(3)-2(-5)-9|}{\sqrt{1^2+2^2+(-2)^2}}\)
= \(\frac{|k+6+10-9|}{\sqrt{1+4+4}}\)
= \(\frac{|k+7|}{3}\)

also given ⊥distance = 3 units
∴ \(\frac{|k+7|}{3}\) = 3
⇒ k + 7 = ± 9
⇒ k = ± 9 – 7
i.e. k = 2, – 16

(ii) Given, ⊥ distance from the point (- 6, 0, 0) to plane
2x – 3y + z – k = 0 be \(\sqrt{14}\)
⇒ \(\frac{|2(-6)-3(0)+0-k|}{\sqrt{2^2+(-3)^2+1^2}}=\sqrt{14}\)
⇒ \(\frac{|-12-k|}{\sqrt{4+9+1}}=\sqrt{14}\)
⇒ |12 + k| = 14
⇒ 12 + k = ± 14
⇒ k = ± 14 – 12 = 2, – 26

Question 5.
(i) Find the distance between the planes 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12.
(ii) Find the distance between the planes \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})\) – 4 = 0 and \(\vec{r} \cdot(6 \hat{i}-9 \hat{j}+18 \hat{k})\) + 30 = 0.
Solution:
(i) Given eqns. of planes are
2x + 3y + 4z = 4 …………………..(1)
and 4x + 6y + 8z = 12 ………………….(2)
Direction numbers of normals to given planes (1) and (2) are
< 2, 3, 4 > and < 4, 6, 8 >
Also their direction numbers are proportional.
Thus, given planes are parallel.
So distance between parallel planes is the distance of any point on one plane from the other plane.
To find the point on plane (1), we put
y = z = 0 in eqn. (1) we have x = 2
∴ (2, 0, 0) be any point on plane (I).
Thus distance between || planes = ⊥ distance of point (2, 0, 0) from plane (2)
= \(\frac{|4(2)+6(0)+8(0)-12|}{\sqrt{4^2+6^2+8^2}}\)
= \(\frac{4}{\sqrt{16+36+64}}\)
= \(\frac{4}{\sqrt{116}}=\frac{2}{\sqrt{29}}\) units.

(ii) The cartesian eqns. of given planes are ;
\((x \hat{i}+y \hat{j}+2 \hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})\) – 4 = 0
⇒ 2x – 3y + 6z – 4 = 0 …………………(1)
and \((x \hat{i}+y \hat{j}+2 \hat{k}) \cdot(6 \hat{i}-9 \hat{j}+18 \hat{k})\) = 30
⇒ 6x – 9y + 18z + 30 = 0 ……………………(2)
Clearly the D’ratios of both planes are < 2, – 3, 6 > and < 6, – 9, 18 >.
Clearly direction numbers of both planes are proportional.
Hence given planes are parallel.
So distance between planes (1) and (2) = ⊥ distance of point on the plane to other plane.
putting y = z = 0 in (1);
2x – 0 – 0 – 4 = 0
⇒ x = 2
Thus the point (2, 0, 0) lies on plane (1)
∴ required distance between them = \(\frac{|6 \times 2-9 \times 0+18 \times 0+30|}{\sqrt{36+81+324}}\)
= \(\frac{42}{\sqrt{441}}=\frac{42}{21}\)
= \(\frac{42}{\sqrt{441}}=\frac{42}{21}\)
= 2 units

Question 6.
If the distance between the planes 2x – 2y + z = 3 and 2x – 2y + z = k is 2 units, find the value(s) of k.
Solution:
Equation of given planes are;
2x – 2y + z – 3 = 0 ……………….(1)
and 2x – 2y + z – k = 0 ………………(2)
Since Direction ratios of normal to both planes are equal.
Thus both planes are parallel.
Now distance between parallel planes is equal to ⊥ distance of any point on (x, y, z) plane (1) from plane (2)
= \(\frac{|2 x-2 y+z-k|}{\sqrt{2^2+(-2)^2+1^2}}\)
= \(\frac{|+3-k|}{3}\) [using (1)]
also given distance between parallel planes = 2 units
∴ \(\frac{|k-3|}{3}\) = 2
⇒ k – 3 = ± 6
⇒ k = ± 6 + 3
i.e. k = + 9, – 3.

Question 7.
If the points (1, 1, p) and (- 3, 0, 1) be equidistant from the plane \(\vec{r} \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})\) + 13 = 0, find the value(s) of p. (NCERT)
Solution:
Given equation of plane be \(\vec{r} \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})\) + 13 = 0
The cartesian equation of plane be
3x + 4y – 12z + 13 = 0
Given, ⊥ distance of (1, 1, p) from plane (1) = ⊥ distance of (- 3, 0, 1) from plane (1)
= \(\frac{|3(1)+4(1)-12 p+13|}{\sqrt{3^2+4^2+(-12)^2}}\)
= \(\frac{|3(-3)+4(0)-12(1)+13|}{\sqrt{3^2+4^2+(-12)^2}}\)
⇒ \(\frac{|3+4-12 p+13|}{13}=\frac{|-9+0-12+13|}{13}\)
⇒ |20 – 12p| = |- 8|
⇒ 20 – 12p = ± 8
⇒ 12p = 20 ± 8
⇒ 12p = 28, 12
⇒ P = \(\frac{28}{12}\), 1 i.e. P = \(\frac{7}{3}\), 1
Thus, required values ofp are 1 and \(\frac{7}{3}\).

Question 8.
Find the distance from the point (7, 2, 4) to the plane determined by the points A (2, 5, – 3), B (- 2, – 3, 5) and C (5, 3, – 3).
Solution:
The equation of plane passing through the point A (2, 5, – 3), B (- 2, – 3, 5) and C (5, 3, – 3) is given by
\(\left|\begin{array}{rrr}
x-2 & y-5 & z+3 \\
-2-2 & -3-5 & 5+3 \\
5-2 & 3-5 & -3+3
\end{array}\right|\) = 0
[Since eqn. of plane througgh the points (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is given by
\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|\) = 0
i.e. \(\left|\begin{array}{ccc}
x-2 & y-5 & z+3 \\
-4 & -8 & 8 \\
3 & -2 & 0
\end{array}\right|\) = 0
expanding along R₁
(x – 2) (0 + 16) – (y – 5) (0 – 24) + (z + 3) (8 + 24) = 0
⇒ 16 (x – 2) + 24 (y – 5) + 32 (z + 3) = 0
⇒ 2 (x – 2) + 3 (y – 5) + 4 (z + 3) = 0
⇒ 2x + 3y + 4z – 7 = 0
be the required eqn. of plane.
∴ required distance from (7, 2, 4) to plane 2x + 3y + 4z – 7 = 0
= \(\frac{|2(7)+3(2)+4(4)-7|}{\sqrt{2^2+3^2+4^2}}\)
= \(\frac{|14+6+16-7|}{\sqrt{4+9+16}}\)
= \(\frac{29}{\sqrt{29}}\)
= \(\sqrt{29}\) units

Question 9.
Find the vector equation of the plane determined by the points A (3, – 1, 2), B (5, 2, 4) and C (- 1, – 1, 6). Hence, find the distance of the plane thus obtained from the origin.
Solution:
The eqn. of plane through the point (3, – 1, 2) be given by
a (x – 3) + b (y + 1) 4 – c (z – 2) = 0 ………………(1)
The point B (5, 2, 4) lies on plane (I)
∴ 2a + 3b + 2c = 0 ………………(2)
plane (1) also pass through the point C (- 1, – 1, 6)
– 4a + 0b + 4c = 0 ………………..(3)
On solving eqn. (2) and eqn. (3)
\(\frac{a}{12}=\frac{b}{-8-8}=\frac{c}{0+12}\)
⇒ \(\frac{a}{3}=\frac{b}{-4}=\frac{c}{3}\)
Thus eqn. (1) becomes;
3 (x – 3)- 4 (y + 1) + 3(z – 2) = 0
⇒ 3x – 4y + 3z = 19 …………………..(4)
⇒ \((x \hat{i}+y \hat{j}+2 \hat{k}) \cdot(3 \hat{i}-4 \hat{j}+3 \hat{k})\) = 19
⇒ \(\vec{r} \cdot(3 \hat{i}-4 \hat{j}+3 \hat{k})\) = 19
where \(\vec{r}\) be the P.V of any point on required plane.
Thus required distance of origin (0, 0, 0) to plane (4).
= \(\frac{|3 \times 0-4 \times 0+3 \times 0-19|}{\sqrt{9+16+9}}\)
= \(\frac{19}{\sqrt{34}}\)

Question 10.
Find the equation of the plane passing through the points (1, 0, – 2), (3, – 1, 0) and perpendicular to the plane 2x – y + z = 8. Also, find the distance of the plane thus obtained from origin.
Solution:
The required eqn. of p lane through the point (1, 0, – 2) be given by
a (x – 1) + b (y – 0) + c (z + 2) = 0 ………………(1)
plane (1) pass through the point (3, – 1, 0)
2a – b + 3c = 0 …………………(2)
Now plane (1) is ⊥ to given plane 2x – y + z = 8
∴ 2a – b + c = 0 …………………(3)
On solving eqn. (2) and eqn. (3) ; we have
\(\frac{a}{-1+2}=\frac{b}{4-2}=\frac{c}{-2+2}\)
i.e. \(\frac{a}{1}=\frac{b}{2}=\frac{c}{0}\)
Thus from (1) ; we have,
x – 1 + 2 (y – 0) + 0 (z + 2) = 0
⇒ x + 2y = 1 …………………(4)
which is the required eqn. of plane.
Thus, required distance of O (0, 0, 0) from plane (4)
= \(\frac{|0+2 \times 0-1|}{\sqrt{1^2+2^2}}\)
= \(\frac{1}{\sqrt{5}}\) units.

Question 11.
Find the equation of the plane passing through the intersection of the planes 4x – y + z = 10 and x + y – z = 4 and parallel to the line with direction ratios < 2, 1, 1 >. Also find the perpendicular distance of the point (1, 1, 1) from this plane.
Solution:
The equation of plane passing through the intersection of the planes
4x – y + z – 10 = 0
and x + y – z – 4 = 0 is given by
4x – y + z – 10 + λ (x + y – z – 4) = 0
⇒ (4 + λ) x + (- 1 + λ) y + (1 – λ) z – 10 – 4λ = 0 ………………(1)
∴ Direction ratios of normal to plane (1) are < 4 + λ, – 1 + λ, 1 – λ >
Since plane (1) is parallel to line with direction ratios < 2, 1, 1 >.
Thus, normal to plane (1) is ⊥ to this line.
(4 + λ) 2 + (- 1 + λ) 1 + (1 – λ) 1 = 0
⇒ 8 + 2λ – 1 + λ + 1 – λ = 0
⇒ 2λ = – 8
⇒ λ = – 4
putting the value of λ = – 4 in eqn. (1) ; we have
(4 – 4) x + (- 1 – 4) y + (1 + 4) z – 10 – 4 (- 4) = 0
⇒ -5y + 5z + 6 = 0
⇒ 5y – 5z – 6 = 0 ………………(2)
which is the required eqn. of plane (2)
Thus, required ⊥distance of the point (1, 1, 1) from plane (2)
= \(\frac{|5(1)-5(1)-6|}{\sqrt{5^2+(-5)^2}}\)
= \(\frac{6}{\sqrt{50}}=\frac{6}{5 \sqrt{2}}\) units
= \(\frac{3 \sqrt{2}}{5}\) units

Question 12.
Find the equations of the planes parallel to the plane x – 2y + 2z = 3 and at a unit distance from the point (1, 1, 1).
Solution:
Equation of given plane be
x – 2y + 2z – 3 = 0 …………………………(1)
Thus eqn. of plane parallel to plane (1) is given by
x – 2y + 2z = k ……………………(2)
⊥ distance of plane (1) from given point (1, 1, 1)
= \(\frac{|1-2(1)+2(1)-k|}{\sqrt{1^2+(-2)^2+2^2}}\)
= \(\frac{|1-k|}{3}\)
also given distance of plane (1) from point (1, 1, 1) be unity.
∴ \(\frac{|1-k|}{3}\) = 1
⇒ |1 – k| = + 3
⇒ 1 – k = ± 3
⇒ k = 4, – 2
putting the value of k in eqn. (1) ; we have
x – 2y + 2z – 4 = 0
and x – 2y + 2z + 2 = 0
are the required equations of planes.

Question 13.
Find the equation of the plane through the point (3, 4, – 1) which is parallel to the plane \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})\) + 7 = 0. Also find the distance between the two planes.
Solution:
eqn. of given plane be,
2x – 3y + 5z + 7 = 0 ………………….(1)
∴ eqn. of any plane || to plane (1) be
2x – 3y + 5z = k ………………..(2)
Since plane (2) passes through the point (3, 4, – 1).
∴ 2 (3) – 3 (4) + 5 (- 1) = k
⇒ k = – 11
∴ from (2) ;
2x – 3y + 5z + 11 = 0 ……………….(3)
be the required eqn. of plane.
Let P (x₁, y₁, z₁) be any point on plane (1)
∴ 2x₁ – 3y₁ + 5z₁ + 7 = 0
Let d be the distance beween parallel planes
∴ d = ⊥ distance of P (x₁, y₁, z₁) from plane (3)
= \(\frac{\left|2 x_1-3 y_1+5 z_1+11\right|}{\sqrt{2^2+(-3)^2+5^2}}\)
= \(\frac{|-7+11|}{\sqrt{4+9+25}}\) [using eqn. (1)]
= \(\frac{|-7+11|}{\sqrt{4+9+25}}\)

Question 12.
Find the equations of the planes parallel to the plane 3x – 6y + 2z = 12 and 6 units away from it.
Solution:
Equation of given plane be
3x – 6y + 2z – 12 = 0 …………………..(1)
∴ eqn. of plane parallel to plane (1) be
3x – 6y + 2z + k = 0 …………………..(2)
So Distance between parallel plane (1) and (2) = ⊥ distance of any point (x, y, z) on plane (1) from plane (2)
= \(\frac{|3 x-6 y+2 z+k|}{\sqrt{3^2+(-6)^2+2^2}}\)
= \(\frac{|+12+k|}{7}\) [using (1)]
also given Distance between || planes = 6 units
∴ \(\frac{|12+k|}{7}\) = 6
⇒ 12 + k = ± 42
⇒ k = ± 42 – 12
∴ k = 30, – 54
∴ from (2) ;
required equations of planes are given by 3x – 6y + 2z + 30 = 0
and 3x – 6y + 2z – 54 = 0

Question 15.
Find the shortest distance between the planes 6x + 2y – 3z = 12 and 12x + 4y – 6z = 17.
Solution:
Given planes are ;
6x + 2y – 3z = 12 ………………………..(1)
and 12x + 4y – 6z = 17 ………………..(2)
Direction ratios of normal to planes (1) and (2) are < 6, 2, – 3 > and < 12, 4, – 6 > and are proportional.
Thus, given planes (1) and (2) are parallel.
So S.D between the given planes = distance between || planes
distance of any point (x, y, z) on plane (1) from plane (2)
= \(\frac{|12 x+4 y-6 z-17|}{\sqrt{12^2+4^2+(-6)^2}}\)
= \(\frac{|2(6 x+2 y-3 z)-17|}{\sqrt{144+16+36}}\)
= \(\frac{|2 \times 12-17|}{\sqrt{196}}\) [using (1)]
= \(\frac{7}{14}=\frac{1}{2}\) units

Question 16.
Find the equation of the planes mid parallel to the planes 2x – 2y + z + 3 = 0 and 2x – 2y + z + 9 = 0.
Solution:
Equations of given planes are
2x – 2y + z + 3 = 0 ……………….(1)
and 2x – 2y + z + 9 = 0 ……………..(2)
Since both planes are parallel.
Let the eqn. of plane parallel to both given planes be 2x – 2y + z + k = 0 ………………….(3)
Let P (x₁, y₁, z₁) be any point on plane (1).
∴ 2x₁ – 2y₁ + z₁ + 3 = 0 ……………………(4)
∴ d₁ = distance between planes (1) and (3)
= ⊥ distance of point P (x₁, y₁, z₁) from plane (3)
= \(\frac{\left|2 x_1-2 y_1+z_1+k\right|}{\sqrt{2^2+(-2)^2+1^2}}\frac{|-3+k|}{3}\)
= \(\frac{|-3+k|}{3}\) [using eqn. (4)]
d₂ = distance between planes (2) and (3)
= ⊥ distance of point P (x₁, y₁, z₁) from plane (3)
= \(\frac{\left|2 x_1-2 y_1+z_1+k\right|}{\sqrt{4+4+1}}\)
= \(\frac{|-9+k|}{3}\)
[∵ 2x₁ – 2y₁ + z₁ + 9 = 0]
Since, plane (3) is mid parallel to eqn. (1) and eqn. (2).
∴ d₁ = d₂
⇒ |- 3 + k| = |- 9 + k|
On squaring ; we have
(k – 3)² = (k – 9)²
⇒ k² – 6k + 9
⇒ k² – 18k + 81
⇒ 12k = 72
⇒ k = 6
putting the value of k in eqn. (3); we get 2x – 2y + z + 6 = 0 be the required eqn. of plane.

Question 17.
Find the equation of the planes passing through the intersection of the planes \(\vec{r} \cdot(2 \hat{i}+6 \hat{j})\) + 12 = 0 and \(\vec{r} \cdot(3 \hat{i}-\hat{j}+4 \hat{k})\) = 0 and at a unit distance from the origin.
Solution:
The cartesian equations of given planes are ;
\(\vec{r} \cdot(2 \hat{i}+6 \hat{j})\) + 12 = 0
⇒ 2x + 6y + 12 = 0 ………………..(1)
and \(\vec{r} \cdot(3 \hat{i}-\hat{j}+4 \hat{k})\) = 0
⇒ 3x – y + 4z = 0 ……………..(2)
The eqn. of plane through the intersection of planes (1) and (2) be given by
2x + 6y + 12 + λ (3x – y + 4z) = 0 …………………(3)
(2 + 3) x + (6 – λ) y + 4λz + 12 = 0
it is given that ⊥ distance from origin to plane (2) = 1
⇒ \(\frac{|(2+3 \lambda) 0+(6-\lambda) 0+4 \lambda \times 0+12|}{\sqrt{(2+3 \lambda)^2+(6-\lambda)^2+(4 \lambda)^2}}\) = 1
⇒ 12 = \(\sqrt{(2+3 \lambda)^2+(6-\lambda)^2+(4 \lambda)^2}\)
⇒ 144 = 26λ² + 40
⇒ 26λ² = 104
⇒ λ² = 4
⇒ λ = ± 2
When ± = 2 ; eqn. (3) becomes;
2x + 6y + 12 + 2 (3x – y + 4z) = 0
⇒ 8x + 4y + 8z + 12 = 0
⇒ 2x + y + 2z + 3 = 0
When λ = – 2 ; eqn. (2) becomes;
2x + 6y + 12 – 2 (3x – y + 4z) = 0
⇒ – 4x + 8y – 8z + 12 = 0
⇒ x – 2y + 2z – 3 = 0

Question 18.
State when the line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is parallel to the plane \(\vec{r} \cdot \vec{n}\) = d. Show that the line \(\vec{r}=\hat{i}+\hat{j}+\lambda(3 \hat{i}-\hat{j}+2 \hat{k})\) is parallel to the plane \(\vec{r} \cdot(2 \hat{j}+\hat{k})\) = 3. Also find the distance between the line and the plane.
Solution:
eqn. of given line be
\(\vec{r}=\vec{a}+\lambda \vec{b}\) …………………..(1)
and eqn. of given plane be
\(\vec{r} \cdot \vec{n}\) = d ……………….(2)
Now line (1) is parallel to plane (2) if normal to plane is ⊥ to the line.
i.e. if \(\vec{n}\) is to \(\vec{b}\) if \(\vec{b} \cdot \vec{n}\) = 0
Now eqn. of given line be
\(\vec{r}=\hat{i}+\hat{j}+\lambda(3 \hat{i}-\hat{j}+2 \hat{k})\) …………………….(3)
and eqn. of given plane be
\(\vec{r} \cdot(2 \hat{j}+\hat{k})\) = 3 …………………(4)
Here \(\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\) ;
\(\vec{n}=2 \hat{j}+\hat{k}\)
Now \(\vec{b} \cdot \vec{n}=(3 \hat{i}-\hat{j}+2 \hat{k}) \cdot(0 \hat{i}+2 \hat{j}+\hat{k})\)
= 3(0) – 1 (2) + 2(1) = 0
Thus line (3) is parallel to plane (4).
∴ required distance between the line and the plane = ⊥ distance from the point with position vector \(\vec{a}=\hat{i}+\hat{j}\) to the plane (4)
= \(\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|}\)
= \(\frac{|(\hat{i}+\hat{j}) \cdot(2 \hat{j}+\hat{k})-3|}{\sqrt{2^2+1^2}}\)
= \(\frac{|1(0)+1(2)+0(1)-3|}{\sqrt{5}}\)
= \(\frac{1}{\sqrt{5}}\) units