ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry Ex 2.7

Practicing ISC Maths Class 12 Solutions Chapter 2 Three Dimensional Geometry Ex 2.7 is the ultimate need for students who intend to score good marks in examinations.

ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry Ex 2.7

Very short answer type questions (1 to 8) :

Question 1.
Find the angle between the planes :
(i) \(\vec{r} \cdot(\hat{i}+\hat{j})\) = 1 and \(\vec{r} \cdot(\hat{i}+\hat{k})\) = 3
(ii) \(\vec{r} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})\) = 6 and \(\vec{r} \cdot(3 \hat{i}+6 \hat{j}-2 \hat{k})\) = 9
(iii) \(\vec{r} \cdot(3 \hat{i}-4 \hat{j}+5 \hat{k})\) = 0 and \(\vec{r} \cdot(2 \hat{i}-\hat{j}-2 \hat{k})\) = 7
(iv) 2x – y + z = 6 and x + y + 2z = 7.
Solution:
(i) Angle between two planes = Angle between their normals
From given eqns.

(ii) eqns. of given planes are \(\vec{r} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})\) = 6
and \(\vec{r} \cdot(3 \hat{i}+6 \hat{j}-2 \hat{k})\) = 9
Here \(\overrightarrow{n_1}=2 \hat{i}-\hat{j}+2 \hat{k}\)
and \(\overrightarrow{n_2}=3 \hat{i}+6 \hat{j}-2 \hat{k}\)
Let θ be the angle between given palnes

(iii) Given \(\overrightarrow{n_1}=3 \hat{i}-4 \hat{j}+5 \hat{k}\)
and \(\overrightarrow{n_2}=2 \hat{i}-\hat{j}-2 \hat{k}\)
∴ \(\left|\vec{n}_1\right|=\sqrt{9+16+25}=\sqrt{50}\)
and \(\left|\overrightarrow{n_2}\right|=\sqrt{4+1+4}\) = 3
and \(\overrightarrow{n_1} \cdot \overrightarrow{n_2}\) = 3 (2) – 4 (- 1) + 5 (- 2)
= 6 + 4 – 10 = 0
∴ cos θ = \(\frac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|}\) = 0
⇒ θ = \(\frac{\pi}{2}\)

(iv) We know that angle θ between given planes
a₁x + b₁y + c₁z + d₁ = 0
and a₂x + b₂y + c₂z + d₂ = 0 is given by
cos θ = \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
Here
a₁ = 2 ;
b₁ = – 1 ;
c₁ = 1 ;
a₂ = 1 ;
b₂ = 1 ;
c₂ = 2
∴ cos θ = \(\frac{2(1)-1(1)+1(2)}{\sqrt{4+1+1} \sqrt{1+1+4}}\)
= \(\frac{3}{6}=\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)

Question 1 (old).
(iv) 3x – 6y + 2z = 7 and 2x + y – 2z = 5 (NCERT)
Solution:

Question 2.
Show that the following pairs of planes are at right angles to each other :
(i) 3x – 5y + 3z – 1 = 0 and 2x + 3y + 3z = 7
(ii) \(\vec{r} \cdot(2 \hat{i}+6 \hat{j}+6 \hat{k})\) = 13 and \(\vec{r} \cdot(3 \hat{i}+4 \hat{j}-5 \hat{k})\) + 7 = 0
Solution:
(i) Given eqn’s of plane are,
3x – 5y + 3z = 1 ………………..(1)
and 2x + 3y + 3z = 7 …………………(2)
D’ numbers of normal to plane (1) are < 3, – 5, 3 >
and D’ numbers of normal to plane (2) are < 2, 3, 3 >
Here 3 (2) – 5 (3) + 3 (3) = 6 – 15 + 9 = 0
Thus plane (1) and (2) are at Ito each other.

(ii) Given vector eqn’s of planes are
\(\vec{r} \cdot(2 \hat{i}+6 \hat{j}+6 \hat{k})\) = 13 …………………(1)
\(\vec{r} \cdot(3 \hat{i}+4 \hat{j}-5 \hat{k})\) + 7 = 0 …………………(2)
Here
\(\overrightarrow{n_{\mathrm{I}}}=2 \hat{i}+6 \hat{j}+6 \hat{k}\)
and \(\overrightarrow{n_2}=+3 \hat{i}+4 \hat{j}-5 \hat{k}\)
Here,
\(\overrightarrow{n_1} \cdot \overrightarrow{n_2}=(2 \hat{i}+6 \hat{j}+6 \hat{k}) \cdot(3 \hat{i}+4 \hat{j}-5 \hat{k})\)
= 2 (3) + 6 (4) + 6 (- 5) = 0
Thus, planes (1) and (2) are at right angles to each other.

Question 3.
Find the value of p for which the following planes are perpendicular to each other :
(i) \(\vec{r} \cdot(p \hat{i}+2 \hat{j}+3 \hat{k})\) = 5 and \(\vec{r} \cdot(\hat{i}+2 \hat{j}-7 \hat{k})\) + 11 = 0
(ii) x – 4y + pz + 3 and 2x + 2y + 3z = 5
(iii) 3x – 6y – 2z = 7 and 2x + y – pz = 5.
Solution:
(i) Given eqn’s of planes are;
\(\vec{r} \cdot(p \hat{i}+2 \hat{j}+3 \hat{k})\) = 5 …………………….(1)
and \(\vec{r} \cdot(\hat{i}+2 \hat{j}-7 \hat{k})\) + 11 = 0 ………………….(2)
Here
\(\overrightarrow{n_1}=p \hat{i}+2 \hat{j}+3 \hat{k}\)
and \(\overrightarrow{n_2}=\hat{i}+2 \hat{j}-7 \hat{k}\)
Since planes (1) and (2) are ⊥ to each other
∴ \(\overrightarrow{n_1} \cdot \overrightarrow{n_2}\) = 0
∴ \((p \hat{i}+2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+2 \hat{j}-7 \hat{k})\) = 0
⇒ p (1) + 2 (2) – 3 (7) = 0
⇒ p + 4 – 21 = 0
⇒ p = 17.

(ii) Given planes in vector form is given by
\(\vec{r} \cdot(\hat{i}-4 \hat{j}+p \hat{k})\) + 3 = 0
⇒ \(\vec{r} \cdot(-\hat{i}+4 \hat{j}-p \hat{k})\) = 3 …………………(1)
and \(\vec{r} \cdot(2 \hat{i}+2 \hat{j}+3 \hat{k})\) = 5
Here
\(\overrightarrow{n_1}=-\hat{i}+4 \hat{j}-p \hat{k}\) ;
\(\overrightarrow{n_2}=2 \hat{i}+2 \hat{j}+3 \hat{k}\)
Since planes (1) and (2) are ⊥ to each other
∴ \(\overrightarrow{n_1} \cdot \overrightarrow{n_2}\) = 0
⇒ \((-\hat{i}+4 \hat{j}-p \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+3 \hat{k})\) = 0
⇒ (- 1) (2) + 4 (2) – p(3) = 0
⇒ 3p = 6
⇒ p = 2

(iii) Given eqns. of planes are
3x – 6y – 2z = 7 ………………(1)
and 2x + y – λ z = 5 ………………(2)
Here
a₁ = 3 ;
b₁ = – 6 ;
c₁ = – 2
a₂ = 2 ;
b₂ = 1 ;
c₂ = – λ
Since, given planes (1) and (2) are ⊥ to each other
∴ 3 (2) + (- 6) 1 – 2 (- λ) = 0
[∵ a₁a₂ + b₁b₂ + c₁c₂ = 0]
⇒ 6 – 6 – 2λ = 0
⇒ λ = 0.

Question 4.
(i) Write the equation of the plane passing through origin and parallel to the plane 5x – 3y + 7z + 13 = 0.
(ii) Find the equation of the plane through the point (1, 4, – 2) and parallel to the plane-2x + y – 3z = 7.
(iii) Find the equation of the plane with intercept 3 on the yaxis and parallel to xz-plane.
Solution:
(i) Equation of given plane be
5x – 3y + 7z + 13 = 0 …………………(1)
∴ equation of plane parallel to plane (1) is given by
5x – 3y + 7z + k = 0 ………………..(2)
Since eqn. (2) passes through the point (0, 0, 0)
∴ 5 (0) – 3 (0) + 7 (0) + k = 0
⇒ k = 0
Thus, eqn. (2) becomes ;
5x – 3y + 7z = 0 be the required eqn. of plane.

(ii) eqn. of given plane be
– 2x + y – 3z = 7 ………………..(1)
∴ equation of plane parallel to plane (1) be given by
– 2x + y – 3z + k = 0 …………………….(2)
Since eqn. (2) passes through the point (1, 4, – 2)
∴ – 2 (1) + 4 – 3 (- 2) + k = 0
⇒ k = – 8
Thus required eqn. of plane becomes ;
– 2x + y – 3z – 8 = 0
⇒ 2x – y + 3z + 8 = 0.

(iii) The eqn. of i.e. y = 0 be y = k .
plane which ¡s parallel to xz-plane
Now required plane (1) has intercept 3 on y-axis
i.e. eqn. (1) pass through (0, 3, 0).
∴ 3 = k
Thus eqn. (1) becomes ;
y – 3 = 0

Question 5.
(i) Find the equation of the plane through the point \(3 \hat{i}+4 \hat{j}-\hat{k}\) and parallel to the plane \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})\) + 5 = 0.
(ii) Find the equation of the plane passing through the point (a, b, c) and parallel to the plane \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 2.
Solution:
(i) Equation of given plane be,
\(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})\) + 2 = 0 ………………(1)
Thus the eqn. of plane parallel to plane (1) be
\(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})\) = λ …………………(2)
Since plane (2) passes through the point (3, 4, – 1).
Thus \(3 \hat{i}+4 \hat{j}-\hat{k}\) satisfies eqn. (2).
∴ \((3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})\) = λ
⇒ 3(2) + 4(- 3) – 1 (5) = λ
⇒ λ = – 11
putting the value of λ in eqn. (2) ; we get
\(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})\) = 0 be the required eqn. of plane.

(ii) Given eqn. of plane be
\(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 2 …………….(1)
The eqn. of any plane || to plane (1) be
\(\) = λ …………………(2)
Since plane (2) passing through the point (a, b, c)
∴ \(a \hat{i}+b \hat{j}+c \hat{k}\) lies on eqn.(2).
∴ \((a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})\) = λ
⇒ a + b + c = λ
putting the value of λ in eqn. (2)
We have \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = a + b + c ……………….(3)
Putting \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) in eqn. (3) ; we get
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})\) = a + b + c
⇒ x + y + z = a + b + c be the required equation of plane.

Question 6.
(i) Find the angle between the line \(\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(3 \hat{i}-\hat{j}+2 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 3.
(ii) Find the angle between the line \(\vec{r}=2 \hat{i}-3 \hat{j}+\lambda(\hat{k})\) and the plane \(\vec{r} \cdot(\hat{j}-\hat{k})\) = 7.
Solution:
(i) eqn. of given line be,
\(\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(3 \hat{i}-\hat{j}+2 \hat{k})\) ………………………..(1)
and eqn. of given plane be,
\(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 3 ……………………..(2)
Here \(\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\)
and \(\vec{n}=\hat{i}+\hat{j}+\hat{h}\)
Let θ be the angle between lin (1) and palne (2)

(ii) eqn. of given line be,
\(\vec{r}=2 \hat{i}-3 \hat{j}+\lambda \hat{k}\) ……………….(1)
and eqn. of palne be,
\(\vec{r} \cdot(\hat{j}-\hat{k})\) = 7 ……………….(2)
Here \(\vec{b}=\hat{k}\)
and \(\vec{n}=\hat{j}-\hat{k}\)
Let θ be the angle between lin (1) and palne (2)
∴ sin θ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)
= \(\frac{|\hat{k} \cdot(\hat{j}-\hat{k})|}{\sqrt{0+0+1} \sqrt{1+1}}\)
= \(\frac{1}{\sqrt{2}}\)
Thus, θ = \(\frac{\pi}{4}\)

Question 6 (old).
(i) Find the angle between the line \(\vec{r}=(\hat{i}+3 \hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and the plane \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 7.
(ii) Find the angle between the line \(\vec{r}=(2 \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}-3 \hat{j}+2 \hat{k})\) and the plane \(\vec{r} \cdot(3 \hat{i}-2 \hat{j}+5 \hat{k})\) = 4.
Solution:
(i) eqn. of given line be
\(\vec{r}=(\hat{i}+3 \hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) ………………..(1)
and eqn. of plane be
\(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 7 ……………….(2)
Let \(\vec{b}\) be the vector along line (1)
and \(\vec{n}\) be the vector normal to given plane (2).
Then \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\) ;
\(\vec{n}=2 \hat{i}-\hat{j}+\hat{k}\)
Let θ be the acute angle between (1) and (2)

(ii) eqn. of given line
\(\vec{r}=(2 \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}-3 \hat{j}+2 \hat{k})\) …………………(1)
and given palne be
\(\vec{r} \cdot(3 \hat{i}-2 \hat{j}+5 \hat{k})\) = 4
Let \(\vec{b}\) be the vector along line (1)
and \(\vec{n}\) be the vector normal to given plane (2).

Question 7.
(i) Find the angle between the line \(\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}\) and the plane 3x + 4y + z + 5 = 0.
(ii) Find the angle between the line \(\frac{x+1}{2}=\frac{3 y+5}{9}=\frac{3-z}{-6}\) and the plane 10x + 2y – 11z = 3.
Solution:
(i) We know that if θ be the angle between the line
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
and plane lx + my + nz = p.
then, sin θ = \(\frac{a l+b m+c n}{\sqrt{a^2+b^2+c^2} \sqrt{l^2+m^2+n^2}}\)
Here,
a = 3 ;
b = – 1 ;
c = 2
and l = 3 ;
m = 4 ;
n = 1
∴ sin θ = \(\frac{3(3)-1(4)+2(1)}{\sqrt{3^2+(-1)^2+2^2} \sqrt{3^2+4^2+1^2}}\)
= \(\frac{7}{\sqrt{14} \sqrt{26}}\)
⇒ sin θ = \(\frac{7}{\sqrt{7} \sqrt{52}}=\sqrt{\frac{7}{52}}\)
⇒ θ = sin^-1 \(\left(\sqrt{\frac{7}{52}}\right)\)

(iii) Given line can be written as
\(\frac{x+1}{2}=\frac{3\left(y+\frac{5}{3}\right)}{9}=\frac{z-3}{6}\)
i.e. \(\frac{x+1}{2}=\frac{y+\frac{5}{3}}{3}=\frac{z-3}{6}\) ……………….(1)
and given plane be 10x + 2y – 11z = 3
∴ D’ ratios of given line (1) are < 2, 3, 6 >
D’ ratios of normal to plane (2) are < 10, 2, – 11 >.
Let 8 be the angle between line (1) and plane (2)

Question 7 (old).
(ii) Find the angle between the line \(\frac{x+1}{3}=\frac{y-1}{2}=\frac{z-2}{4}\) and the plane 2x + y – 3z + 4 = 0.
Solution:
The direction numbers of given line are < 3, 2, 4 >
and the direction numbers of normal to plane are < 2, 1, – 3 >
We know that,
if θ be the acute angle between line,
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
and given plane Ax + By + Cz + D = 0
Then sin θ = \(\frac{|a \mathrm{~A}+b \mathrm{~B}+c \mathrm{C}|}{\sqrt{a^2+b^2+c^2} \sqrt{\mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2}}\)
Let θ be the angle between given line and given plane.
Then
sin θ = \(\frac{|3(2)+2(1)+4(-3)|}{\sqrt{3^2+2^2+4^2} \sqrt{2^2+1^2+(-3)^2}}\)
= \(\frac{|-4|}{\sqrt{29} \sqrt{14}}\)
= \(\frac{4}{\sqrt{406}}\)
∴ θ = sin^-1 (\(\frac{4}{\sqrt{406}}\)).

Question 8.
(i) If the plane 2x – 3y – 6z = 13 makes an angle sin^-1 (λ) with the x-axis, then find the value of λ.
(ii) If the angle between the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) and the plane 2x – y + √λz = – 4 is sin^-1 (\(\frac{1}{3}\)), find the value of λ.
Solution:
(i) D’ numbers of x-axis arc < 1, 0, 0 >
and D’ numbers of normal to given plane are < 2, – 3, – 6 >
Let θ be the angle between given plane and x-axis
∴ sin θ = \(\frac{2}{7}\)
Also it is given that, the given plane makes
an angle sin^-1 λ with x-axis
∴ sin^-1 \(\frac{2}{7}\) = sin^-1 λ
⇒ λ = \(\frac{2}{7}\).

(ii)) The direction numbers of given line are < 1, 2, 2 >
and direction numbers of normal to given plane are < 2, – 1, √λ >
If θ be the angle between given line and given plane
∴ sin θ = \(\frac{|1 \times 2+2 \times(-1)+2 \sqrt{\lambda}|}{\sqrt{1^2+2^2+2^2} \sqrt{2^2+(-1)^2+(\sqrt{\lambda})^2}}\)
= \(\frac{|2 \sqrt{\lambda}|}{3 \sqrt{5+\lambda}}\) ……………………(1)
also given angle between given line and plane be
sin θ = \(\frac{1}{3}\)
∴ from (1) ;
\(\frac{1}{3}=\frac{|2 \sqrt{\lambda}|}{3 \sqrt{5+\lambda}}\)
⇒ \(\sqrt{5+\lambda}=|2 \sqrt{\lambda}|\)
On squaring both sides ; we get
5 + λ = 4λ
⇒ 3λ = 5
⇒ λ = \(\frac{5}{3}\)

Question 9.
In the following problems, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the (acute) angle between them:
(i) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0 (NCERT)
(ii) 4x + 8y + z – 8 = 0 and y + z – 4 = 0 (NCERT)
(iii) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0 (NCERT)
(iv) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0 (NCERT)
Solution:
(i) The given planes are
2x + y + 3z – 2 = 0 ………………….(1)
and x – 2y + 5 = 0 …………….(2)
The D’ ratios of normal to planes (1) and (2) are < 2, 1, 3 > and < 1, – 2, 0 >
Here 2(1) + 1 (- 2) + 3(0) = 0
∴ planes (1) and (2) are perpendicular.

(ii) The given planes are
4x + 8y + z – 8 = 0 ………………..(1)
and y + z – 4 = 0 …………………(2)
The D’ratios of planes (1) and (2) are < 4, 8, 1 > and < 0, 1, 1 >
Clearly direction ratios of the normal to planes are not proportional.
∴ both planes are not parallel.
Also 4 . 0 + 8 . 1 + 1 . 1 = 9 ≠ 0
∴ both planes are not ⊥.
If θ be the angle between them
∴ cos θ = \(\frac{4 \cdot 0+8 \cdot 1+8 \cdot 1}{\sqrt{16+64+1} \sqrt{0+1+1}}\)
∴ cos θ = \(\frac{9}{9 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
∴ θ = 45°

(iii) The given planes are
7x + 5y + 6z + 30 = 0 ………………..(1)
and 3x – y – 10z + 4 = 0 …………………(2)
D’ ratios of Normal to planes (1) and (2) are not proportional.
∴ Planes (1) and (2) are not parallel.
Also, 7 . 3 + 5 (- 1) + 6 (- 10) ≠ 0
planes (1) and (2) are not ⊥.
If θ be the angle between them
Then cos θ = \(\frac{|7(3)+5(-1)+6(-10)|}{\sqrt{49+25+36} \sqrt{9+1+100}}\)
= \(\frac{|-44|}{\sqrt{110} \sqrt{110}}\)
= + \(\frac{2}{5}\)

(iv) The given planes are
2x – 2y + 4z + 5 = 0 ………………..(1)
and 3x – 3y + 6z – 1 = 0 ……………..(2)
The direction ratios of (1) and (2) are < 2, – 2, 4 > and < 3, – 3, 6 >
Here \(\frac{2}{3}=\frac{-2}{-3}=\frac{4}{6}\)
i.e. D’ratios of normal to planes are proportional.
∴ both planes are parallel.

Question 9 (old).
(ii) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(vi) 2x – y + 4z = 5 and 5x – 25y + 10z = 6.
Solution:
(ii) The given planes are.
2x + y + 3z – 1 = 0 …………………..(1)
and 2x – y + 3z + 3 = 0 ……………….(2)
∴ D’ratios of plane (1) and (2) are < 2, 1, 3 > and < 2, – 1, 3 >
Clearly direction ratios of the normal to planes are not proportional.
∴ (1) and (2) are not parallel.
also 2 (2) + 1 (- 1) + 3(3) = 12 0
∴ both planes are not perpendicular.
If θ be the angle between them
∴ cos θ = \(\frac{2 \cdot 2+1(-1)+3(3)}{\sqrt{4+1+9} \sqrt{4+1+9}}\)
= \(\frac{12}{14}=\frac{6}{7}\)
θ = cos^-1 \(\left(\frac{6}{7}\right)\)

(vi) Given eqn’s of planes are;
2x – y + 4z = 5 …………………(1)
and 5x – 25y + 10z = 6 …………………(2)
eqn. (2) can be written as ;
25 [2x – y + 4z] = 6
⇒ 2x – y + 4z = 2 . 4 …………………..(3)
Thus Direction ratios of normals to both planes (1) and (3) are equal i.e. < 2, – 1, 4 >
Hence both given planes are parallel.

Question 10.
Prove that the plane 2x + 3y – 4z = 11 is perpendicular to each of the planes x + 2y + 2z – 7 = 0 and 5x + 6y + 7z = 23.
Solution:
The equations of given planes are;
2x + 3y – 4z = 11 ………………….(1)
x + 2y + 2z – 7 = 0 …………………(2)
5x + 6y + 7z = 23 …………………….(3)
The plane (1) ⊥ to plane (2)
if 2 (1) + 3 (2) – 4 (2) = 0
[∵ a₁a₂ + b₁b₂ + c₁c₂ = 0]
if 2 + 6 – 8 0 if 0 = 0, which is true.
The plane (1) is ⊥ to plane (3) if
2 (5) + 3 (6) – 4 (7) = 0
[∵ a₁a₂ + b₁b₂ + c₁c₂3 = 0]
if 10 + 18 – 28 = 0 if 0 = 0,which is true.
Thus, the plane (1) is perpendicular to each of the two planes (2) and (3).

Question 11.
Find the equation of the plane passing through the points (1, – 1, 2), (2, – 2, 2) and perpendicular to the plane 6x – 2y + 2z = 9.
Solution:
The eqn. of any plane passing through the point (1, – 1, 2) is
a (x – 1) + b (y + 1) + c (z – 2) = 0 ……………….(1)
Since the plane (1) passes through (2, – 2, 2).
So the point (2, – 2, 2) lies on eqn. (1).
∴ a (2 – 1) + b (- 2 + 1) + c (2 – 2) = 0
i.e. a – b + 0c = 0 ………………..(2)
Also plane (1) is ⊥ to given plane
6x – 2y + 2z = 9
∴ 6a – 2b + 2c = 0
On solving eqn. (2) and eqn. (3) by cross-multiplication,
we get \(\frac{a}{-2-0}=\frac{b}{0-2}=\frac{c}{-2+6}\)
i.e. \(\frac{a}{-2}=\frac{b}{-2}=\frac{c}{4}\)
⇒ \(\frac{a}{1}=\frac{b}{1}=\frac{c}{-2}\) = k (say)
i.e. a = k ;
b = k ;
c = – 2k
putting the values of a, b and c in eqn. (1) we get
k (x – 1) + k (y + 1) – 2k (z – 2) = 0
⇒ x + y – 2z + 4 = 0 be the required eqn. of plane.

Question 12.
(i) Find the equation of the plane through the point (1, 2, 3) and perpendicular to each of the planes x + y + 2z = 3 and 3x + 2y + z = 4.
(ii) Find the equation of the plane passing through the point \(2 \hat{i}-3 \hat{j}+\hat{k}\), \(5 \hat{i}+2 \hat{j}-\hat{k}\) and perpendicular to the plane \(\vec{r} \cdot(\hat{i}-4 \hat{j}+5 \hat{k})\) + 3 = 0.
Solution:
(i) The eqn. of plane through (1, 2, 3) is given by
a (x – 1) + b (y – 2) + c(z – 3) = 0 ………………..(3)
Now plane (3) is ⊥ to both planes (1) and (2)
∴ their normals are ⊥ to plane (3).
∴ a + b + 2c = 0 ………………..(4)
and 3a + 2b + c = 0 ………………..(5)
On solving (4) and (5) ; we get
\(\frac{a}{-3}=\frac{b}{5}=\frac{c}{-1}\) = k (say), where k ≠ 0
∴ a = – 3k ;
b = 5k
and c = – k
Putting all these values in eqn. (3) ; we have
– 3k (x – 1) + 5k (y – 2) + (- k) (z – 3) = 0
⇒ – 3x + 5y – z – 4 = 0
⇒ 3x – 5y + z + 4 = 0 is the required equation of plane.

(ii) Let the eqn. of plane passing through the point \(2 \hat{i}-3 \hat{j}+\hat{k}\)
i.e. (2, – 3, 1) is given by
a (x – 2) + b (y + 3) + c(z – 1) = 0 …………………(1)
Also plane (1) passes through the point
\(5 \hat{i}+2 \hat{j}-\hat{k}\) i.e. (5, 2, – 1).
∴ a (5 – 2) + b (2 + 3) + c (- 1 – 1) = 0
i.e. 3a + 5b – 2c = 0 ………………………(2)
Also, eqn. of given plane be
\(\vec{r} \cdot(\hat{i}-4 \hat{j}+5 \hat{k})\) + 3 = 0
In cartesian form, eqn. of plane be
x – 4y + 5z + 3 = 0 ……………………(3)
Since plane (1) is ⊥ plane (3).
∴ a – 4b + 5c = 0 …………………(4)
On solving (2) and (4) ; we have
\(\frac{a}{25-8}=\frac{b}{-2-15}=\frac{c}{-12-5}\)
i.e. \(\frac{a}{17}=\frac{b}{-17}=\frac{c}{-17}\)
i.e. \(\frac{a}{1}=\frac{b}{-1}=\frac{c}{-1}\) = k (say)
∴ a = k,
b = – k ;
c = – k
Thus eqn. (1) becomes ;
k (x – 2) – k (y + 3) – k (z – 1) = 0
⇒ x – y – z – 4 = 0
⇒ x – y – z = 4 ………………(5)
[∵ k ≠ 0]
Let ‘ be the position vector of point P(x, y. z)
Then \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
Then eqn. (5) becomes ;
\(\vec{r} \cdot(\hat{i}-\hat{j}-\hat{k})\) = 4
be the required vector eqn. of plane.

Question 13.
(i) Find the equation of the plane through the point (1, 2, 3) and perpendicular to the ptanes x + y + 2z = 3 and 3x + 2y + z = 4. (ISC 2007)
(ii) Find the equation of the plane through the point (- 1, – 1, 2) and perpendicular to the planes 3x + 2y – 3z = 1 and 5x – 4y + z = 5. (ISC 2004)
Solution:
(i) Given eqns. of planes be x + y + 2z = 3 …………………(1)
3x + 2y + z = 4 …………………(2)
The eqn. of plane through (1, 2, 3) is given by
a(x – 1) + b(y – 2) + c (z – 3) = 0 …………………(3)
Now plane (3) is ⊥ to both planes (1) and (2)
∴ their normals are ⊥ to plane (3).
∴ a + b + 2c = 0
and 3a + 2b + c = 0
On solving (4) and (5) ; we get
\(\frac{a}{-3}=\frac{b}{5}=\frac{c}{-1}\) = k (say), where k ≠ 0
∴ a = – 3k ;
b = 5k
and c = – k
Putting all these valuesin eqn. (3) ; we have
– 3k( x – 1) + 5k (y – 2) + (- k) (z – 3) = 0
– 3x + 5y – z -4 = 0
⇒ 3x – + z – 4 = 0 is the required equation of plane.

(ii) The equation of any plane through the point (- 1, – 1, 2) is
a (x + 1) + b (y + 1) + c (z – 2) = 0
where < a, b, c > are the D’ numbers of normal to plane (1).
Since the plane (1) is perpendicular to each of the given planes
3x + 2y – 3z = 1
and 5x – 4y + z = 5
3a + 2b – 3 c= 0
and 5a – 4b + c = 0
Solving eqns. (2) and (3) by cross multiplication, we have
\(\frac{a}{2-12}=\frac{b}{-15-3}=\frac{c}{-12-10}\)
i.e. \(\frac{a}{5}=\frac{b}{9}=\frac{c}{11}\) = k (say)
∴ a = 5k ;
b = 9k
and c = 11k
putting the values of a, b and e in eqn. (1) ; we get
5k (x + 1) + 9k (y + 1) + 11k (7 – 2) = 0
⇒ 5x + 9y + 11z = 8 be the required eqn. of plane.

Question 14.
Find the equation of the plane through the point \(\hat{i}-3 \hat{j}-2 \hat{k}\) and perpendicular to the planes \(\vec{r} \cdot(\hat{i}+2 \hat{j}+2 \hat{k})\) = 5 and \(\vec{r} \cdot(3 \hat{j}+3 \hat{j}+2 \hat{k})\) + 7 = 0.
Solution:
The equation of plane through the point \(\hat{i}-3 \hat{j}-2 \hat{k}\) i.e. (1, – 3, – 2) is given by
a (x – 1) + b (y + 3) + c (z + 2) = 0 …………………(1)
eqn’s of given planes in cartesian form are
x + 2y + 2z = 5 …………………(2)
and 3x + 3y + 2z + 7 = 0 ……………….(3)
Since plane (1) given by eqn. (1) is perpendicular to planes given by eqn. (2) and eqn. (3).
∴ a + 2b + 2c = 0 ……………..(4)
3a + 3b + 2c = 0 …………………(5)
On solving (4) and (5) ; we have
\(\frac{a}{4-6}=\frac{b}{6-2}=\frac{c}{3-6}\)
i.e. \(\frac{a}{-2}=\frac{b}{4}=\frac{c}{-3}\) = k (say) where k ≠ 0
∴ a = – 2k ;
b = 4k ;
c = – 3k
∴ eqn. (1) becomes ;
– 2k (x – 1) + 4k (y + 3) – 3k (z + 2) = 0
⇒ – 2x + 4y – 3z + 8 = 0 [∵ k ≠ 0]
⇒ 2x – 4y + 3z – 8 = 0
which gives the required eqn. of plane ¡n cartesian form.
Clearly its vector eqn. be \(\vec{r} \cdot(2 \hat{i}-4 \hat{j}+3 \hat{k})\) = 8
where \(\vec{r}\) be the P.V of the point P x, y, z) on the plane.

Question 15.
Find the vector equation of the plane passing through the point \(\) and perpendicular to the line of intersection of the planes \(\vec{r} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})\) = 0 and \(\vec{r} \cdot(\hat{i}+3 \hat{j}-5 \hat{k})\) + 7 = 0.
Solution:
Since the line of intersection of both planes
\(\vec{r} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})\) = 0
and \(\vec{r} \cdot(\hat{i}+3 \hat{j}-5 \hat{k})\) + 7 = 0
lies in both the planes. The normals to these planes along the vectors
\(\overrightarrow{n_1}=2 \hat{i}-\hat{j}+2 \hat{k}\) = 0
and \(\overrightarrow{n_2}=\hat{i}+3 \hat{j}-5 \hat{k}\)
Let \(\vec{n}\) be the normal to the required plane.
So \(\vec{n}\) is ⊥ to both \(\overrightarrow{n_1} \text { and } \overrightarrow{n_2}\)
also \(\overrightarrow{n_1} \times \overrightarrow{n_2}\) is ⊥ to both \(\overrightarrow{n_1} \text { and } \overrightarrow{n_2}\)
Thus \(\vec{n}\) is parallel to \(\overrightarrow{n_1} \times \overrightarrow{n_2}\)
∴ \(\overrightarrow{n_1} \times \overrightarrow{n_2}\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 2 \\
1 & 3 & -5
\end{array}\right|\)
= \(\hat{i}(5-6)-\hat{j}(-10-2)+\hat{k}(6+1)\)
= \(-\hat{i}+12 \hat{j}+7 \hat{k}\)
Thus, the required plane passing through the point \(5 \hat{i}+2 \hat{j}-3 \hat{k}\) and is ⊥ to the vector
\(-\hat{i}+12 \hat{j}+7 \hat{k}\) and its eqn. isgiven by \(\vec{r} \cdot(-\hat{i}+12 \hat{j}+7 \hat{k})\)
= \((5 \hat{i}+2 \hat{j}-3 \hat{k}) \cdot(-\hat{i}+12 \hat{j}+7 \hat{k})\)
[∵ \((\vec{r}-\vec{a}) \cdot \vec{n}\) = 0
i.e. \(\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}\)]
⇒ \(\vec{r} \cdot(-\hat{i}+12 \hat{j}+7 \hat{k})\) = 5 (- 1) + 2 (12) – 3 (7)
⇒ \(\vec{r} \cdot(-\hat{i}+12 \hat{j}+7 \hat{k})\) = – 2
⇒ \(\vec{r} \cdot(\hat{i}-12 \hat{j}-7 \hat{k})\) = 2,
which is the required eqn. of plane.

Aliter :
Let the equation of plane through the point \(5 \hat{i}+2 \hat{j}-3 \hat{k}\)
i.e. (5, 2, – 3) be given by
a (x – 5) + b (y – 2) + c (z – 3) = 0 ……………….(1)
where < a, b, c > are the direction numbers of normal to given plane.
Since plane (1) is perpendicular to line of intersection of both given planes
i.e. 2x – y + 2z = 0 ………………..(2)
and x + 3y – 5z + 7 = 0 ………………….(3)
∴ plane (1) is ⊥ to both planes (2) and (3).
∴ 2a – b + 2c = 0 …………………..(4)
a + 3b – 5c = 0 …………………(5)
On solving (4) and (5) ; we have
\(\frac{a}{5-6}=\frac{b}{2+10}=\frac{c}{6+1}\)
i.e. \(\frac{a}{-1}=\frac{b}{12}=\frac{c}{7}\) = k (say) ; k ≠ 0
∴ a = – k ;
b = 12k
and c = 7k
Thus eqn. (1) becomes ;
– k (x – 5) + 12k (y – 2) + 7k (z + 3) = 0
– x + 12y + 7z + 2 = 0
⇒ x – 12y – 7z = 2
⇒ \(\vec{r} \cdot(\hat{i}-12 \hat{j}-7 \hat{k})\) = 2
be the required vector equation of plane, where \(\vec{r}\) be the P.V of the point P (x, y, z) on given plane.

Question 16.
Find the equation of the plane passing through the points (2, 3, 1) and (4, – 5, 3) and parallel to x-axis.
Solution:
The eqn. of plane passing through the point (2, 3, 1) is given by
a (x – 2) + b (y – 3) + c (z – 1) = 0 ………………(1)
where < a, b, c > are the D’ ratios of the normal to plane (t).
Also, eqn. (1) passes through the point (4, – 5 3).
∴ a (4 – 2) + b (- 5 – 3) + c (3 – 1) = 0
i.e. 2a – 8b + 2c = 0
⇒ a – 4b + c = 0 …………………..(2)
Also, D’ Numbers of x-axis are < 1, 0, 0 >
Since plane (1) is || to x-axis
∴ normal to plane (1) is ⊥ to x-axis.
∴ a + 0b + 0c = 0 ………………….(3)
On solving eqn. (2) and (3) ; we have
\(\frac{a}{0}=\frac{b}{1-0}=\frac{c}{4}\) = k (say), k ≠ 0
∴ a = 0 ;
b = k ;
c = 4k
Thus eqn. (1) becomes ;
0 (x – 2) + k (y – 3) + 4k (z – 1) = 0
⇒ y + 4z – 7 = 0 [∵ k ≠ 0]
which is the required eqn. of plane.

Question 17.
Show that the equation ax + by + d = 0 represents a plane parallel to the z-axis. hence find the equation to a plane through the points (2, – 3, 1) and (- 4, 7, 6) and parallel to z-axis.
Solution:
Given eqn. of plane be
ax + by + d = 0 …………………(1)
∴ D’ ratios of normal to plane (1) are < a, b, 0 >
also. D’ ratios of z-axis are < 0, 0, 1 >
Now eqn. (1) i.e. given plane is parallel to z- axis if normal to plane is ⊥ to the z-axis
if a (0) + b (0) + 0 (1) = 0
[∵ a₁a₂2 + b₁b₂ + c₁c₂ = 0]
if0 + 0 + 0 = 0 if 0 = 0, which is true.
Let the eqn. of plane through the point (2,- 3, 1) be given by
a (x – 2) + b (y + 3) + c (z – 1) = 0 ……………..(2)
where < a, b, c > are the Direction numbers of the normal to plane.
Since plane (2) passes through the point (- 4, 7, 6).
∴ a (- 4 – 2) + b (7 + 3) + c (6 – 1) = 0
i.e. – 6a + 10b + 5c = 0 …………………….(3)
also, given plane (2) is parallel to z-axis.
∴ 0a + 0b + 1c = 0 …………………(4)
On solving (3) and (4) ; we have
\(\frac{a}{10-0}=\frac{b}{0+6}=\frac{c}{0}\)
⇒ \(\frac{a}{5}=\frac{b}{3}=\frac{c}{0}\) = k (say) ; k ≠ 0
∴ a = 5k ;
b = 3k ;
c = 0
Thus, eqn. (2) becomes ;
5k (x – 2) + 3k (y + 3) + 0 (z – 1) = 0
⇒ 5x + 3y – 1 =0 [∵ k ≠ 0]
which is the required eqn. of plane.

Question 18.
Show that the equation ax + cz + d = 0 represents a plane parallel to y-axis. Hence find the equation to a plane through the point (3, – 5, – 2), which makes intercept – 4 on x-axis and is parallel to y-axis.
Solution:
Given equation of plane be
ax + cz + d = 0 ………………….(1)
∴ Direction ratios of normal to plane (1) are < a, 0, c >
also D’ ratios of y-axis are < 0, 1, 0 >
Now given plane is || to y-axis if normal to plane (1) is ⊥ to the y-axis.
i.e. if a(0) + 0(1) + c(0) = 0
[a₁a₂2 + b₁b₂ + c₁c₂ = 0]
if 0 = 0, which is true.
Let the eqn of plane through the point (3, – 5, – 2) be
a (x – 3) + b (y + 5) + c (z + 2) = 0 …………………….(2)
where < a, b, c > are the direction numbers of normal to plane (1).
Since the plane makes intercepts – 4 on x-axis
∴ plane (2) passes through the point (- 4, 0, 0).
a (- 4 – 3) + b (0 + 5) + c (0 + 2) = 0
i.e. – 7a + 5b + 2c = 0 …………………(3)
Also plane (2) is parallel to y-axis whose d’numbers are < 0, 1, 0 >
∴ 0a + 1b + 0c = 0 …………………..(4)
On solving eqn. (3) and eqn. (4) ; we have
\(\frac{a}{0-2}=\frac{b}{0-0}=\frac{c}{-7-0}\)
i.e. \(\frac{a}{-2}=\frac{b}{0}=\frac{c}{-7}\) = k (say) ; k ≠ 0
∴ a = – 2k ;
b = 0 ;
c = – 7k
Thus, eqn. (2) becomes ;
– 2k (x – 3) + 0 (y + 5) – 7k (z + 2) = 0
⇒ – 2x – 7z – 8 = 0 [∵ k ≠ 0]
i.e. 2x + 7z + 8 = 0
which is the required eqn. of plane.

Question 19.
Find the equation of a plane passing through the point (2, – 1, 5), perpendicular to the plane x + 2y – 3z = 7 and parallel to the line \(\frac{x+5}{3}=\frac{y+1}{-1}=\frac{z-2}{1}\).
Solution:
Given eqn. of plane be
x + 2y – 3z = 7 ……………………..(1)
and eqn. of given line be
\(\frac{x+5}{3}=\frac{y+1}{-1}=\frac{z-2}{1}\) …………………(2)
The eqn. of plane through the point (2, – 1, 5) is given by
a (x – 2) + b (y + 1) + c (z – 5) = 0 ……………..(3)
where < a, b, c > are the direction ratios of normal to plane (3).
Now plane (3) is ⊥ to plane (1).
∴ a + 2b – 3c = 0 ……………….(4)
Also it is given that, plane (3) is parallel to line (2)
∴ Normal to plane is ± to line (2).
∴ 3a – b + c = 0 ……………….(5)
On solving eqn. (4) and eqn. (5); we have
\(\frac{a}{2-3}=\frac{b}{-9-1}=\frac{c}{-1-6}\)
i.e. \(\frac{a}{-1}=\frac{b}{-10}=\frac{c}{-7}\)
i.e. \(\frac{a}{1}=\frac{b}{10}=\frac{c}{7}\) = k (say) [where k ≠ 0]
a = k ;
b = 10k;
c = 7k
Thus eqn. (3) becomes ;
k (x – 2) + 10 k (y + 1) + 7k (z – 5) = 0
⇒ x + 10y + 7z – 27 = 0
which is the required eqn. of plane.

Question 20.
Find the cartesian equation of the plane passing through the intersection of the planes x + 2y – 3 = 0 and 2x – y + z = 1 and the origin. Also, write the equation of the plane so obtained in vector form.
Solution:
Given eqns. of planes are ;
x + 2y – 3 = 0 ……………………(1)
and 2x – y + z = 1 ……………….(2)
The eqn. of plane through the intersection of planes (1) and (2) be given by
(x + 2y – 3) + (2x – y + z – 1) = 0 ………………..(3)
Now plane (3) pass through the point (0, 0, 0)
∴ (0 + 0 – 3) + (0 – 0 + 0 – 1) = 0
⇒ λ = – 3
Thus eqn. (3) becomes;
(x + 2y – 3) – 3 (2x – y + z – 1) = 0
⇒ – 5x + 5y – 3z = 0
⇒ 5x – 5y + 3z = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(5 \hat{i}-5 \hat{j}+3 \hat{k})\) = 0
⇒ \(\vec{r} \cdot(5 \hat{i}-5 \hat{j}+3 \hat{k})\) = 0
where \(\vec{r}\) be the P.V of any point on required plane.

Question 20 (old).
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 and perpendicular to the plane x – y + z = 0. (NCERT)
Solution:
The eqn. of any plane passing through the line of intersection of given planes
x + y + z – 1 = 0
and 2x + 3y +4z – 5 = 0is given by
x + y + z – 1 + λ (2x + 3y + 4z – 5) = 0
i.e. (1 + 2λ) x + (1 + 3λ) y + (1 + 4λ) z – 1 – 5λ = 0 ………………….(1)
Since the plane(1) is ⊥ to given plane x – y + z = 0
Thus, (1 + 2λ) 1 + (1 + 3λ) (- 1) + (1 + 4λ) 1 = 0
⇒ 3λ + 1 = 0
⇒ λ = – 1/3
putting the value of λ in eqn. (1) ; we get
x + 0y – z + 2 = 0
i.e. \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+0 \hat{j}-\hat{k})\) + 2 = 0
⇒ \(\vec{r} \cdot(\hat{i}-\hat{k})\)+ 2 = 0 be the required eqn. of plane.

Question 21.
Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
Solution:
The equation of plane through the Line of intersection of the planes x + + 3z – 4 = 0
and 2x + y – z + 5 = 0 is given by
(x + 2y + 3z – 4) + λ (2x + y – z + 5) = 0
where λ be the parameter.
⇒ (1 + 2λ) x + (2 + λ) y + (3 – λ) z + (- 4 + 5λ) = 0 …………………..(1)
∴ Direction numbers of normal to plane (1) are < 1 + 2λ, 2 + λ, 3 – λ >
Also plane (1) is ⊥ to given plane
5x + 3y + 6z + 8 = 0 …………………….(2)
where d1 ratios of normal to plane (2) are < 5, 3, 6 >
∴ (1 + 2λ) 5 + (2 + λ) 3 + (3 – λ) 6 = 0
⇒ 5 + 10 λ + 6 + 3λ + 18 – 6 = 0
⇒ 7λ + 29 = 0
⇒ λ = – \(\frac{29}{7}\)
putting the value of λ in eqn. (1) ; we have
\(\left(1-\frac{58}{7}\right) x+\left(2-\frac{29}{7}\right) y+\left(3+\frac{29}{7}\right) z+\left(-4-\frac{145}{7}\right)\) = 0
⇒ – 51x – 15y + 50z – 173 = 0
⇒ 51x + 15y – 50z + 173 = 0
which is the required equation of plane.

Question 22.
Find the equation of the plane passing through the line of intersection of the planes 2x + 3y – z + 1 = 0 and x + y – 2z + 3 = 0 and perpendicular to the plane 3x – y – 2z – 4 = 0. Also find the inclination of this plane with the xy-plane.
Solution:
The given planes are,
2x + 3y + z + 1 = 0 …………………(1)
and x + y – 2z + 3 = 0 …………………..(2)
eqn. of plane through the intersection of planes (1) and (2) is given by
i.e. (2x + 3y – z + 1) + k (x + y – 2z + 3) = 0 ………………..(3)
i.e. (2 + k) x + (3 + k) y + z (- 1 – 2k) + (1 + 3k) = 0
Now plane (3) is ⊥ to given plane 3x – y – 2z – 4 = 0
∴their normals are ⊥
i.e. 3 (2 + k) – 1 (3 + k) – 2(- 1 – 2k) = 0
⇒ 6k + 5 = 0
⇒ k = – \(\frac{5}{6}\)
∴ From (3) ; we have
2x + 3y – z + 1 – \(\frac{5}{6}\) (x + y – 2z + 3) = 0
⇒ 7x + 13y + 4z – 9 = 0 is the required eqn. of planes.
D’ No’s of normal to plane are < 7, 13, 4 >
also D’ Nos of normal to xy-plane are < 0, 0, 1 >
If θ be the angle between them
Then cos θ = \(\frac{|7 \cdot 0+13 \cdot 0+4 \cdot 1|}{\sqrt{49+169+16}}\)
= \(\frac{4}{\sqrt{234}}\)

Question 23.
Find the equation of the plane through the line of intersection of \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})\) = 1 and \(\vec{r} \cdot(\hat{i}-\hat{j})\) + 4 = 0 and perpendicular to the plane \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) + 8 = 0. Hence, find whether the plane so obtained contains the line x – 1 = 2y – 4 = 3z – 12.
Solution:
In cartesian form, given planes are \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})\) = 1
⇒ 2x – 3y + 4z = 1 …………………(1)
and \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-\hat{j})\) + 4 = 0
x – y + 4 = 0 ……………………..(2)
Now eqn. of plane through the line of intersection of planes (1) and (2) is given by
(2x – 3y + 4z – 1) + (x – y + 4) = 0
⇒ (2 + λ) x + (- 3 – λ) y + 4z + (4λ – 1) = 0 ………………….(3)
also eqn. of plane \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) + 8 = 0 in cartesian form
it is given by 2x – y + z + 8 = 0 …………………(4)
Now direction numbers of normals to plane (3) and (4) are
Since it is given that plane (3) ¡s ! to plane (4)
∴ (2 + λ) 2 + (- 3 – λ) (- 1) + 4 × 1 = 0
⇒ 4 + 2λ + 3 + λ + 4 = 0
⇒ 3λ = – 11
λ = – \(\frac{11}{3}\)
putting the value of λ in eqn. (3); we get
\(\left(2-\frac{11}{3}\right) x+\left(-3+\frac{11}{3}\right) y+4 z+\left(-\frac{44}{3}-1\right)\) = 0
⇒ – 5x + 2y + 12z – 47 = 0
⇒ – 5x + 2y + 12z = 47
⇒ \(\vec{r} \cdot(-5 \hat{i}+2 \hat{j}+12 \hat{k})\) = 47 ………………..(5)
which is the required vector eqn. of plane.
Given line be \(\frac{x-1}{1}=\frac{2(y-2)}{1}\) = 3 (z – 4)
i.e. \(\frac{x-1}{1}=\frac{y-2}{\frac{1}{2}}=\frac{z-4}{\frac{1}{3}}\)
⇒ \(\frac{x-1}{6}=\frac{y-2}{3}=\frac{z-4}{2}\)
Its vector eqn. be
\(\vec{r}=\hat{i}+2 \hat{j}+4 \hat{k}+\mu(6 \hat{i}+3 \hat{j}+2 \hat{k})\) ………………….(6)
Comparing eqn. (5) with
\(\vec{r} \cdot \vec{n}\) = d
we have \(\vec{n}=-5 \hat{i}+2 \hat{j}+12 \hat{k}\) and d = 47
Comparing eqn. (6) with \(\vec{r}=\vec{a}+\mu \vec{b}\)

Question 24.
Find the equation of the plane passing through the line of intersection of the planes 2x + y – z = 3, 5x – 3y + 4z + 9 = O and parallel to the line \(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}\).
Solution:
The eqn. of plane passing through the line of intersection of given planes
2x + y – z = 3,
5x – 3y + 4z + 9 = 0
is given by (2x + y – z – 3) + (5x – 3y + 4z + 9) = 0
⇒ (2 + 5λ) x + (1 – 3λ) y + (- 1 + 4λ) z + (- 3 + 9λ) = 0 ………………….(1)
where λ be the parameter
∴ Direction ratios of normal to plane (1) are < 2 + 5λ, 1 – 3λ, – 1 + 4λ >
also eqn. of given line be
\(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}\) ………………….(2)
∴ Direction ratios of given line (2) are < 2, 4, 5 >
Since plane (1) is parallel to line 2.
∴ normal to plane is ⊥ to the line (2).
∴ (2 + 5λ) 2 + (1 – 3λ) 4 + (- 1 + 4λ) 5 = 0
[∵ a₁ a₂ + b₁ b₂ + c₁ c₂ = 0]
⇒ 4 + 10λ + 4 – 12λ – 5 + 20λ = 0
⇒ 18λ + 3 = 0
⇒ λ = – \(\frac{3}{18}=-\frac{1}{6}\)
putting the value of λ = – \(\frac{1}{6}\) in eqn. (1) ; we have
\(\left(2-\frac{5}{6}\right) x+\left(1+\frac{1}{2}\right) y+\left(-1-\frac{2}{3}\right) z+\left(-3-\frac{3}{2}\right)\) = 0
⇒ \(\frac{7 x}{6}+\frac{3}{2} y-\frac{5}{3} z-\frac{9}{2}\) = 0
⇒ 7x + 9y – 10z – 27 = 0
which is the required equation of plane.

Question 24 (old).
Find the equation of the plane through the line of intersection of the planes 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 and parallel to the line x = 2y = 3z.
Solution:
The equation of any plane through the line of intersection of given planes
3x – 4y + 5z – 10 = 0 and 2x + 2y – 3z – 4 = 0 is given by
(3x – 4y + 5z – 10) + λ (2x + 2y – 3z – 4) = 0
i.e. (3 + 2λ) x + (- 4 + 2λ) + (5 -3λ) z – 10 – 4λ = 0 ……………….(1)
D’ ratios of normal to plane (1) are proportional to, < 2λ + 3, 2λ – 4, 5 – 3λ >
and eqn. of given line be x = 2y = 3z
⇒ \(\frac{x}{1}=\frac{y}{1 / 2}=\frac{z}{1 / 3}\)
i.e. \(\frac{x}{6}=\frac{y}{3}=\frac{z}{2}\) ……………(2)
∴ D’ ratios of given line are proportional to < 6, 3, 2 >.
Since the plane (1) is parallel to line (2).
∴ Normal to plane is ⊥ to line (2).
Thus (2λ + 3) 6 + (2λ – 4) 3 + (5 – 3λ) 2 = 0
⇒ 12λ + 16 = 0
⇒ λ = – \(\frac{4}{3}\)
putting the value of λ in eqn. (1) ; we get
x – 20y + 27z = 14 be the required eqn. of plane.

Question 25.
Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0and whose x-intercept is twice its z-intercept. Hence, find the vector equation of the plane passing through the point (2, 3, – 1) and parallel to the plane obtained above.
Solution:
eqns. of given planes are
x + 2y + 3z – 4 = 0 ……………….(1)
2x + y – z + 5 = 0 ………………….(2)
eqn. of plane passing through the line of intersection of planes (1) and (2) is given by
(x + 2y + 3z – 4) + λ (2x + y – z + 5) = 0
(1 + 2λ) x + (2 + λ) y + (3 – λ) z = 4 – 5λ …………………(3)
For x-intercept of plane (3) ; we have
y = z = 0
∴ x-intercept of plane (3) = \(\frac{4-5 \lambda}{1+2 \lambda}\)
For z-intercept of plane (3); we have
x = y = 0
∴ z-intercept of plane (3) = \(\frac{4-5 \lambda}{3-\lambda}\)
according to given condition, we have
\(\frac{4-5 \lambda}{1+2 \lambda}=2\left(\frac{4-5 \lambda}{3-\lambda}\right)\)
⇒ (4 – 5λ) [2 (1 + 2λ) – 3 (3 – λ)] = 0
⇒ (4 – 5λ) (2 + 4λ – 3 + λ) = 0
⇒ (4 – 5λ) (- 1 + 5λ) = 0
⇒ λ = \(\frac{4}{5}, \frac{1}{5}\)
When λ = – \(\frac{4}{5}\) ;
x-intercept = z-intercept = 0
which is not possible.
∴ λ = \(\frac{1}{5}\),
putting the value of λ in eqn. (3) ; we have
\(\left(1+\frac{2}{5}\right) x+\left(2+\frac{1}{5}\right) y+\left(3-\frac{1}{5}\right)\) = 4 – 5 × \(\frac{1}{5}\)
⇒ \(\frac{7}{5} x+\frac{11}{5} y+\frac{14 z}{5}\) = 3
⇒ 7x + 11y + 14z = 15 …………………(4)
which is the required eqn. of plane.
Now eqn. of plane through the point (2, 3, – 1) is given by
a (x 2) + b (y – 3) + c (z + 1) = 0 ……………(5)
where < a, b, c > are the direction numbers of the normal to plane.
D’ ratios of normal to plane (4) are < 7, 11, 14 >
also plane (4) is parallel to plane (5).
So normal to plane (4) is normal to plane (5)
\(\frac{a}{7}=\frac{b}{11}=\frac{c}{14}\) = k (say)
i.e. a = 7k ;
b = 11k
and c = 14k
∴ from (5) ; we have
7k (x – 2) + 11k (y – 3) + 14k (z + 1) = 0
⇒ 7x + 11y + 14z = 33
i.e. \(\vec{r} \cdot(7 \hat{i}+11 \hat{j}+14 \hat{k})\) = 33 be the required plane.