OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(c)

Utilizing ISC Class 12 Maths OP Malhotra Solutions Chapter 24 The Plane Ex 24(c) as a study aid can enhance exam preparation.

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

Question 1.
Find the angle between the planes
(i) \(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) = 6 and
\(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)) = 7
(ii) \(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) = 6 and
\(\vec{r}\) (3 \(\hat{i}\) + 6 \(\hat{j}\) – 2 \(\hat{k}\)) = 9 (CB)
(iii) x + y + 2 z = 9 and 2 x – y + z = 15
(iv) 2 x – 3 y + 4 z = 1 and -x + y = 4
Answer:
(i) Given plane are
\(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) ) = 6
and
\(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)) = 7
Given planes (1) and (2) are of the form
\(\vec{r}\) \(\vec{n}_1\) = d₁ and \(\vec{r}\) \(\vec{n}_2\) = d₂
i.e. \(\vec{n}_1\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)
and \(\vec{n}_2\) = \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
Let θ be the angle between the given planes.
∴ cos θ = \(\vec{n}_1\) \(\vec{n}_2\)\(|\vec{n}_1||\vec{n}_2|\)
= \(\frac{(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}+\hat{j}+2 \hat{k})}{|2 \hat{i}-\hat{j}+\hat{k}||\hat{i}+\hat{j}+2 \hat{k}|}\)
= \(\frac{2(1)-1(1)+1(2)}{\sqrt{4+1+1} \sqrt{1+1+4}}\)
= \(\frac{3}{\sqrt{6} \cdot \sqrt{6}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ θ = \(\frac{\pi}{3}\)

(ii) eqns. of given pianes are
\(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) = 6
and \(\vec{r}\) (3 \(\hat{i}\) + 6 \(\hat{j}\) – 2 \(\hat{k}\)) = 9
Here ⇒ n₁ = 2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)
and ⇒ n₂ = 3 \(\hat{i}\) + 6 \(\hat{j}\) – 2 \(\hat{k}\)
Let θ be the angle between given planes
Then cos θ = \(\frac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}\)
= \(\frac{(2 \hat{i}-\hat{j}+2 \hat{k}) (3 \hat{i}+6 \hat{j}-2 \hat{k})}{\sqrt{4+1+4} \sqrt{9+36+4}}\)
∴ cos θ = \(\frac{2(3)-1(6)+2(-2)}{3 \times 7}\) = \(-\frac{4}{21}\)
Then θ = cos^-1 (\(-\frac{4}{21}\))

(iii) Given eqn. of planes are
and
x + y + 2 z = 9
2 x – y + z = 15
∴ D ratios of normal to plane (1) are < 1, 1, 2 > and D’ ratios of normal to plane (2) are < 2, -1, 2 >
Here a₁ = 1 ; b₁ = 1 ; c₁ = 2
a₂ = 2 ; b₂ = -1 ; c₂ = 1
Let θ be the angle between given planes Then
cos θ = \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
= \(\frac{1(2)+1(-1)+2(1)}{\sqrt{1+1+4} \sqrt{4+1+1}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
Then θ = \(\frac{\pi}{3}\)

(iv) Given eqns. of planes are
2 x – 3 y + 4 z = 1
-x + y = 4
and
D ratios of normal to plane (1) are < 2, -3, 4 > and D’ ratios of normal to plane (2) are < -1, 1, 0 >
Let θ be the angle between planes Then
cos θ = \(\frac{2(-1)-3(1)+4(0)}{\sqrt{2^2+(-3)^2+4^2} \sqrt{(-1)^2+1^2+0}}\)
= \(frac{-5}{\sqrt{29} \sqrt{2}}\) = \(\frac{-5}{\sqrt{58}}\)
∴θ = cos^-1 ( \(\frac{-5}{\sqrt{58}}\))

Question 2.
Show that the following planes are at right angles
(i) \(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) = 5 and
\(\vec{r}\) (\(-\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) = 3
(ii) 3 x – 2 y + z + 17 = 0 and
4 x + 3 y – 6 z – 25 = 0
Answer:
(i) We know that, planes
\(\vec{r}\) \(\overrightarrow{n_1}\) = d₁ and \(\vec{r}\) \(\overrightarrow{n_2}\) = d₂
are perpendicular if \(\overrightarrow{n_1}\) \(\overrightarrow{n_2}\) = 0
Here \(\vec{n}_1\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)
and
\(\overrightarrow{n_2}\) = (\(-\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
Here
\(\overrightarrow{n_1}\) \(\overrightarrow{n_2}\) = (2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) ( \(-\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
= 2(-1) – 1(-1) + 1(1)
= -2 + 1 + 1 = 0
Thus given planes are at right angles.

(ii) Equation of given planes are
3 x – 2 y + z + 17 = 0
and 4 x + 3 y – 6 z – 25 = 0
D ratios of normal to plane (1) and (2) are < 3,-2,1 > and < 4,3,-6 >
Let θ be the angle between given planes
Then cosθ = \(\frac{3(4)-2(3)+1(-6)}{\sqrt{9+4+1} \sqrt{16+9+36}}\) = 0
⇒ θ = \(\frac{\pi}{2}\)
Thus both planes are at right angles.

Question 3.
If the planes \(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + λ \(\hat{k}\)) = 5 and \(\vec{r}\) (3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)) = 5 are perpendicular, find the value of λ.
Answer:
Given planes are \(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + λ \(\hat{k}\)) = 5 and
\(\vec{r}\) (3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)) = 5
On comparing with
\(\vec{r}\) \(\vec{n}_1\) = d₁ and \(\vec{r}\) \(\vec{n}_2\) = d₂
Here
\(\vec{n}_1\) = 2 \(\hat{i}\) – \(\hat{j}\) + λ \(\hat{k}\)
\(\vec{n}_2\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)
and
Since both planes are perpendicular so their normals are also \perp.
∴ \(\vec{n}_1\) \(\vec{n}_2\) = 0
⇒ (2 \(\hat{i}\) – \(\hat{j}\) + λ \(\hat{k}\)) (3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)) = 0
⇒ 2(3) – 1(2) + λ = 0
⇒ 4 + 2 λ = 0
⇒ λ = -2

Question 4.
The planes \(\vec{r}\) (2 \(\hat{i}\) – λ \(\hat{j}\) + \(\hat{k}\)) = 3 and. \(\vec{r}\) (4 \(\hat{i}\) + \(\hat{j}\) – µ \(\hat{k}\)) = 5 are parallel. Determine λ and µ.
Answer:
Given eqn.’s of planes are
and
\(\vec{r}\) (2 \(\hat{i}\) – λ \(\hat{j}\) + \(\hat{k}\)) = 3
\(\vec{r}\) (4 \(\hat{i}\) + \(\hat{j}\) – µ \(\hat{k}\)) = 5
on comparing with
\(\vec{r}\) \(\overrightarrow{n_1}\) = d and \(\vec{r}\) \(\overrightarrow{n_2}\) = d₂
\(\vec{n}_1\) = 2 \(\hat{i}\) – λ \(\hat{j}\) + \(\hat{k}\)
and \(\vec{n}_2\) = 4 \(\hat{i}\) + \(\hat{j}\) – µ \(\hat{k}\)
Since the given planes are parallel
∴\(\vec{n}_1\) = t \(\vec{n}_2\)
(2 \(\hat{i}\) – λ \(\hat{j}\) + \(\hat{k}\)) = t(4 \(\hat{i}\) + \(\hat{j}\) – µ \(\hat{k}\))
for some non-zero scalor.
⇒ 2 = 4 t ⇒ t = \(\frac{1}{2}\)
-λ = t = \(\frac{1}{2}\)
and 1 = -µ t ⇒ µ = \(-\frac{1}{t}\) = -2
Thus λ = \(\frac{-1}{2}\) and µ = -2

Question 5.
Find the angle between
(i) the line \(\frac{x-1}{1}\) = \(\frac{y-2}{-1}\) = \(\frac{z+1}{1}\) and the plane 2 x + y – z = 4
(ii) the line \(\frac{x-2}{3}\) = \(\frac{y+1}{-1}\) = \(\frac{z-3}{2}\) and the plane 3 x + 4 y + z + 5 = 0
Answer:
(i) Clearly the given line passes through the point whose position vector \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\) and to vector \(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) and the given plane is normal to vector \(\vec{n}\) = 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
Let θ be the angle between given line and plane.
Then sin θ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)
= \(\frac{(\hat{i}-\hat{j}+\hat{k}) \cdot(2 \hat{i}+\hat{j}-\hat{k})}{\sqrt{1+1+1} \sqrt{4+1+1}}\)
= \(\frac{2-1-1}{\sqrt{3} \sqrt{6}}\) = 0
∴ θ = 0

(ii) 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 l x + m y + n z = p.
then, sinθ = \(\frac{a l+b m+c n}{\sqrt{a^2+b^2+c^2} \sqrt{l^2+m^2+n^2}}\)
Here, a = 3 ; b = -1 ; c = 2
and l = 3 ; m = 4 ; n = 1
∴ sinθ = \(\frac{3(3)-1(4)+2(1)}{\sqrt{3^2+(-1)^2+2^2} \sqrt{3^2+4^2+1^2}}\)
= \(\frac{7}{\sqrt{14} \sqrt{26}}\)
⇒ sinθ = \(\frac{7}{\sqrt{7} \sqrt{52}}\) = \(\sqrt{\frac{7}{52}}\)
⇒ θ = sin^-1 (\(\sqrt{\frac{7}{52}}\))

Question 6.
Find the angle between the line
\(\vec{r}\) = (2 \(\hat{i}\) + 3 \(\hat{j}\) + 9 \(\hat{k}\)) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\))
and the plane \(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) = 5
Answer:
We know that, angle between the line
\(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\)
and the plane \(\vec{r}\) \(\vec{n}\) = d is given by
sinθ = \(\frac{\vec{b}\vec{n}}{|\vec{b}||\vec{n}|}\)
Here \(\vec{b}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) ; \(\vec{n}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
∴ sinθ = \(\frac{(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})}{\sqrt{2^2+3^2+4^2} \sqrt{1^2+1^2+1^2}}\)
= \(\frac{2+3+4}{\sqrt{29} \sqrt{3}}\) = \(\frac{3 \sqrt{3}}{\sqrt{29}}\)
∴θ = sin^-1 (\(\frac{3 \sqrt{3}}{\sqrt{29}}\))

Question 7.
Find the angle between the line
\(\vec{r}\) = \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\) + λ(2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)) and the plane \(\vec{r}\) (3 \(\hat{i}\) – 2 \(\hat{j}\) + 6 \(\hat{k}\)) = 14
Answer:
Eqn. of given line be
\(\vec{r}\) = \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\) + λ(2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\))
and eqn. of given plane be
\(\vec{r}\) (3 \(\hat{i}\) – 2 \(\hat{j}\) + 6 \(\hat{k}\)) = 14
We know that if θ be the angle between line \(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\) and plane \(\vec{r}\) \(\vec{n}\) = d then 90° – θ be the angle between \(\vec{n}\) and \(\vec{b}\)
∴ cos(90° – θ) = sinθ = \(\frac{\vec{b}\cdot \vec{n}}{|\vec{b}||\vec{n}|}\)
Here \(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\) and \(\vec{n}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) + 6 \(\hat{k}\)
∴ \(\vec{b}\) \(\vec{n}\) = (2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)) (3 \(\hat{i}\) – 2 \(\hat{j}\) + 6 \(\hat{k}\))
= 2(3) + 1(-2) + 2(6)
= 6 – 2 + 12 = 16
\(|\vec{b}|\) and \(|\vec{n}|\) = \(\sqrt{4+1+4}\) = 3
and \(|\vec{n}|\) = \(\sqrt{9+4+36}\) = 7
∴ from (3); sin θ = \(\frac{16}{3 \times 7}\) = \(\frac{16}{21}\)
⇒ θ = sin^-1 \(\frac{16}{21}\)

Question 8.
If the line \(\vec{r}\) = (\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)) + λ(2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)) is parallel to the plane \(\vec{r}\) (3 \(\hat{i}\) – 2 \(\hat{j}\) + m \(\hat{k}\)) = 14, find the value of m.
Answer:
Eqn. of given line be
\(\vec{r}\) = (\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)) + λ(2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\))
Here given line passes through a point whose P.V. \(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)
and || to \(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
Also eqn. of given plane to
\(\vec{r}\) (3 \(\hat{i}\) – 2 \(\hat{j}\) + m \(\hat{k}\)) = 14
Thus the plane is normal to
\(\vec{n}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) + m \(\hat{k}\)
Since the line (1) is parallel to plane (2). ∴ normal to plane (2) is \perp to line (1)
∴ \(\vec{b}\) \(\vec{n}\) = 0
⇒ (2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)) (3 \(\hat{i}\) – 2 \(\hat{j}\) + m \(\hat{k}\)) = 0
⇒ 2(3) + 1(-2) + 2(m) = 0
⇒ 4 + 2 m = 0 ⇒ m = -2

Question 9.
Find the equation of the line passing through the point (3, 0, 1) and parallel to the planes x + 2 y = 0 and 3 y – z = 0.
Answer:
The eqn. of line passing through the point (3, 0, 1) and having direction ratios < a, b, c > is given by
\(\frac{x-3}{a}\) = \(\frac{y-0}{b}\) = \(\frac{z-1}{c}\)
Now line (1) is parallel to plane
x + 2 y = 0
D ratios of normal to plane (2) are < 1, 2, 0 >
∴a + 2 b + 0 c = 0
Also eqn. of given plane be
3y – z = 0
and direction ratios of normal to plane (4) are < 0, 3, -1 > and plane (4) is parallel to line (1)
∴ normal to plane (4) is ⊥ to line (1)
∴ 0 a + 3 b – c = 0
On solving eqn. (3) (5) by cross-multiplication method.
\(\frac{a}{-2-0}\) = \(\frac{b}{0+1}\) = \(\frac{c}{3-0}\) = k (say)
⇒ a = -2 k ; b = k andc = 3 k
putting all these values of a, b and c in eqn. (1); we get
\(\frac{x-3}{-2 k}\) = \(\frac{y-0}{k}\) = \(\frac{z-1}{3 k}\)
i.e. \(\frac{x-3}{-2}\) = \(\frac{y-0}{1}\) = \(\frac{z-1}{3}\) be the reqd. line.

Question 10.
Find the vector equation of the line passing through the point with position vector 2 \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\) and perpendicular to the plane \(\vec{r}\) (6 \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\)) + 2 = 0.
Answer:
Given eqn. of plane be
\(\vec{r}\) (6 \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\)) + 2 = 0
since the required line is ⊥ to the plane (1)
∴ line is parallel to normal
\(\vec{n}\) = 6 \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\)
Hence the required eqn. of line passing through the point whose P.V. is
\(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\) and || to
\(\vec{n}\) = 6 \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\)
is given by \(\vec{r}\) = \(\vec{a}\) + λ \(\vec{n}\)
i.e. \(\vec{r}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\) + λ(6 \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\)) ne the required vector eqn. of line.

Question 11.
Find the equation of the plane
(i) through the points (1, 0, -1),(3, 2, 2) and parallel to the line
\(\frac{x-1}{1}\) = \(\frac{y-1}{-2}\) = \(\frac{z-2}{3}\) .
(ii) Through the points (2, 2, -1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.
Answer:
(i) Let the eqn. of plane through the point (1, 0, -1) be given by a(x – 1) + b(y – 0) + c(z + 1) = 0 …………….. (1) Where < a, b, c > are the direction ratios of normal to plane.
Also plane (1) is parallel to given line
\(\frac{x-1}{1}\) = \(\frac{y-1}{-2}\) = \(\frac{z-2}{3}\)
∴ normal to plane (1) is ⊥ to given line
∴ a – 2 b + 3 c = 0
Also given plane (1) passes through the point (3, 2, 2)
∴ a(3 – 1) + b(2 – 0) + c(2 + 1) = 0
⇒ 2 a + 2 b + 3 c = 0
On solving eqn. (2) and (3) by crossmultiplication method, we have
\(\frac{a}{-6-6}\) = \(\frac{b}{6-3}\) = \(\frac{c}{2+4}\)
⇒ \(\frac{a}{-12}\) = \(\frac{b}{3}\) = \(\frac{c}{6}\)
i.e. \(\frac{a}{4}\) = \(\frac{b}{-1}\) = \(\frac{c}{-2}\) = k (say)
where k ≠ 0
∴ a = 4 k ; b = -k ; c = -2 k
putting the values of a, b c in eqn. (1); we get
4 k(x – 1) – k y – 2 k(z + 1) = 0
⇒ 4 x – y – 2 z – 6 = 0
be the reqd. eqn. of plane.
(ii) The eqn. of plane through the point (2, 2, -1) is given by
a(x – 2) + b(y – 2) + c(z + 1) = 0
Now plane (1) passes through the point (3, 4, 2).
∴the point (3,4,2) satisfies eqn. (1). a(3-2)+b(4-2)+c(2+1)=0 i.e. a+2 b+3 c=0
where the direction ratios of normal to plane are proportional to < a, b, c >
Since the plane parallel to line having d’ ratios are proportional to < 7, 0, 6 >
∴7 a + 0 b + 6 c = 0
On solving eqn. (2) and eqn. (3); we get
\(\frac{a}{12-0}\) = \(\frac{b}{21-6}\) = \(\frac{c}{0-14}\)
⇒ \(\frac{a}{12}\) = \(\frac{b}{15}\) = \(\frac{c}{-14}\) = k (say)
∴ a = 12 k ; b = 15 k ; c = -14 k
putting the values of a, b and c in eqn. (1); we get
12 k(x – 2) + 15 k(y – 2) – 14 k(z + 1) = 0
⇒ 12 x + 15 y – 14 z – 68 = 0
be the required equation of plane.

Question 12.
Find the vector equation of the line through the origin which is perpendicular to the plane \(\vec{r}\) (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) = 3.
Answer:
Given eqn. of plane be
\(\vec{r}\) (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) = 3
Since the required line is ⊥ to plane (1) thus reqd. line is parallel to the normal to plane (1) i.e. \(\vec{n}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
Hence the required vector eqn. of line passing through point whose P.V. is
\(\vec{a}\) = 0 \(\hat{i}\) + 0 \(\hat{j}\) + 0 \(\hat{k}\)
and parallel to \(\vec{n}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\) is given by
\(\vec{r}\) = \(\vec{a}\) + λ \(\vec{n}\)
\(\vec{r}\) = λ(\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\))
i.e. \(\vec{r}\) = λ(\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\))
where λ be any colar.

Question 13.
Find the equation of the plane through (2, 3, -4) and (1, -1, 3) and parallel to x axis.
Answer:
The equation of any plane through the point (2, 3, -4) is given by
a(x – 2) + b(y – 3) + c(z + 4) = 0
Since the plane (1) passes through (1, -1, 3) and the point (1, -1, 3) lies on eqn. (1)
a(1 – 2) + b(-1 – 3) + c(3 + 4) = 0
i.e. -a – 4 b + 7 c = 0
Now direction ratios of x-axis are proportional to < 1, 0, 0 >
Now the plane (1) is parallel to x-axis
∴Normal to plane is ⊥ to x-axis
∴a + 0 b + 0 c = 0
On solving (2) and (3); we have
\(\frac{a}{0}\) = \(\frac{b}{7-0}\) = \(\frac{c}{4}\) = k (say)
∴ a = 0 ; b = 7 k ; c = 4 k
putting the values of a, b, c in eqn. (1); we have
0(x – 2) + 7 k(y – 3) + 4 k(z + 4) = 0
⇒ 7 y + 4 z = 5 be the required eqn. of plane.

Question 14.
Find the vector equation of the line passing through the point (1, -1, 2) and perpendicular to the plane
2x – y + 3 z – 5 = 0
Answer:
eqn. of given plane be
2 x – y + 3 z – 5 = 0
i.e. its vector eqn. be
(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)) = 5
i.e. \(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)) = 5
So the given plane be normal to
\(\vec{n}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)
Also the required line is ⊥ to plane (1).
∴required line is parallel to vector \(\vec{n}\).
Hence the equation of required line passing through the point having position vector
\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)
and || to vector \(\vec{n}\) is \(\vec{r}\) = \(\vec{a}\) + λ \(\vec{n}\)
i.e. \(\vec{r}\) = \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\) + λ(2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\))

Question 15.
Find the equation of the plane passing through the points (-1, 1, 1) and (1, -1, 1) and perpendicular to the plane
x + 2 y + 2 z = 5
Answer:
The eqn. of any plane passing through the point (-1, 1, 1) be given by
a(x + 1) + b(y – 1) + c(z – 1) = 0
Since plane (1) passes through (1, -1, 1). Thus (1, -1, 1) lies on eqn. (1).
i.e. a(1 + 1) + b(-1 – 1) + c(1 – 1) = 0
i.e. 2 a – 2 b + 0 c = 0
Now plane (1) is ⊥ to given plane
i.e. x + 2 y + 2 z = 5
a + 2 b + 2 c = 0
On solving eqn. (2) and (3); we get
\(\frac{a}{-4-0}\) = \(\frac{b}{0-4}\) = \(\frac{c}{4+2}\)
i.e. \(\frac{a}{-4}\) = \(\frac{b}{-4}\) = \(\frac{c}{6}\)
i.e. \(\frac{a}{2}\) = \(\frac{b}{2}\) = \(\frac{c}{-3}\) = k (say)
⇒ a = 2 k ; b = 2 k and c = -3 k
putting the values of a, b and c in eqn. (1) we get
2 k(x + 1) + 2 k(y – 1) – 3 k(z – 1) = 0
⇒ 2 x + 2 y – 3 z + 3 = 0
be the required eqn. of plane.

Question 16.
Find the equation of the plane passing through the point (1, 1, 1) and perpendicular to each of the planes x + 2 y + 3 z = 7 and 2 x – 3 y + 4 z = 0.
Answer:
Let the eqn. of plane through the point (1, 1, 1) is given by
a(x – 1) + b(y – 1) + c(z – 1) = 0
Where < a, b, c > be the direction ratios of normal to plane (1).
Since the plane (1) is ⊥ to plane
x + 2 y + 3 z = 7
∴ a + 2 b + 3 c = 0
Also the plane (1) is ⊥ to the plane
2 x – 3 y + 4 z = 0
i.e. 2 a – 3 b + 4 c = 0
On solving eqn. (2) and (3) by crossmultiplication method, we have
\(\frac{a}{8+9}\) = \(\frac{b}{6-4}\) = \(\frac{c}{-3-4}\)
i.e. \(\frac{a}{17}\) = \(\frac{b}{2}\) = \(\frac{c}{-7}\) = k(say);
where k ≠ 0
∴ a = 17 k ; b = 2 k and c = -7 k
putting all these values in eqn. (1); we have
17 k(x – 1) + 2 k(y – 1) – 7 k(z – 1) = 0
⇒ 17 x + 2 y – 7 z – 12 = 0
be the required eqn. of plane.

Question 17.
Find the vector equation in the scalar product form of the plane through the point (4, 2, 4) and perpendicular to the
planes \(\vec{r}\) (2 \(\hat{i}\) + 5 \(\hat{j}\) + 4 \(\hat{k}\)) = -1 and
\(\vec{r}\) (4 \(\hat{i}\) + 7 \(\hat{j}\) + 6 \(\hat{k}\)) = -2 .
Answer:
Let \(\vec{n}\) be a vector normal to required plane. Also vector normal to given planes are
and
\(\vec{n}_1\) = 2 \(\hat{i}\) + 5 \(\hat{j}\) + 4 \(\hat{k}\)
\(\vec{n}_2\) = 4 \(\hat{i}\) + 7 \(\hat{j}\) + 6 \(\hat{k}\)
Now \(\vec{n}_1\) × \(\vec{n}_2\) be a vector normal to both \(\vec{n}_1\) and \(\vec{n}_2\)
Since the required plane is ⊥ to two given planes.
∴ \(\vec{n}\) is ⊥ to each of two vectors \(\vec{n}_1\) and \(\vec{n}_2\)
Thus \(\vec{n}\) = \(\vec{n}_1\) × \(\vec{n}_2\) = \(|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
2 & 5 & 4 \\
4 & 7 & 6
\end{array}|\)

= \(\hat{i}\)(30 – 28) – \(\hat{j}\) (12 – 16) + \(\hat{k}\) (14 – 20)
= 2 \(\hat{i}\) + 4 \(\hat{j}\) – 6 \(\hat{k}\)
∴required eqn. of plane be given by
(\(\vec{r}\) – \(\vec{a}\)) \(\vec{n}_1\) × \(\vec{n}_2\) = 0
⇒ [\(\vec{r}\) – (4 \(\hat{i}\) + 2 \(\hat{j}\) + 4 \(\hat{k}\))] (2 \(\hat{i}\) + 4 \(\hat{j}\) – 6 \(\hat{k}\)) = 0
⇒ \(\vec{r}\) (2 \(\hat{i}\) + 4 \(\hat{j}\) – 6 \(\hat{k}\))
= (4 \(\hat{i}\) + 2 \(\hat{j}\) + 4 \(\hat{k}\)) (2 \(\hat{i}\) + 4 \(\hat{j}\) – 6 \(\hat{k}\))
= 4(2) + 2(4) – 4(6) = -8
Which is the reqd. eqn. of plane.
Aliter : Let the eqn. of plane through the point (4, 2, 4) is given by
a(x – 4) + b(y – 2) + c(z – 4) = 0
Where < a, b, c > be the direction ratios of normal to plane (1).
eqn. of given planes in cartesian form are :
2 x + 5 y + 4 z = -1
4 x + 7 y + 6 k = -2
and
Since the plane (1) is ⊥ to plane (2).
∴normal to plane (1) is ⊥ to normal to plane (2).
2 a + 5 b + 4 c = 0
Also plane (1) is ⊥ to plane (3)
∴normal to plane (1) is ⊥ to normal to plane (3)
∴ 4 a + 7 b + 6 c = 0
On Solving eqn. (4) and (5) sumltancously using cross multiplication method, we have
\(\frac{a}{30-28}\) = \(\frac{b}{16-12}\) = \(\frac{c}{14-20}\)
i.e. \(\frac{a}{2}\) = \(\frac{b}{4}\) = \(\frac{c}{-6}\)
i.e. \(\frac{a}{1}\) = \(\frac{b}{2}\) = \(\frac{c}{-3}\) = k (say); where k ≠ 0
∴ a = k ; b = 2 k and c = -3 k
putting the values of a, b c in eqn. (1) ; we have
k(x – 4) + 2 k(y – 2) – 3 k(z – 4) = 0
⇒ x + 2 y – 3 z + 4 = 0
In vector form, \(\vec{r}\) (\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) = -4 be the reqd. plane.
Question 18.
Find the vector equation of the plane through the point (2 \(\hat{i}\) – \(\hat{j}\) – 4 \(\hat{k}\)) and plane parallel to the plane
\(\vec{r}\) (2 \(\hat{i}\) + 5 \(\hat{j}\) + 4 \(\hat{k}\)) = -1
Answer:
eqn. of given plane be
\(\vec{r}\) (4 \(\hat{i}\) – 12 \(\hat{j}\) – 3 \(\hat{k}\)) – 7 = 0
Thus the eqn. of plane || to plane (1) be given by
\(\vec{r}\) (4 \(\hat{i}\) – 12 \(\hat{j}\) – 3 \(\hat{k}\)) = d
Since eqn. (2) passes throguh the point (2 \(\hat{i}\) – \(\hat{j}\) – 4 \(\hat{k}\)).
∴(2 \(\hat{i}\) – \(\hat{j}\) – 4 \(\hat{k}\)) (4 \(\hat{i}\) – 12 \(\hat{j}\) – 3 \(\hat{k}\)) = d
⇒ 2(4) – 1(-12) – 4(-3) = d
⇒ 8 + 12 + 12 = d
⇒ d = 32
∴from (1) ; \(\vec{r}\) (4 \(\hat{i}\) – 12 \(\hat{j}\) – 3 \(\hat{k}\)) = 32 be the required vector eqn. of plane.
Aliter : Eqn. of plane in cartesian eqn. be
(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (4 \(\hat{i}\) – 12 \(\hat{j}\) – 3 \(\hat{k}\)) – 7 = 0
⇒ 4 x – 12 y – 3 z = 7
∴Required eqn. of plane which is parallel to plane (1) be
4 x – 12 y – 3 z = k
Since it is given that eqn. (2) passes through the point (2 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\)) i.e. (2, -1, -4)
∴ 4(2) – 12(-1) – 3(-4) = k
⇒ k = 32
∴ from (2) ;
4 x – 12 y – 3 z = 32
be the reqd. eqn. of plane.