Angle Between Two Planes Class 12 OP Malhotra Exe-24C ISC Maths Solutions Ch-24 The Plane. In this article you would learn how to solve questions / Problems on vector equation of a plane in normal form with answer. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

The Plane Class 12 OP Malhotra Exe-24C ISC Maths Solutions Ch-24 The Plane

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-24	The Plane
Writer	OP Malhotra
Exe-24(c)	Angle Between Two Planes

Angle Between Two Planes

The Plane Class 12 OP Malhotra Exe-24C Solutions

Que-1: Find the angle between the planes
(i) r→ (2î – ĵ + k̂) = 6 and
r→ (î + ĵ + 2k̂) = 7
(ii) r→ (2î – ĵ + 2k̂) = 6 and
r→ (3î + 6 ĵ – 2k̂) = 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

Sol: (i) Given plane are
r→ (2î – ĵ + k̂ ) = 6
and
r→ (î + ĵ + 2k̂) = 7
Given planes (1) and (2) are of the form
r→  n1→ = d1 and r⃗ n⃗ 2 = d2
i.e. n1→ = 2î – ĵ + k̂
and n→2 = î + ĵ + 2k̂
Let θ be the angle between the given planes.
∴ cos θ = n1→ n→2/|n1→||n→2|
= (2î−ĵ+k̂)⋅(î+ĵ+2k̂)|2î−ĵ+k̂||î+ĵ+2k̂|
= 2(1)−1(1)+1(2)/√4+1+1√1+1+4
= 3/√6⋅√6 = 3/6 = 1/2
∴ θ = π/3

(ii) eqns. of given pianes are
r→  (2î – ĵ + 2k̂) = 6
and r→  (3î + 6ĵ – 2k̂) = 9
Here ⇒ n1→ = 2î – ĵ + 2k̂
and ⇒ n→2 = 3î + 6ĵ – 2k̂
Let θ be the angle between given planes
Then cos θ = n1→⋅n→2/|n1→||n→2|
= (î−ĵ+2k̂)(î+6ĵ−2k̂)/√4+1+4√9+36+4
∴ cos θ = 2(3)−1(6)+2(−2)/3×7 = −4/21
Then θ = cos^-1(−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 =a₁a₂+b₁b₂+c₁c₂/√(a₁)²+(b₁)²+(c₁)² √(a₂)²+(b₂)²+(c₂)²
= 1(2)+1(−1)+2(1)/√ 1+1+4 √4+1+1= 3/6 = 1/2
Then θ = π/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 = 2(−1)−3(1)+4(0)/√2²+(−3)²+4²√(−1)²+1²+0
= -5/√29√2 = −5/√58
∴θ = cos^-1( −5/√58)

Que-2: Show that the following planes are at right angles
(i) r→(2î – ĵ + k̂) = 5 and
r→ (−î – ĵ + k̂) = 3
(ii) 3 x – 2 y + z + 17 = 0 and
4 x + 3 y – 6 z – 25 = 0

Sol: (i) We know that, planes
r→ n1→ = d₁ and r→  n→2 = d₂
are perpendicular if n1→ n→2 = 0
Here n1→ = 2î – ĵ + k̂
and
n→2 = (−î – ĵ + k̂)
Here
n1→ n→2 = (2î – ĵ + k̂) ( −î – ĵ + k̂)
= 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 = 3(4)−2(3)+1(−6)/√9+4+1√16+9+36 = 0
⇒ θ = π/2
Thus both planes are at right angles.

Que-3: If the planes r→ (2î – ĵ + λk̂) = 5 and r→ (3î+ 2ĵ + 2k̂) = 5 are perpendicular, find the value of λ.

Sol: Given planes are r→ (2î – ĵ + λk̂) = 5 and
r→ (3î + 2ĵ + 2k̂) = 5
On comparing with
r→  n1→ = d₁ and r→  n→2 = d₂
Here
n1→ = 2î – ĵ + λk̂
n→2 = 3î + 2ĵ + 2k̂
and
Since both planes are perpendicular so their normals are also \perp.
∴ n1→ n→2 = 0
⇒ (2î –ĵ + λk̂) (3î + 2ĵ + 2k̂) = 0
⇒ 2(3) – 1(2) + λ = 0
⇒ 4 + 2 λ = 0
⇒ λ = -2

Que-4: The planes r→(2î – λĵ + k̂) = 3 and. r→ (4î + ĵ – µk̂) = 5 are parallel. Determine λ and µ.

Sol: Given eqn.’s of planes are
and
r→ (2î – λĵ + k̂) = 3
r→ (4î + ĵ – µk̂) = 5
on comparing with
r→ n1→ = d and r→ n→2 = d₂
n1→ = 2î – λĵ + k̂
and n→2 = 4î + ĵ – µ k̂
Since the given planes are parallel
∴n1→ = t n→2
(2î – λĵ + k̂) = t(4î + ĵ – µk̂)
for some non-zero scalor.
⇒ 2 = 4 t ⇒ t = 1/2
-λ = t = 12
and 1 = -µ t ⇒ µ = −1/t = -2
Thus λ = −1/2 and µ = -2

Que-5: Find the angle between
(i) the line x−1/1 = y−2/−1 = z+1/1 and the plane 2 x + y – z = 4
(ii) the line x−2/3 = y+1/−1 = z−3/2 and the plane 3 x + 4 y + z + 5 = 0

Sol: (i) Clearly the given line passes through the point whose position vector î + 2ĵ – k̂ and to vector b→ = î – ĵ + k̂ and the given plane is normal to vector n→ = 2î + ĵ – k̂
Let θ be the angle between given line and plane.
Then sin θ = b→ ⋅n→ |b→ ||n→|
= (î−ĵ+k̂)⋅(2î+ĵ−k̂)/√1+1+1√4+1+1
= 2−1−1/√3√6 = 0
∴ θ = 0

(ii) We know that if θ be the angle between the line x−x₁/a = y−y₁/b = z−z₁/c and plane l x + m y + n z = p.
then, sinθ = al+bm+cn/√a²+b²+c² √l²+m²+n²
Here, a = 3 ; b = -1 ; c = 2
and l = 3 ; m = 4 ; n = 1
∴ sinθ = 3(3)−1(4)+2(1)/√3²+(−1)²+2²√3²+4²+1²
= 7/√14√26
⇒ sinθ = 7/√7√52 = √7/52
⇒ θ = sin^-1  (√7/52)

Que-6: Find the angle between the line
r→ = (2î + 3ĵ + 9k̂) + λ(2î + 3ĵ + 4k̂)
and the plane r→ (î + ĵ + k̂) = 5

Sol: We know that, angle between the line
r→ = a→ + λ b→
and the plane r→  n → = d is given by
sinθ = b→ n → /|b→ ||n → |
Here b→ = 2î + 3ĵ + 4k̂ ; n → = î + ĵ + k̂
∴ sinθ = (2î+3ĵ+4k̂)⋅(î+ĵ+k̂)/√2²+3²+4²√1²+1²+1²
= 2+3+4/√29√3 = 3√3/√29
∴θ =  sin^-1 (3√3/√29)

Que-7: Find the angle between the line
r→ = î – 2ĵ + k̂ + λ(2î +ĵ + 2k̂) and the plane r→(3î – 2ĵ + 6 k^) = 14

Sol: Eqn. of given line be
r→ = î – 2ĵ + k̂ + λ(2î + ĵ + 2k̂)
and eqn. of given plane be
r→ (3î – 2ĵ + 6k̂) = 14
We know that if θ be the angle between line r→ = a→ + λ b→ and plane r→  n→ = d then 90° – θ be the angle between n→ and b→
∴ cos(90° – θ) = sinθ = b→ ⋅n→ |b→ ||n→ |
Here b→ = 2î + ĵ + 2k̂ and n→ = 3 î – 2ĵ + 6k̂
∴ b→ n→ = (2î + ĵ + 2k̂) (3î – 2ĵ + 6k̂)
= 2(3) + 1(-2) + 2(6)
= 6 – 2 + 12 = 16
|b→ | and |n→ | = √4+1+4 = 3
and |n→ | = √9+4+36 = 7
∴ from (3); sin θ = 16/3×7 = 16/21
⇒ θ =  sin^-1 (16/21)

Que-8: If the line r→ = (î – 2ĵ + k̂) + λ(2î+ ĵ + 2k̂) is parallel to the plane r→ (3î – 2 ĵ + mk̂) = 14, find the value of m.

Sol: Eqn. of given line be
r→ = (î – 2ĵ + k̂) + λ(2î +ĵ + k̂)
Here given line passes through a point whose P.V. a→ =î – 2ĵ + k̂
and || to b→ = 2î + ĵ + 2k̂
Also eqn. of given plane to
r→ (3î – 2ĵ + mk̂) = 14
Thus the plane is normal to
n→ = 3î – 2ĵ + mk̂
Since the line (1) is parallel to plane (2). ∴ normal to plane (2) is \perp to line (1)
∴ b→ n→ = 0
⇒ (2î + ĵ + 2k̂) (3î – 2ĵ + mk̂) = 0
⇒ 2(3) + 1(-2) + 2(m) = 0
⇒ 4 + 2 m = 0 ⇒ m = -2

Que-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.

Sol: The eqn. of line passing through the point (3, 0, 1) and having direction ratios < a, b, c > is given by
x−3/a = y−0/b = 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.
a/−2−0 = b/0+1 = 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
x−3/−2k = y−0/k = z−1/3k
i.e. x−3/−2 = y−0/1 = z−1/3 be the reqd. line

Que-10: Find the vector equation of the line passing through the point with position vector 2î – 3ĵ – 5k̂ and perpendicular to the plane r→ (6î – 3ĵ + 5k̂) + 2 = 0.

Sol: Given eqn. of plane be
r→ (6î – 3ĵ + 5k̂) + 2 = 0
since the required line is ⊥ to the plane (1)
∴ line is parallel to normal
n→ = 6î – 3ĵ + 5k̂
Hence the required eqn. of line passing through the point whose P.V. is
a→ = 2î – 3ĵ – 5k̂ and || to
n→ = 6î – 3ĵ + 5k̂
is given by r→ = a→ + λ n→
i.e. r→ = 2î – 3ĵ – 5k̂ + λ(6î – 3ĵ + 5k̂) ne the required vector eqn. of line

Que-11: Find the equation of the plane
(i) through the points (1, 0, -1),(3, 2, 2) and parallel to the line
x−1/1 = y−1/−2 = 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.

Sol: (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
x−1/1 = y−1/−2 = 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 cross multiplication method, we have
a/−6−6 = b/6−3 = c/2+4
⇒ a/−12 = b/3 = c/6
i.e. a/4 = b/−1 = 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
a/12−0 = b/21−6 = c/0−14
⇒ a/12 = b/15 = 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.

Que-12: Find the vector equation of the line through the origin which is perpendicular to the plane r→ (î + 2ĵ + 3k̂) = 3.

Sol: Given eqn. of plane be
r→ (î + 2ĵ + 3k̂) = 3
Since the required line is ⊥ to plane (1) thus reqd. line is parallel to the normal to plane (1) i.e. n→ = î + 2ĵ + 3k̂
Hence the required vector eqn. of line passing through point whose P.V. is
a→ = 0î + 0ĵ + 0k̂
and parallel to n→ = î + 2ĵ + 3k̂ is given by
r→ = a→ + λ n→
r→ = λ(î + 2ĵ + 3k̂)
i.e. r→ = λ(î + 2ĵ + 3k̂)
where λ be any colar.

Que-13: Find the equation of the plane through (2, 3, -4) and (1, -1, 3) and parallel to x axis.

Sol: 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
a/0 = b/7−0 = 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.

Que-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

Sol: eqn. of given plane be
2 x – y + 3 z – 5 = 0
i.e. its vector eqn. be
(xî + yĵ + zk̂) (2î – ĵ + 3k̂) = 5
i.e. r→ (2î – ĵ + 3k̂) = 5
So the given plane be normal to
n→ = 2î – ĵ + 3k̂
Also the required line is ⊥ to plane (1).
∴required line is parallel to vector n→ .
Hence the equation of required line passing through the point having position vector
î – ĵ + 2k̂
and || to vector n→ is r→ = a→ + λ n→
i.e. r→ = î – ĵ + 2k̂ + λ(2î – ĵ + 3k̂)

Que-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

Sol: 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
a/−4−0 = b/0−4 = c/4+2
i.e. a/−4 = b/−4 = c/6
i.e. a/2 = b/2 = 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.

Que-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.

Sol: 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
a/8+9 = b/6−4 = c/−3−4
i.e. a/17 = b/2 = 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.

Que-17: Find the vector equation in the scalar product form of the plane through the point (4, 2, 4) and perpendicular to the planes  r→ (2î + 5ĵ + 4k̂) = -1 and
r→ (4î + 7ĵ + 6k̂) = -2 .

Sol: Let n→ be a vector normal to required plane. Also vector normal to given planes are
and
n1→ = 2î + 5ĵ + 4k̂
n2→ = 4î + 7ĵ + 6k̂
Now n1→ × n2→ be a vector normal to both n1→ and n→2
Since the required plane is ⊥ to two given planes.
∴ n→ is ⊥ to each of two vectors n1→ and n2→
î    ĵ    k̂
Thus n→ = n1→ × n→2 = | 2  5   4 |
4  7   6
= î(30 – 28) – ĵ (12 – 16) + k̂ (14 – 20)
= 2î + 4ĵ – 6k̂
∴required eqn. of plane be given by
(r→ – a→ ) n1→ × n2→ = 0
⇒ [r→ – (4î + 2ĵ + k̂)] (2î + 4ĵ – 6k̂) = 0
⇒ r→ (2î + 4ĵ – 6k̂)
= (4î + 2ĵ + 4k̂) (2î + 4ĵ – 6k̂)
= 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
a/30−28 = b/16−12 = c/14−20
i.e. a/2 = b/4 = c/−6
i.e. a/1 = b/2 = 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, r→ (î + 2ĵ – 3k̂) = -4 be the reqd. plane.

Que-18: Find the vector equation of the plane through the point (2î – ĵ – 4k̂) and plane parallel to the plane
r→ (2î + 5ĵ + 4k̂) = -1

Sol: eqn. of given plane be
r→ (4î – 12ĵ – 3k̂) – 7 = 0
Thus the eqn. of plane || to plane (1) be given by
r→ (4î – 12ĵ – 3k̂) = d
Since eqn. (2) passes throguh the point (2î – ĵ – 4k̂).
∴(2î – ĵ – 4k̂) (4î – 12ĵ – 3k̂) = d
⇒ 2(4) – 1(-12) – 4(-3) = d
⇒ 8 + 12 + 12 = d
⇒ d = 32
∴from (1) ; r→ (4î – 12ĵ – 3k̂) = 32 be the required vector eqn. of plane.
Aliter : Eqn. of plane in cartesian eqn. be
(xî + yĵ + zk̂) (4î – 12ĵ – 3k̂) – 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 (2î + ĵ – 4k̂) 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.

–: End of Angle Between Two Planes Class 12 OP Malhotra Exe-24C ISC Maths Ch-24  Solutions :–

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