Angle Between Two Planes Class 12 OP Malhotra Exe-24E ISC Maths Solutions Ch-24 The Plane. In this article you would learn how to determine a plane under given conditions and equations of some particular planes practice questions / example / problems with step by step solutions / answer. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

The Plane Class 12 OP Malhotra Exe-24F 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(f)	Determination of Plane Under Given Conditions and Equations of Some Particular Planes

Determination of Plane Under Given Conditions and Equations of Some Particular Planes

The Plane Class 12 OP Malhotra Exe-24F Solutions

Que-1: Find the equation of the plane which
(i) passes through P(3, -2, 4) and is perpendicular to a line whose direction ratios are 2, 2, -3;
(ii) passes through P(2, -3, 5) and has the line joining A(1, -3, -5) and B(2, 2, 3) as a normal;
(iii) bisects the line joining (5, -2, 6) and (7, 2, 0) at right angles;
(iv) passes through P(1, -2, -4) and is parallel to the plane 7 x – 4 y + 6 z + 2 = 0;
(v) passes through the points (-8, 6, 0),(0, 12, 0), and (-10, 0, -9);
(vi) passes through the points (6, 2, 3),(3, 3, -2),(2, -2, -1);
(vii) passes through the y-axis and the point (4, 2, -3).

Sol: (i) Given direction ratios of normal to plane are ∴ < 2, 2, 3 >.
Thus, eqn. of plane through the point (3, -2, 4) and having direction ratios of normal to plane are
∴ < 2, 2, -3 > is given by 2(x – 3) + 2(y + 2) – 3(z – 4) = 0
⇒ 2 x + 2 y – 3 z + 10 = 0 be the required eqn. of plane.

(ii) ∴ directon ratios of normal to required plane are
∴< 2 – 1, 2 + 3, 3 + 5 >
i.e. ∴ < 1, 5, 8 > Thus required eqn. of plane through the point P(2, -3, 5) is given by
1(x – 2) + 5(y + 3) + 8(z – 5) = 0
⇒ x + 5 y + 8 z – 27 = 0

(iii) D ratios of line AB are
< 7 – 5, 2 + 2, 0 – 6 > i.e.
< 2, 4, -6 > Thus A B be normal to required plane. Also the required plane passes through the mid point of AB
i.e. 5+7/2, −2+2/2, 6+0/2 i.e. (6, 0, 3).
∴ eqn. of plane through the point (6, 0, 3) is given by
1(x – 6) + 2(y – 0) – 3(z – 3) = 0
⇒ x + 2 y – 3 z + 3 = 0

(iv) eqn. of plane parallel to given plane
7 x – 4 y + 6 z + 2 = 0 be given by 7 x – 4 y + 6 z + k = 0
Since eqn. (1) passes through the point P(1,-2,-4).
∴ 7 × 1 – 4 × (-2) + 6(-4) + k = 0
⇒ 7 + 8 – 24 + k = 0 ⇒ k = 9
putting the value of k in eqn. (1) ; we have
7 x – 4 y + 6 z + 9 = 0 be the reqd. eqn. of plane.

(v) Let the eqn. of plane through the point (-8, 6, 0) be given by
a(x + 8) + b(y – 6) + c(z – 0) = 0
Now eqn. (1) passes through the point (0, 12, 0).
∴ 8 a + 6 b + 0 c = 0
⇒ 4 a + 3 b + 0 c = 0
Also, plane (1) passes through the point (-10, 0, -9)
∴ a(-10 + 8) + b(0 – 6) + c(-9 – 0) = 0
⇒ -2 a – 6 b – 9 c = 0
on solving eqn. (2) and eqn. (3); we have
a−2/7−0 = b/0+36
= c/−24+6
⇒ a/−27
= b/36 = c/−18
⇒ a/3 = b/−4
= c/2 = k (say) where k ≠ 0
∴ a = 3 k ; b = -4 k and c = 2 k
putting the values of a, b and c in eqn. (1); we have
3 k(x + 8) – 4 k(y – 6) + 2 k(z) = 0
⇒ 3 x – 4 y + 2 z + 48 = 0

(vi) Let the eqn. of plane through the point (6, 2, 3) is given by
a(x – 6) + b(y – 2) + c(z – 3) = 0
where ∴ < a, b, c > are the direction rates of normal to plane (1). eqn. (1) passes through the point (3, 3, -2).
a(3 – 6) + b(3 – 2) + c(-2 – 3) = 0
⇒ -3 a + b – 5 c = 0
Also, plane (1) passes through the point (2, -2, -1).
a(2 – 6) + b(-2 – 2) + c(-1 – 3) = 0
a + b + c = 0
⇒ -4 a – 4 b – 4 c = 0
By cross-multiplication method; we have
a/1+5 = b/−5+3
= c/−3−1
⇒ a/6 = b/−2
= c/−4
⇒ a/3 = b/−1 = c−2 = k (say); k ≠ 0
∴ a = 3 k ; b = -k ; c = -2 k
putting the values of a, b and c in eqn. (1); we have
3 k(x – 6) – k(y – 2) – 2 k(z – 3) = 0
⇒ 3 x – y – 2 z – 10 = 0 be the required eqn. of plane

(vii) since y-axis be the line of intersection of x o y plane (i.e. z = 0 ) and y o z plane ( i.e. x = 0 ) is given by
z + k x = 0
Since plane (1) passes through the point (4, 2, -3)
∴ -3 + 4 k = 0
⇒ k = 3/4
∴ from (1); z + 3x/4 = 0
⇒ 3 x + 4 z = 0 be the reqd. plane.

Que-2: Find the equation of the plane
(i) parallel to the plane 4 x – 4 y + 7 z – 3 = 0 and distant 4 units from the point (4, 1, -2);
(ii) which passes through the point (3, -2, 4) and is perpendicular to each of the planes 7 x – 3 y + z – 5 = 0 and 4 x – y – z + 9 = 0.
(iii) perpendicular to each of the planes 3 x – y + z = 0 and x + 5 y + 3 z = 0 and is at a distance of √6 from the origin ;
(iv) through (2, 2, 2) and (0, -2, 0) and perpendicular to the plane x – 2 y + 3 z – 7 = 0.

Sol: (i) eqn. of given plane be
4 x – 4 y + 7 z – 3 = 0 …………………….. (1)
Thus eqn. of plane parallel to plane (1) be given by
4 x – 4 y + 7 z + k = 0 …………………….. (2)
also it is given that ⊥ distance of point (4, 1, -2) from given plane (2) = 4 units
|4×4−4×1+7×(−2)+k|/√4²+(−4)²+7² = 4
⇒ |16−4−14+k|/√16+16+49 = 4
⇒ |k−2|/9 = 4
⇒ |k – 2| = 36 ⇒ k – 2 = ± 36
⇒ k = ± 36 + 2
⇒ k = 38, – 34
putting the values of k in eqn. (2); we have
4 x – 4 y + 7 z + 38 = 0 and 4 x – 4 y + 7 z – 34 = 0 be the required eqns. of planes.

(ii) The equations of given planes are ;
7 x – 3 y + z – 5 = 0 …………………….. (1)
4 x – y – z + 9 = 0 …………………….. (2)
Thus the eqn. of plane through the point (3, -2, 4) is given by
a(x – 3) + b(y + 2) + c(z – 4) = 0 …………………….. (3)
where < a, b, c > are the direction ratios of normal to plane (3).
Since the required plane (3) is ⊥ to plane (1) and (2).
∴ 7 a – 3 b + c = 0 …………………….. (4)
4 a – b – c = 0 …………………….. (5)
on solving eqn. (4) and (5) simultaneously using cross multiplication method, we have
a/3+1 = b/4+7
= c/−7+12
i.e. a/4 = b/11 = c/5 = k (say)
⇒ a = 4 k ; b = 11 k and c = 5 k
putting the values of a, b, c in eqn. (3); we have
4 k(x – 3) + 11 k(y + 2) + 5 k(z – 4) = 0
⇒ 4 x + 11 y + 5 z – 10 = 0 be the reqd. plane.

(iii) Let the eqn. of required plane be
a x + b y + c z + d = 0 …………………….. (1)
where < a, b, c > be the direction ratios of normal to plane (1).
The eqns. of given planes are
3 x – y + z = 0 …………………….. (2)
x + 5 y + 3 z = 0 …………………….. (3)
and
hiven planes.
Since the plane (1) is ⊥ to both given planes.
∴ 3 a – b + c = 0 …………………….. (4)
a + 5 b + 3 c = 0 …………………….. (5)
on solving (4) and (5) simultaneously
∴ a/−8 = b/1−9 = c/15+1
⇒ a/1 = b/1
= c/−2 = k (say) ; where k ≠ 0
∴ a = k ; b = k ; c = -2 k
putting the values of a, b, c in eqn. (1); we have
x + y – 2 z + d/k = 0 …………………….. (6)
⇒ x + y – 2 z + d = 0
Also it is given that ⊥ distance from (0, 0, 0) to given plane (6) = √6
∣0+0−2×0+d′∣/√1²+1²+(−2)²
= √6
⇒ d = ± 6 .
∴ from eqn. (6); we have x + y – 2 z ± 6 = 0 be the reqd. equations of planes.

(iv) Let the eqn. of plane through the point (2, 2, 2) is given by
a(x – 2) + b(y – 2) + c(z – 2) = 0 …………………….. (1)
plane (1) passes through the point (0, -2, 0).
∴ a(0 – 2) + b(-2 – 2) + c(0 – 2) = 0
⇒ a + 2 b + c = 0 …………………….. (2)
⇒ -2 a – 4 b – 2 c = 0
Since eqn. (1) is ⊥ to plane
x – 2 y + 3 z – 7 = 0
a – 2 b + 3 c = 0 …………………….. (3)
a/6+2 = b/1−3
= c/−2−2
i.e. a/8 = b/−2 = c/−4
i.e. a/4 = b/−1 = c/−2 = k
∴ a = 4 k ; b = -k ; c = -2 k ; k ≠ 0
putting the values of a, b and c in eqn. (1); we have
4 k(x – 2) – k(y – 2) – 2 k(z – 2) = 0
⇒ 4 x – y – 2 z – 2 = 0 be the required eqn. of plane.

Que-3: Find the equation of the plane which contains the line of intersection of the planes x + 2 y + 3 z – 4 = 0 and 2 x + y – z + 5 = 0 and is perpendicular to the plane 5 x + 3 y – 6 z + 8 = 0.

Sol: Given eqns. of planes are ; and
x + 2 y + 3 z – 4 = 0 …………………….. (1)
2 x + y – z + 5 = 0 …………………….. (2)
Thus the eqn. of any plane through the line of intersection of given planes be
(x + 2 y + 3 z – 4) + k(2 x + y – z + 5) = 0
(1 + 2 k) x + (2 + k) y + (3 – k) z – 4 + 5 k = 0 …………………….. (3)
Now plane (3) is ⊥ to given plane
5 x + 3 y – 6 z + 8 = 0
∴ (1 + 2 k) 5 + ( 2 + k) 3 + (3 – k)(-6) = 0 …………………….. (4)
⇒ 5 + 10 k + 6 + 3 k – 18 + 6 k = 0
⇒ 19 k – 7 = 0
⇒ k = 7/19
putting the value of k = 719 in eqn. (3) ; we get
(1 + 14/19) x + (2 + 7/19) y + (3 – 7/19) z – 4 + 35/19 = 0
⇒ 33 x + 45 y + 50 z – 41 = 0 be the reqd. plane.

Que-4: Find the equation of the plane through the intersection of the planes x + y + z = 1 and 2 x + 3 y – z + 4 = 0 and parallel to the x-axis.

Sol: The eqn. of any palen through the line if intersection of two given planes x + y + z – 1 = 0 and 2 x + 3 y – z + 4 = 0 is given by
(x + y + z – 1) + k(2 x + 3 y – z + 4) = 0
⇒ (1 + 2 k) x + (1 + 3 k) y + (1 – k) z – 1 + 4 k = 0 …………………….. (1)
Now plane (1) is parallel to x-axis whose direction ratios are < 1, 0, 0 >
∴ Normal to plane (1) is ⊥ to x-axis.
Thus, (1 + 2 k) 1 + (1 + 3 k) 0 + (1 – k) 0 = 0
⇒ 2 k = -1
⇒ k = −1/2
∴ from (1); we have
−1/2 y + 3/2 z – 1 – 2 = 0
⇒ -y + 3 z – 6 = 0
⇒ y – 3 z + 6 = 0 be the required eqn. of plane.

Que-5: (i) Find the equation of the plane through (2, 3, -4) and (1, -1, 3) and parallel to the x-axis.
(ii) Find the equation of the plane passing through the points (2, 3, 1) and (4, -5, 3) and parallel to the x-axis.

Sol: (i) The eqn. of any plane through the point (2, 3, -4) is given by
a(x – 2) + b(y – 3) + c(z + 4) = 0 …………………….. (1)
where < a, b, c > are the direction ratios of normal to plane (1).
Also paine (1) passes through the point (1, -1, 3).
∴ a(1 – 2) + b(-1 – 3) + c(3 + 4) = 0
⇒ -a – 4 b + 7 c = 0 ⇒ a + 4 b – 7 c = 0 …………………….. (2)
Also plane (1) is parallel to x-axis.
∴ Normal to plane (1) is ⊥ to x-axis.
∴ a + 0 b + 0 c = 0 …………………….. (3)
on solving eqn. (2) and eqn. (3) simultaneously by cross-multiplication method, we have
a/0−0 = b/−7−0 = c/0−4
⇒ a/0 = b/−7 = c/−4 = k (say)
∴ a = 0 ; b = -7 k ; c = -4 k ; k ≠ 0
∴ from (1); we have
0(x – 2) – 7 k (y – 3) – 4 k(z + 4) = 0
⇒ -7 y + 21 – 4 z – 16 = 0
⇒ -7 y – 4 z + 5 = 0
⇒ 7 y + 4 z – 5 = 0 which is the required plane.

(ii) eqn. of any plane through the point (2, 3, 1) is given by
a(x – 2) + b(y – 3) + c(z – 1) = 0 …………………….. (1)
where < a, b, c > be the d ratios of normal to plane (1).
Now plane (1) passes through the point (4, -5, 3).
a(4 – 2) + b(-5 – 3) + c(3 – 1) = 0
⇒ 2 a – 8 b + 2 c = 0
⇒ a – 4 b + c = 0 …………………….. (2)
Since plane (1) is parallel to x-axis.
∴ Normal to plane (1) is ⊥ to x-axis whose direction ratios are < 1, 0, 0 > i.e. a + 0 b + 0 c = 0 …………………….. (3)
on solving eqn. (2) and eqn. (3) simultaneously
a/0 = b/1−0 = c/0+4 = k (say); k ≠ 0
⇒ a = 0 k = 0 ; b = k ; c = 4 k
putting the values of a, b and c in eqn. (1); we get
0(x – 2) + k(y – 3) + 4 k(z – 1) = 0
⇒ y + 4 z – 7 = 0
which is the required eqn. of plane.

Que-6: Find the equation of a plane which is perpendicular to the plane 2 x – 3 y + 6 z + 8 = 0 and passes through the intersection of the planes x + 2 y + 3 z – 4 = 0 and 2 x – y – z + 5 = 0.

Sol: eqns. of given planes are ;
x + 2 y + 3 z – 4 = 0 …………………….. (1)
2 x – y – z + 5 = 0  …………………….. (2)
and
Thus the eqn. of any plane through the line of intersection of given planes is given by
x + 2 y + 3 z – 4 + k(2 x – y – z + 5) = 0
⇒ (1 + 2 k) x + (2 – k) y + (3 – k) z – 4 + 5 k = 0 …………………….. (1)
Now plane (1) is normal to plane 2 x – 3 y + 6 z + 8 = 0
∴ 2(1 + 2 k) – 3(2 – k) + 6(3 – k) = 0
⇒ 2 + 4 k – 6 + 3 k + 18 – 6 k = 0
⇒ k + 14 = 0
⇒ k = -14
putting the value of k in eqn. (1); we have
-27 x + 16 y + 17 z – 74
⇒ 27 x – 16 y – 17 z + 74 = 0
⇒ 27, which is the reqd. plane.

Que-7: (i) Find the equation of the plane passing through A(-1, 1, 1) and B(1, 1, 1) and perpendicular to the plane x – 2 y + 2 z = 3.
(ii) Also, find the distance of the point A from the plane x – 2 y + 2 z = 3.

Sol: (i) Any plane through the point A(-1, 1, 1) be given by
a(x + 1) + b(y – 1) + c(z – 1) = 0 …………………….. (1)
where < a, b, c > are the direction ratios of normal to plane (1).
The point B(1, -1, 1) lies an eqn. (1); we have
a(1 + 1) + b(-1 – 1) + c(1 – 1) = 0
⇒ 2 a – 2 b + 0 c = 0
⇒ a – b + 0 c = 0 …………………….. (2)
Now plane (1) ⊥ to given plane
x – 2 y + 2 z = 3
a – 2 b + 2 c = 0 …………………….. (3)
∴ on solving eqn. (2) and eqn. (3) simultaneously by cross multiplication method, we have
a/−2−0 = b/0−2 = c/−2+1
i.e. a/−2 = b/−2 = c/−1
i.e. a/2 = b/2 = c/1 = k (say) ; k ≠ 0
∴ a = 2 k ; b = 2 k and c = k
putting the value of a, b and c in eqn. (1); we have
2 k(x + 1) + 2 k(y – 1) + k(z – 1) = 0
⇒ 2 x + 2 y + z – 1 = 0 be the reqd. plane

(ii) Required distance of A(-1, 1, 1) from x – 2 y + 2 z – 3 = 0
= |−1−2×1+2×1−3|/√1²+(−2)²+2²
= |−1−2+2−3|/3 = 4/3 units

Que-8: A plane meets the plane x = 0, where x = 0, 2 y – 3 z = 5, and the plane z = 0 where z = 0, 7 x + 4 y = 10. Find the equation to the plane.

Sol: The eqn. of any plane through the line of intersection of planes x = 0 and 2 y – 3 z – 5 = 0 be given by
2 y – 3 z – 5 + k x = 0 …………………….. (1)
Now plane (1) meets the planes z = 0 and 7 x + 4 y = 10
∴ from (1); 2 y – 5 + k (10−4y/7) = 0
i.e. 14 y – 35 + 10 k – 4 k y = 0
i.e. (14 – 4 k) y + 10 k – 35 = 0
∴ 14 – 4 k = 0 and 10 k – 35 = 0
i.e. k = 7/2 and k = 7/2
∴ from (1); 2 y – 3 z – 5 + 7/2 x = 0
⇒ 4 y – 6 z – 10 + 7 x = 0
which is the required eqn. of plane.

Que-9: Prove that the plane 2 x + y – 3 z + 5 = 0, 5 x – 7 y + 2 z + 3 = 0, 5 and x + 10 y – 11 z + 12 = 0 have a line in common.

Sol: The eqn. of plane through the line of intersection of first two planes is given by
2 x + y – 3 z + 5 + k(5 x – 7 y + 2 z + 3) = 0
⇒ (2 + 5 k) x + (1 – 7 k) y + (-3 + 2 k) z + 5 + 3 k = 0 …………………….. (1)
Now plane (1) is identical to given plane
if
x + 10 y – 11 z + 12 = 0
2−15k/1 = 1−7k/10
= −3−12k/−11 = 5+3k/12 …………………….. (2)
From first two fractions ; 20 + 50 k = 1 – 7 k
⇒ 57 k = -19 ⇒ k = −1/3
putting k = −1/3 in last two fractions ; we have
−3−(2/3)/−11 = 5−1/12
⇒ 1/3 = 1/3 which is true.
Thus the given three planes have same line of intersection.

Que-10: Find the equation of the plane passing through the intersection of the plane.
r→ (î + ĵ + k̂) = 1 and r→ (2î + 3ĵ – k̂) + 4 = 0 and parallel to x-axis.

Sol: The eqn. of plane passing through the line of intersection of given planes
r→ (î +ĵ + k̂) = 1 and r→ (2î + 3ĵ – k̂) + 4 = 0
be given by r→ (î + ĵ + k̂) – 1] + λ[r⃗ (2î + 3ĵ – k̂) + 4] = 0
⇒ r→ [(1 + 2λ)î + (1 + 3λ)ĵ + (1 – λ) k̂] + 4λ – 1 = 0 …………………….. (1)
Since eqn. (1) is parallel to x-axis.
∴ Normal to plane is ⊥ to x-axis.
∴(1 + 2 λ) 1 + (1 + 3λ) 0 + (1 – λ) 0 = 0
⇒ 1 + 2λ = 0
⇒ λ = −1/2
∴ from (1) ; we have
r⃗ [−ĵ/2 + 3/2 k̂] – 3 = 0

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

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