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

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

Question 1.
Find the distance from the point.
(i) P(-4, 3, 7) to the plane 2 x + 6 y – 9 z = 2
(ii) P(2, 1, -1) to the plane x – 2 y + 4 z = 9.
(iii) 2 \(\hat{i}\) – \(\hat{j}\) – 4 \(\hat{k}\) from the plane \(\vec{r}\) (4 \(\hat{i}\) – 12 \(\hat{j}\) – 3 \(\hat{k}\)) – 7 = 0
Answer:
(i) eqn. of given plane be 2 x + 6 y – 9 z = 2 …………….(1)
∴ required ⊥ distance from point P(-4,3,7) to palne (1)
= \(\frac{|2(-4) + 6 \times 3 – 9 \times 7-2|}{\sqrt{2^2+6^2+(-9)^2}}\)
= \(\frac{|-8+18-63-2|}{\sqrt{4+36+81}}\)
= \(\frac{|-55|}{\sqrt{121}}\)
= \(\frac{55}{11}\)
= 5 units

(ii) eqn. of given plane be x – 2 y + 4 z – 9 = 0 …………..(1)
∴ reqd. ⊥ distance from P (2, 1, -1) to plane (1)
= \(\frac{|2-2 \times 1+4(-1)-9|}{\sqrt{1^2+(-2)^2+4^2}}\) = \(\frac{|2-2-4-9|}{\sqrt{1+4+16}}\) = \(\frac{13}{\sqrt{21}}\) units

(iii) eqn. of given plane be \(\vec{r}[latex] (4 [latex]\hat{i}\) – 12 \(\hat{j}\) – 3 \(\hat{k}\)) – 7 = 0
∴ reqd. ⊥ distance from point P} whose P.V 2 \(\hat{i}\) – \(\hat{j}\) – 4 \(\hat{k}\) to plane (1)
= \(\frac{|(2 \hat{i}-\hat{j}-4 \hat{k}) (4 \hat{i}-12 \hat{j}-3 \hat{k})-7|}{\sqrt{4^2+(-12)^2+(-3)^2}}\) = \(\frac{|2(4)-1(-12)-4(-3)-7|}{\sqrt{16+144+9}}\)
= \(\frac{|8+12+12-7|}{13}\) = \(\frac{25}{13}\) units

Question 2.
Find the distance of the point
(i) (3, 3, 3) from the plane \(\vec{r}\) (5 \(\hat{i}\) + 2 \(\hat{j}\) – 7 \(\hat{k}\)) + 9 = 0
(ii) \(\hat{i}\) – 2 \(\hat{j}\) – 3 \(\hat{k}\) from the plane \(\vec{r}\) (2 \(\hat{i}\) + 5 \(\hat{j}\) – \(\hat{k}\)) = 0
Answer:
(i) We know that, the ⊥ distance of point with position vector \(\vec{a}\) from given plane \(\vec{r}\) \(\vec{n}\) – d = 0 is given by \(\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|}\)
Given eq̣n. of plane be \(\vec{r}\) (5 \(\hat{i}\) + 2 \(\hat{j}\) – 7 \(\hat{k}\)) + 9 = 0 ……………(1)
Here \(\vec{a}\) = 3 \(\hat{i}\) + 3 \(\hat{j}\) + 3 \(\hat{k}\) ; \(\vec{n}\) = 5 \(\hat{i}\) + 2 \(\hat{j}\) – 7 \(\hat{k}\) ; d = -9
∴ required ⊥ distance of point (3, 3, 3) from given plane (1)
= \(\frac{|(3 \hat{i}+3 \hat{j}+3 \hat{k}) (5 \hat{i}+2 \hat{j}-7 \hat{k})+9|}{\sqrt{5^2+2^2+(-7)^2}}\)
= \(\frac{|3(5)+3(2)+3(-7)+9|}{\sqrt{25+4+49}}\) = \(\frac{9}{\sqrt{78}}\)

(ii) eqn. of given plane be \(\vec{r}\) (2 \(\hat{i}\) + 5 \(\hat{j}\) – \(\hat{k}\)) = 4 ……………….(2)
∴ required distance of point \(\hat{i}\) – 2 \(\hat{j}\) – 3 \(\hat{k}\) from given plane (1)
= \(\frac{|(\hat{i}-2 \hat{j}-3 \hat{k}) \cdot(2 \hat{i}+5 \hat{j}-\hat{k})-4|}{\sqrt{2^2+5^2+(-1)^2}}\)
= \(\frac{|1(2)-2(5)-3(-1)-4|}{\sqrt{4+25+1}}\)
= \(\frac{|-9|}{\sqrt{30}}\) = \(\frac{9}{\sqrt{30}}\) units

Question 3.
Find the equation of the planes parallel to the plane x – 2 y + 2 z – 3 = 0 which is at a unit distance from the points (1, 2, 3).
Answer:
The eqn. of given plane be
x – 2 y + 2 z – 3 = 0 …………….(1)
Thus eqn. of plane parallel to plane (1) be given by
x – 2 y + 2 z + k = 0 ………………..(2)
Now plane (2) is at a unit distance from the point (1, 2, 3).
\(\frac{|1-2(2)+2(3)+k|}{\sqrt{1^2+(-2)^2+2^2}}\) = 1
⇒ \(\frac{|3+k|}{3}\) = 1
⇒ |3+k| = 3 ⇒ 3 + k = ± 3 ⇒ k = 0,-6
∴ from (2); x – 2 y + 2 z = 0 and x – 2 y + 2 z – 6 = 0 be the required eqns. of planes.

Question 4.
Find the distance between the parallel planes x + y – z + 4 = 0 and x + y – z + 5 = 0.
Answer:
Given eqns. of given planes are
x+y-z+4=0
x+y-z+5=0
∴ required distance between parallel planes =⊥ distance of any point on plane (1) from plane (2)
= \(\frac{|x+y-z+5|}{\sqrt{1^2+1^2+(-1)^2}}\) = \(\frac{|-4+5|}{\sqrt{1+1+1}}\) = \(\frac{1}{\sqrt{3}}\)
[using eqn. (1); x + y – z = -4 ]

Question 5.
Find the shortest distance between the planes 2 x – y + 3z – 4 = 0 and 6 x – 3 y + 9 z + 13 = 0.
Answer:
Let P(x₁, y₁, z₁) be any point on plane
2 x – y + 3 z – 4 = 0 i.e. 2 x₁ – y₁ + 3 z₁ – 4 = 0
Let d be the distance between given parallel planes.
∴ d =⊥ distance of P(x₁, y₁, z₁) from 6 x – 3 y + 9 z + 13 = 0
= \(\frac{|6 x_1-3 y_1+9 z_1+13|}{\sqrt{6^2+(-3)^2+9^2}}\)
= \(\frac{|3(2 x_1-y_1+3 z_1)+13|}{\sqrt{36+9+81}}\)
= \(\frac{|3(4)+13|}{\sqrt{126}}\)
= \(\frac{25}{\sqrt{126}}\) = \(\frac{25}{3 \sqrt{14}}\)

Question 6.
Show that the two points \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) are – 3 \(\hat{i}\) + \(\hat{k}\) equidistant from the plane \(\vec{r}\) (3 \(\hat{i}\) + 4 \(\hat{j}\) – 12 \(\hat{k}\)) + 13 = 0 and lie on opposite side of the plane.
Answer:
eqn. of given plane be
\(\vec{r}\) (3 \(\hat{i}\) + 4 \(\hat{j}\) – 12 \(\hat{k}\)) + 13 = 0
⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (3 \(\hat{i}\) + 4 \(\hat{j}\) – 12 \(\hat{k}\)) + 13 = 0
⇒ 3 x + 4 y – 12 z + 13 = 0
Given \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) and -3 \(\hat{i}\) + \(\hat{k}\) are P.V’s of points (1, 1, 1) and Q(-3, 0, 1).
∴ ⊥ distance of P(1, 1, 1) from plane (1) = \(\frac{|3+4-12+13|}{\sqrt{3^2+4^2+(-12)^2}}\) = \(\frac{8}{13}\)
∴ ⊥ distance of Q(-3, 0, 1) from plane (1) = \(\frac{|3 \times(-3) + 4 \times – 12 \times 1+13|}{\sqrt{3^2+4^2+(-12)^2}}\)
= \(\frac{8}{13}\)
Thus both points P and Q are equidistant from given plane (1). putting P(1, 1, 1) in L.H.S. of eqn. (1) = 3 + 4 – 12 + 13 = 8 > 0
and putting Q(-3, 0, 1) in L.H.S. of eqn. (1) = 3(-3) + 4 × 0 – 12 × 1 + 13
= -21 + 13 = -8 < 0
Thus both points P and Q lies on opposite side of given plane (1).

Question 7.
Find the equation of the plane through the point (3, 4, -1) which is parallel to the plane \(\vec{r}\) (3 \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\)) + 7 = 0. Also, find the distance between the two planes.
Answer:
eqn. of given plane in cartesian form be given by
(x \(\hat{i}\) + y \(\hat{j}\) + 7 \(\hat{k}\)) (2 \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\)) + 7 = 0
⇒ 2 x – 3 y + 5 z + 7 = 0
⇒ Thus, the eqn. of plane parallel to plane (1) be given by
2 x – 3 y + 5 z + k = 0
Now plane (2) passes through the given point (3, 4, -1).
∴ 2 × 3 – 3 × 4 + 5 × (-1) + k = 0 ⇒ k = 11
putting the value of k in eqn. (2); we get
2 x – 3 y + 5 z + 11 = 0
Let P(x₁, y₁, z₁) be any point on plane (1)
∴ 2 x₁ – 3 y₁ + 5 z₁ + 7 = 0
Thus reqd. distance between planes = Length of ⊥ from P(x₁, y₁, z₁) on plane (4)
= \(\frac{\left|2 x_1-3 y_1+5 z_1+11\right|}{\sqrt{2^2+(-3)^2+5^2}}\)
= \(\frac{|-7+11|}{\sqrt{4+9+25}}\)
= \(\frac{4}{\sqrt{38}}\) units

Question 8.
A plane is at a constant distance p from the origin and meets the coordinate axes in A, B, C. Show that the locus of the centroid of triangle A B C is x^-2 + y² + z^-2 = 9 p^-2.
Answer:
Let the eqn. of plane be \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
where the plane (1) meets x-axis at A(0, 0, 0) ; B(0, b, 0) and C(0, 0, c).
∴ Centroid of ∆ABC be (\(\frac{a+0+0}{3}\), \(\frac{0+b+0}{3}\), \(\frac{0+0+c}{3}\)) i.e. (\(\frac{a}{3}\), \(\frac{b}{3}\), \(\frac{c}{3}\))
Let the centroid of ∆ABC be Q(α, β, γ).
∴ α = \(\frac{a}{3}\) ; β = \(\frac{b}{3}\) and γ = \(\frac{c}{3}\)
⇒ a = 3 α ; b = 3β and c = 3γ
given p =⊥ distance from (0,0,0) to plane (1)
p = \(\frac{\left|\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1\right|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\)
= \(\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\)
⇒ \(\sqrt{\frac{1}{a^2} + \frac{1}{b^2}+\frac{1}{c^2}}\)
= \(\frac{1}{p}\) ; on squaring both sides; we have
\(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) + \(\frac{1}{c^2}\)
= \(\frac{1}{p^2}\)
⇒ \(\frac{1}{9\alpha^2}\) + \(\frac{1}{9\beta^2}\) + \(\frac{1}{9\gamma^2}\)
= \(\frac{1}{p^2}\)
⇒ \(\frac{1}{\alpha^2}\) + \(\frac{1}{\beta^2}\) + \(\frac{1}{\gamma^2}\)
= \(\frac{9}{p^2}\)
∴ Locus of centroid Q(α, β , γ) be given by
\(\frac{1}{x^2}\) + \(\frac{1}{y^2}\) + \(\frac{1}{z^2}\) = \(\frac{9}{p^2}\)
⇒ x^-2 + y^-2 + z^-2 = 9 p^-2

Question 9.
Find the distance of the point (2, 3, 5) from the x y-plane.
Answer:
Equation of x y plane be z = 0.
⊥ distance of point (2, 3, 5) from plane z = 0 = \(\frac{|5-0|}{\sqrt{1^2}}\) = 5

Question 10.
Find the equation of the plane mid-parallel to the planes
(i) 2 x – 2 y + z + 3 = 0 and 2 x – 2 y + z + 9 = 0
(ii) 2 x – 3 y + 6 z + 21 = 0 and 2 x – 3 y + 6 z – 14 = 0.
Answer:
(i) Equations of given planes are
and
2 x – 2 y + z + 3 = 0
2 x – 2 y + z + 9 = 0
Since both planes are parallel.
Let the eqn. of plane parallel to both given planes be
2 x – 2 y + z + k = 0
Let P(x₁, y₂, z₁) be any point on plane (1).
∴ 2 x₁ – 2 y ₁ + z₁ + 3 = 0
∴ d₁ = distance between planes (1) and (3)
=⊥ distance of point P (x₁, y₁, z₁) from plane (3)
= \(\frac{\left|2 x_1-2 y_1+z_1+k\right|}{\sqrt{2^2+(-2)^2+1^2}}\) = \(\frac{|-3+k|}{3}\)
d₂ = distance between planes (2) and (3)
= ⊥ distance of point P (x₁, y₁, z₁) from plane (3)
= \(\frac{|2 x_1-2 y_1+z_1+k|}{\sqrt{4+4+1}}\)
= \(\frac{|-9+k|}{3}[latex]
[∵ 2 x₁ – 2 y₁ + 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 – 6 k + 9 = k² – 18 k + 81
⇒ 12 k = 72 ⇒ k = 6
putting the value of k in eqn. (3); we get
2 x – 2 y + z + 6 = 0 be the required eqn. of plane.

(ii) Eqns. of given planes are
2 x – 3 y + 6 z + 21 = 0
2 x – 3 y + 6 z – 14 = 0
Thus eqn. of plane parallel to given planes (1) and (2) be given by
2 x – 3 y + 6 z + k = 0
Let P(x₁, y₁, z₁) be any point on plane (3)
2 x₁ – 3 y₁ + 6 z₁ + k = 0
Since the plane (3) is mide parallel to plane (1) and (2)
∴ ⊥ distance from P(x₁, y₁, z₁) to plane (1)
=⊥ distance from P(x₁, y₁, z₁) to plane (2)
[latex]\frac{|2 x_1-3 y_1+6 z_1+21|}{\sqrt{2^2+(-3)^2+36}}\)
= \(\frac{|2 x_1-3 y_1+6 z_1-14|}{\sqrt{2^2+(-3)^2+6^2}}\)
⇒ \(\frac{|-k+21|}{7}\) = \(\frac{|-k-14|}{7}\)
⇒(-k + 21) = ±(k + 14)
i.e. -k + 21 = k + 14 or -k + 21 = -k – 14
⇒ 2 k = 7 or 21 = -14 , which is false
⇒ k = \(\frac{7}{2}\)
putting the value of k in eqn. (3); we get
2 x – 3 y + 6 z + \(\frac{7}{2}\) = 0
⇒ 4 x – 6 y + 12 z + 7 = 0
be the reqd. plane.

Question 11.
Show that the line whose vector equation is \(\vec{r}\) = 2 \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\) + λ(\(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\)) parallel to the plane whose vector equation is \(\vec{r} ([latex]\hat{i}\) + 5 \(\hat{j}\) + \(\hat{k}\)) = 5. Also, find the distance between them.
Answer:
The given eqn. of line in vector form be
\(\vec{r}\) = 2 \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\) + λ(\(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\))
Thus its cartesian eqn. be
\(\frac{x-2}{1}\) = \(\frac{y+2}{-1}\) = \(\frac{z-3}{4}\)
Also, given vector eqn. of plane be \(\vec{r}\) (\(\hat{i}\) + 5 \(\hat{j}\) + \(\hat{k}\)) = 5
So its cartesian eqn. be
x + 5 y + z = 5
d ratios of normal to plane (1) are < 1, 5, 1 >
Now line is parallel to plane (2)
if 1(1) – 1(5) + 4(1) = 0 and (2, -2, 3) does not lies on plane (2)
if 0 = 0, which is true and 2 + (-2) + 3 = 5
⇒-5 = 5
Thus, line (1) is parallel to plane (2).
∴ required distance between line (4) and plane (2) = distance of point (2, -2, 3) from plane (2)
=\(\frac{|2+5(-2)+3-5|}{\sqrt{1^2+5^2+1^2}}\)
= \(\frac{10}{\sqrt{27}}\) units

Question 12.
From the point (1, 2, 4) a perpendicular is drawn on the plane 2 x + y – 2 z + 3 = 0. Find the equation, the length and the coordinates of the foot of the perpendicular.
Answer:
Given eqn. of plane be 2 x + y – 2 z + 3 = 0
Thus eqn. of ⊥ PM which passes through the point P(1, 2, 4) and having direction ratios < 2, 1, -2 > is given by
\(\frac{x-1}{2}\) = \(\frac{y-2}{1}\)
= \(\frac{z-4}{-2}\) = t (say)
So any point on line ( 2) be given by (2 t + 1, t + 2, -2 t + 4).
This point be the required foot of ⊥ i.e. M if it lies on plane (1).
∴ 2(2 t + 1) + t + 2 – 2(-2 t + 4) + 3 = 0
⇒ 4 t + 2 + t + 2 + 4 t – 8 + 3 = 0
⇒ 9 t = 1 ⇒ t = \(\frac{1}{9}\)

∴ required coordinates of foot of ⊥ be given by M(\(\frac{11}{9}\), \(\frac{19}{9}\), \(\frac{34}{9}\))
∴|PM| = length of ⊥ from P(1, 2, 4) to plane (1)
\(\frac{|2 \times 1+2-2 \times 4+3|}{\sqrt{2^2+1^2+(-2)^2}}\) = \(\frac{1}{3}\) units.

Question 13.
Find the image of the point P(1, 3, 4) in the plane 2 x – y + z + 3 = 0. Alternatively. Find the image of the point having position vector \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) in the plane \(\vec{r}\)(2\hat{i}[/latex] – \(\hat{j}\) + \(\hat{k}\)) + 3 = 0.
Answer:
Given eqn. of plane be 2 x – y + z + 3 = 0
Let M be the foot of ⊥ drawn from given point P(1, 3, 4).
∴ eqn. of line PM. i.e. passes through the point P(1, 3, 4) and normal to given plane be given by
\(\frac{x-1}{2}\) = \(\frac{y-3}{-1}\) = \(\frac{z-4}{1}\) = t (say)
Any point on line (2) be M(2 t + 1, -t + 3, t + 4) and this point be the foot of ⊥ M, if M(2 t + 1, -t + 3, t + 4) lies on plane (2).
∴ 2(2 t + 1) – (-t + 3) + t + 4 + 3 = 0
⇒ 6 t + 6 = 0 ⇒ t = -1
∴ Coordinates of point M are (-1, 4, 3).
Let Q(α, β , γ) be the image of point P if M be the mid point of the segment PQ

∴(\(\frac{\alpha+1}{2}\), \(\frac{\beta+3}{2}\), \(\frac{\gamma+4}{2}\)) = (-1, 4, 3)
i.e. \(\frac{\alpha+1}{2}\) = -1
⇒ α = -3 ; \(\frac{\beta+3}{2}\) =4
⇒ β = 5 and \(\frac{\gamma+4}{2}\) = 3
⇒ γ = 2
Thus the required image of point (1, 3, 4) be (-3, 5, 2).

Question 14.
Find the equation of the planes which are perpendicular to each of the planes 3 x – y + z = 0 and x + 5 y + 3 z = 0 and at a distance of \(\sqrt{6}\) units from the origin.
Answer:
Let the eqn. of required plane be a x + b y + c z + d = 0 where < a, b, c > be the direction ratios of normal to plane (1).
The eqns. of given planes are
3 x – y + z = 0
x + 5 y + 3 z = 0
and
Since the plane (1) is ⊥ to both given planes.
∴ 3 a – b + c = 0
a + 5 b + 3 c = 0
on solving (4) and (5) simultaneously
∴ \(\frac{a}{-8}\) = \(\frac{b}{1-9}\) = \(\frac{c}{15+1}\)
⇒ \(\frac{a}{1}\) = \(\frac{b}{1}\) = \(\frac{c}{-2}\) = 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 + \(\frac{d}{k}\) = 0 ⇒ x + y – 2 z + d = 0
Also it is given that ⊥ distance from (0,0,0) to given plane (6) = \(\sqrt{6}\)
\(\frac{|0+0-2 \times 0+d^{\prime}|}{\sqrt{1^2+1^2+(-2)^2}}\)
= \(\sqrt{6}\) ⇒ d = ± 6
∴ from eqn. (6) ; we have x + y – 2 z ± 6 = 0 be the reqd. equations of planes.

Question 15.
Find the locus of a point whose distance from the origin is three times its distance from the plane 2 x – y + 2 z = 3.
Answer:
Let P (x, y, z) be any point whose locus is to be find out.
It is given that distance of point P from origin O = 3 × distance of point P(x, y, z) from plane 2 x – y + 2 z – 3 = 0
\(\sqrt{(x-0)^2+(y-0)^2+(z-0)^2}\) = \(\frac{3|2 x-y+2 z-3|}{\sqrt{2^2+(-1)^2+2^2}}\)
⇒ \(\sqrt{x^2+y^2+z^2}\) = \(\frac{3(2 x-y+2 z-3)}{3}\)
On squaring both sides ; we have
x² + y² + z² = 4 x² + y² + 4 z² – 4 x y + 9 – 12 z + 2(2 x – y)(2 z – 3)
⇒ x² + y² + z² = 4 x² + y² + 4 z² – 4 x y + 9 – 12 z + 8 x z – 12 x – 4 y z + 6 y
⇒ 3 x² + 3 z² – 4 x y + 8 x z – 4 y z + 6 y – 12 z + 9 = 0
which is the required eqn. of locus.