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Mathematics Question on Three Dimensional Geometry

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x – 3y + 5z = 8. If the mirror image of the point
(2,−12,2)(2,−\frac{1}{2},2)
in the rotated plane is B( a, b, c),then

A

a8=b5=c−4\frac{a}{8}=\frac{b}{5}=\frac{c}{-4}

B

a4=b−5=c−2\frac{a}{4}=\frac{b}{-5}=\frac{c}{-2}

C

a8=b−5=c4\frac{a}{8}=\frac{b}{-5}=\frac{c}{4}

D

a4=b5=c2\frac{a}{4}=\frac{b}{5}=\frac{c}{2}

Answer

a8=b5=c−4\frac{a}{8}=\frac{b}{5}=\frac{c}{-4}

Explanation

Solution

The correct answer is (A) : a8=b5=c−4\frac{a}{8}=\frac{b}{5}=\frac{c}{-4}
Consider the equation of plane,
P : (2x + 3y + z + 20) + λ(x – 3y + 5z – 8) = 0
P : (2 + λ)x + (3 – 3λ)y + (1 + 5λ)z + (20 – 8λ) = 0
∵ Plane P is perpendicular to 2x + 3y + z + 20 = 0
So, 4 + 2λ + 9 – 9λ + 1 + 5λ = 0 ,vso λ=7
P :9x – 18y + 36z – 36 = 0
Or P :x – 2y + 4z = 4
If image of (2,−12,2)(2,−\frac{1}{2},2)
in plane P is (a, b, c) then
(a−2)1=(b+12)−2=(c−2)4\frac{(a−2)}{1}=\frac{(b+\frac{1}{2})}{−2}=\frac{(c−2)}{4 }
and (a+2)2−2(b−122)+4(c+22)=4\frac{(a+2)}{2}−2(\frac{b−\frac{1}{2}}{2})+4(\frac{c+2}{2})=4
Clearly
a=43,b=56anda=\frac{4}{3},b=\frac{5}{6} and c=−23c=−\frac{2}{3}
So, a :b : c = 8 : 5 : – 4