Let the plane P: 4 x-y+z=10 be rotated by an angle π/2 about its line of intersection with the plane x+y-z=4. If α is the distance of the point (2,3,-4) from the new position of the plane P, then 35 α is equal to :

Solution

Maths sir · 4 steps

Every plane through that line is P + λ P'; pick the λ that makes it perpendicular to P.

1. Family of planes. Rotating P about its line of intersection with x+y-z=4 keeps the plane passing through that line. So the new plane has the form (4x-y+z-10)+λ(x+y-z-4)=0, with normal (4+λ, λ-1, 1-λ).
2. Rotation by 90°. A rotation of π/2 makes the new plane perpendicular to P. So its normal is perpendicular to P's normal (4,-1,1): 4(4+λ)-(λ-1)+(1-λ)=18+2λ=0, which gives λ=-9.
3. New plane. Put λ=-9: -5x-10y+10z+26=0, that is, 5x+10y-10z-26=0. Its normal has length √(25+100+100)=15.

   α=(|5(2)+10(3)-10(-4)-26|)/15=(|10+30+40-26|)/15=54/15=18/5

4. So. 35α=35×18/5=126, which is option (a).

The idea to keep. Rotating a plane about a line means the new plane still passes through that line, so write it as P₁+λ P₂=0. The angle condition then fixes λ.

Answer from the NTA final answer key. Solution by PrepWiser.