Let P be the plane containing the straight line (x-3)/9=(y+4)/-1=(z-7)/-5 and perpendicular to the plane containing the straight lines x/2=y/3=z/5 and x/3=y/7=z/8. If d is the distance of P from the point (2,-5,11), then d² is equal to :

Solution

Maths sir · 5 steps

The normal of P is perpendicular to both the given line and the other plane's normal, so cross them.

1. Normal of second plane. Both lines x/2=y/3=z/5 and x/3=y/7=z/8 pass through the origin. So their plane has normal (2,3,5)×(3,7,8) = (-11,-1,5).
2. Normal of P. P contains the line with direction (9,-1,-5), so its normal is perpendicular to that direction. P is also perpendicular to the second plane, so its normal is perpendicular to (-11,-1,5) as well. Cross the two vectors.

   (9,-1,-5)×(-11,-1,5) = (-10, 10, -20) ∥ (1,-1,2)

3. Equation of P. P passes through (3,-4,7), a point on the given line. So (x-3)-(y+4)+2(z-7)=0, which is x-y+2z-21=0.
4. Distance. For (2,-5,11): d = (|2+5+22-21|)/√(1+1+4) = 8/√6.
5. So. d² = 64/6 = 32/3, which is option (c).

The idea to keep. If a plane contains a line and is perpendicular to another plane, its normal is the cross product of the line's direction and that plane's normal.

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