Let P be the plane, passing through the point (1,-1,-5) and perpendicular to the line joining the points (4,1,-3) and (2,4,3). Then the distance of P from the point (3,-2,2) is :

Solution

Maths sir · 3 steps

The line joining the two points is the plane's normal, so read the plane straight off that direction.

1. Find the normal. The plane is perpendicular to the line through (4,1,-3) and (2,4,3). So that line's direction is the plane's normal: (2-4, 4-1, 3-(-3)) = (-2, 3, 6).
2. Write the plane. Use the point (1,-1,-5): -2(x-1) + 3(y+1) + 6(z+5) = 0. This expands to -2x + 3y + 6z + 35 = 0.

   d = (|-2(3) + 3(-2) + 6(2) + 35|)/√((-2)² + 3² + 6²) = (|-6 - 6 + 12 + 35|)/√49 = 35/7

3. So. The distance of P from (3,-2,2) is 35/7 = 5, which is option (a).

The idea to keep. A plane perpendicular to a line takes that line's direction ratios as its normal. Then use (|ax₁+by₁+cz₁+d|)/√(a²+b²+c²) for the distance.

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