The square of the distance of the point of intersection
of the line (x - 1)/2 = (y - 2)/3 = (z + 1)/6 and the plane 2x - y + z = 6 from the point (-1, -1, 2) is __________.

Solution

Maths sir · 3 steps

Write the line as one moving point with a parameter, drop it into the plane, and t falls out.

1. General point. Set each fraction equal to t. Any point on the line is then (1 + 2t, 2 + 3t, -1 + 6t).
2. Meet the plane. Substitute into 2x - y + z = 6: 2(1 + 2t) - (2 + 3t) + (-1 + 6t) = 6, which gives 7t - 1 = 6, so t = 1. The point of intersection is (3, 5, 5).

   d² = (3 + 1)² + (5 + 1)² + (5 - 2)² = 16 + 36 + 9

3. So. The square of the distance from (-1, -1, 2) is 61.

The common slip: 29. Subtracting 1 instead of -1 for the x and y coordinates gives 2² + 4² + 3² = 29, not 61. Subtracting a negative coordinate means adding.

The idea to keep. Line meets plane? Write the line's general point with one parameter, put it into the plane equation, and solve a single linear equation for the parameter.

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