The distance of the point P(3, 4, 4) from the point of intersection of the line joining the points. Q(3, -4, -5) and R(2, -3, 1) and the plane 2x + y + z = 7, is equal to ______________.

Solution

Maths sir · 3 steps

Put the line QR in terms of one parameter, drop it into the plane, then measure from P.

1. Line QR. The direction is R - Q = (2-3, -3+4, 1+5) = (-1, 1, 6). A general point on the line is (3 - t, -4 + t, -5 + 6t).
2. Meet the plane. Put this point into 2x + y + z = 7: 2(3 - t) + (-4 + t) + (-5 + 6t) = 7. This gives 5t - 3 = 7, so t = 2 and the meeting point is A(1, -2, 7).

   PA = √((3-1)² + (4+2)² + (4-7)²) = √(4 + 36 + 9) = √49

3. So. The distance from P(3, 4, 4) to the point of intersection is √49, which is 7.

The common slip: 49. Stopping at the sum of squares gives 49. Take the square root to get the distance, 7.

The idea to keep. To find where a line meets a plane, write the line as one moving point with parameter t, substitute it into the plane, and solve for t. One linear equation does the whole job.

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