First slide

Planes in 3D

Question

$The distance of the point (1, 0, - 3) from the plane x - y - z = 9 measured parallel to the$ $line \frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{- 6}$

Moderate

A

6
B

7
C

8
D

9

Solution

$\begin{array}{l} Let \frac{x - 1}{2} = \frac{y - 0}{3} = \frac{z + 3}{- 6} = t (say) \Rightarrow x = 1 + 2 t, y = 3 t, z = - 3 - 6 t \\ Let P be a point in the plane x - y - z = 9 \\ P = (3, 3, - 9) \Rightarrow A P = 7 \end{array}$

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