First slide

Planes in 3D

Question

If the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{- 1}$ intersects the plane $2 x + 3 y - z + 13 = 0$ at a point P then PQ is equal to

Moderate

A

14
B

$\sqrt{14}$
C

2 $\sqrt{7}$
D

$2 \sqrt{14}$

Solution

$\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{- 1} = t$
$P (3 t_{1} + 2, 2 t_{1} - 1, - t_{1} + 1) 1$ lies on $2 x + 3 y - z + 13 = 0$
$t_{1} = - 1 \Rightarrow P (- 1, - 3, 2)$
$Q (3 t_{2} + 2, 2 t_{2} - 1, - t_{2} + 1)$ lies on $3 x + y + 4 z = 16 \Rightarrow t_{2} = 1 \Rightarrow Q (5, 1, 0)$

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