First slide

Planes in 3D

Question

If the points $(1, 1, p)$ and $(- 3, 0, 1)$ be equidistant from the plane $\bar{r} . (3 \bar{i} + 4 \bar{j} + 12 \bar{k}) + 13 = 0$ then the value of P $(p \neq 1)$ =

Moderate

Solution

$A (1, 1, 0) B (- 3, 0, 1)$
$\frac{| 3 + 4 - 12 p + 13 |}{\sqrt{9 + 6 + 144}} = \frac{| - 9 - 12 + 13 |}{\sqrt{9 + 16 + 144}}$
$| 20 - 12 p | = 8; | 5 - 3 p | = 2; 5 - 3 p = \pm 2$

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