First slide

Planes in 3D

Question

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

Moderate

A

10 units
B

9 units
C

8 units
D

7units

Solution

$The given plane is x - y - z = 9 \dots \dots . (1)$
$The given line AB is \frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{6} \dots \dots \dots \dots \dots \dots \dots \dots$ (2)
$The equation of the line passing through (1, 0, - 3) and parallel to \frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{6} i.$
$\frac{x - 1}{2} = \frac{y - 0}{3} = \frac{z + 3}{- 6} = r \dots \dots \dots \dots (3)$
Any point on (3) is
$p (2 r + 1, 3 r, - 6 r - 3)$
$P lies on x - y - z = 9$
$2 r + 1 - 3 r + 6 r + 3 = 9$
$\Rightarrow r = 1$
$p (3, 3, - 9)$
$Q (1, 0, - 3)$
$P Q = \sqrt{4 + 9 + 36} = 7$

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