First slide

Straight lines in 3D

Question

$The distance of the point (1, - 2, 3) from the plane x - y + z - 5 = 0, parallel to the line$ $\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{- 6} is$

Moderate

A

1
B

2
C

3
D

4

Solution

$Equation of line through A (1, - 2, 3) and parallel to the line \frac{x}{2} = \frac{y}{3} = \frac{z - 1}{- 6} is$
$\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 6} = r \dots \dots \dots (1)$
$Any point P on this line is P (2 r + 1, 3 r - 2, - 6 r + 3) .$
$If P is the point of intersection of line (1) and plane x - y + z - 5 = 0, then$
$\begin{array}{l} 2 r + 1 - 3 r + 2 - 6 r + 3 - 5 = 0 \\ \Rightarrow r = \frac{1}{7} \\ ∴ P (\frac{9}{7}, \frac{- 11}{7}, \frac{15}{7}) \\ Required distance = A P = \sqrt{{(1 - \frac{9}{7})}^{2} + {(- 2 + \frac{11}{7})}^{2} + {(3 - \frac{15}{7})}^{2}} \end{array}$
$= \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = 1$

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