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

4

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3

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2

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1

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Solution

The correct option is D $1$
$\begin{matrix} \frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 6} B \equiv (2 r + 1, 3 r - 2, - 6 r + 3) 2 r + 1 - {3 r - 2} - 6 r + 3 = 5 - 7 r + 6 = 5 7 r = 1 ∴ r = \frac{1}{7} A B = \sqrt{{(2 r)}^{2} + {(3 r)}^{2} + {(- 6 r)}^{2}} = \sqrt{49 r^{2}} = \sqrt{49 \times \frac{1}{49}} = 1 H e n c e, t h e o p t i o n D i s t h e c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

(1, - 2, 3)

x - y + z = 5

\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{- 6}

\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}

(1, - 2, 3)

x - y + z = 5

\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{- 6}

(1, - 2, 3)

x - y + z = 5

\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 6} .

\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}

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