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Find the distance of the point (0, -3, 2) from the plane x+2y-

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Find the distance of the point $(0, - 3, 2)$ from the plane $x + 2 y - z = 1$ , measure parallel to the line $\frac{x + 1}{3} = \frac{y + 1}{2} = \frac{z}{3}$ .

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Find the distance of the point

(0, - 3, 2)

from the plane

x + 2 y - z = 1

, measure parallel to the line

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

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Text solutionVerified

We have the plane

p : x + 2 y - z = 1

point

A (0, - 3, 2)

L in e 1 : \frac{x + 1}{3} = \frac{y + 1}{2} = \frac{z}{3}

Equation of line passing through point

A

and parallel to line

1

is,

L in e 2 : \frac{x}{3} = \frac{y + 3}{2} = \frac{z - 2}{3}

Let,

\frac{x}{3} = \frac{y + 3}{2} = \frac{z - 2}{3} = a

\Rightarrow x = 3 a, y = 2 a - 3, z = 3 a + 2

Let, this be a point of this line lying on given plane

As,

x + 2 y - z = 1

\Rightarrow 3 a + 2 a - 3 + 3 a + 2 = 1

\Rightarrow 8 a = 2

\Rightarrow a = \frac{1}{4}

So,

x = 3 a = \frac{3}{4}

y = 2 a - 3 = - \frac{5}{2}

z = 3 a + 2 = \frac{11}{4}

So, distance of the point from the plane is,

d = (\frac{3}{4})^{2} + (- \frac{5}{2} + 3)^{2} + (\frac{11}{4} - 2)^{2}

= \frac{22}{4}

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Question Text	Find the distance of the point $(0, - 3, 2)$ from the plane $x + 2 y - z = 1$ , measure parallel to the line $\frac{x + 1}{3} = \frac{y + 1}{2} = \frac{z}{3}$ .
Answer Type	Text solution:1
Upvotes	150