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Find the equation of the plane passing through the points (1,

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Find the equation of the plane passing through the points $(1, 2, 3)$ and $(0, - 1, 0)$ and parallel to the line $\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z}{- 3}$ .

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Camila Discussed

Find the equation of the plane passing through the points

(1, 2, 3)

and

(0, - 1, 0)

and parallel to the line

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

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

Any plane through the point

(1, 2, 3)

a (x - 1) + b (y - 2) + c (z - 3) = 0

...........(i)

Since it passes through the point

(0, - 1, 0)

, we have

a (0 - 1) + b (- 1 - 2) + c (0 - 3) = 0 \Rightarrow a + 3 b + 3 c = 0

.........(ii)

This plane is parallel to the line

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

, so

2 a + 3 b - 3 c = 0

.........(iii)

On solving (ii) and (iii), we get

\frac{a}{( 9 + 9 )} = \frac{b}{( - 3 - 6 )} = \frac{c}{( 6 - 3 )} \Rightarrow \frac{a}{18} = \frac{b}{- 9} = \frac{c}{3}

\Rightarrow \frac{a}{6} = \frac{b}{- 3} = \frac{c}{1} = λ

(say)

Hence, the required equation of the plane is

6 λ (x - 1) - 3 λ (y - 2) + λ (z - 3) = 0

\Rightarrow 6 x - 3 y + z = 3

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Question Text	Find the equation of the plane passing through the points $(1, 2, 3)$ and $(0, - 1, 0)$ and parallel to the line $\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z}{- 3}$ .
Answer Type	Text solution:1
Upvotes	150