The equation of the plane which passes through the point of intersection of lines x-13-y-21-z-32 and x-31-y-12-z-23 and at greatest distance from origin is

Let a point (3λ+1,λ+2,2λ+3) of the first line also lies on the second line. Then, 3λ+1 31=λ+2 12=2λ+3 23⇒λ=1 Hence, the point of intersection P of the two lines ...

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Standard XII

Physics

Multiplication with Vectors

The equation ...

Question

The equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and at greatest distance from origin is

4 x + 3 y + 5 z = 40

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4 x + 3 y + 5 z = 50

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2 x + 3 y + 5 z = 50

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x + 3 y + 5 z = 35

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Solution

The correct option is B $4 x + 3 y + 5 z = 50$
Let a point $(3 λ + 1, λ + 2, 2 λ + 3)$ of the first line also lies on the second line.
Then, $\frac{3 λ + 1 - 3}{1} = \frac{λ + 2 - 1}{2} = \frac{2 λ + 3 - 2}{3} \Rightarrow λ = 1$
Hence, the point of intersection $P$ of the two lines is $P (4, 3, 5)$
Since, the plane is at greatest distance from origin $(O)$ , the line $- - \to O P$ is perpendicular to the plane.
$∴$ D.R's of $- - \to O P = (4, 3, 5)$
The equation of plane is $4 (x - 4) + 3 (y - 3) + 5 (z - 5) = 0$
$\Rightarrow 4 x + 3 y + 5 z = 50$

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The equation of the plane which passes through the point of intersection of lines x−13=y−21=z−32 and x−31=y−12=z−23 and at greatest distance from origin is

The equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and at greatest distance from origin is