The equation of a plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{\mathcal{z}–3}{2}$, and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{\mathcal{z} - 2}{3}$ and at greatest distance from point (0, 0, 0) is

Question

Dr. Abubacker · Accepted Answer

(b)
Let a point $(3\lambda + 1,\ \lambda + 2,\ 2\lambda + 3)$ of the first line also lies on the second line
Then $\frac{3\lambda + 1 - 3}{1} = \frac{\lambda + 2 - 1}{2} = \frac{2\lambda + 3 - 2}{3} \Rightarrow \lambda = 1$
Hence, the point of intersection $P$ of the two lines (4, 3, 5)
Equation of plane perpendicular to $OP$, where $O$ is (0, 0, 0) and passing through $P$ is
$4x + 3y + 5\mathcal{z} = 50$

Q. The equation of a plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{\mathcal{z}–3}{2}$, and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{\mathcal{z} - 2}{3}$ and at greatest distance from point (0, 0, 0) is

Explanation

Reviewed by Dr. Abubacker

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