jee-main 2017 Q88

jee-main · India · 02apr Vectors 3D & Lines Distance from a Point to a Line (Show/Compute)

The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$, having normal perpendicular to both the lines $\dfrac{x-1}{1} = \dfrac{y+2}{-2} = \dfrac{z-4}{3}$ and $\dfrac{x-2}{2} = \dfrac{y+1}{-1} = \dfrac{z+7}{-1}$, is:
(1) $\dfrac{20}{\sqrt{74}}$
(2) $\dfrac{10}{\sqrt{83}}$
(3) $\dfrac{5}{\sqrt{83}}$
(4) $\dfrac{10}{\sqrt{74}}$

The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$, having normal perpendicular to both the lines $\dfrac{x-1}{1} = \dfrac{y+2}{-2} = \dfrac{z-4}{3}$ and $\dfrac{x-2}{2} = \dfrac{y+1}{-1} = \dfrac{z+7}{-1}$, is:\\
(1) $\dfrac{20}{\sqrt{74}}$\\
(2) $\dfrac{10}{\sqrt{83}}$\\
(3) $\dfrac{5}{\sqrt{83}}$\\
(4) $\dfrac{10}{\sqrt{74}}$

Paper Questions

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