If L₁ is the line of intersection of the plane 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 and L₂ is the line of intersection of the plane x + 2y – z – 3 = 0, 3x – y + 2z – 1 = 0, then the distance of the origin from the plane containing the lines L₁
and L₂ is :

$\frac{1}{\sqrt{2}}$

$\frac{1}{4 \sqrt{2}}$

$\frac{1}{3 \sqrt{2}}$

$\frac{1}{2 \sqrt{2}}$

Solution

$\frac{1}{3 \sqrt{2}}$

$L_{1} i s p a r a l l e l t o |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 2 & 3 \\ 1 & - 1 & 1 \end{matrix}| = \hat{i} + \hat{j} L_{2} i s p a r a l l e l t o |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 1 \\ 3 & - 1 & 2 \end{matrix}| = 3 \hat{i} - 5 \hat{j} - 7 \hat{k} A l s o, L_{2} p a s s e s t h r o u g h (\frac{5}{7}, \frac{8}{7}, 0) S o r e q u i r e d p l a n e i s |\begin{matrix} x - \frac{5}{7} & y - \frac{8}{7} & z \\ 1 & 1 & 0 \\ 3 & - 5 & - 7 \end{matrix}| = 0 \Rightarrow 7 x - 7 y + 8 z + 3 = 0 N o w, p e r p e n d i c u l a r d i s \tan c e = \frac{3}{\sqrt{162}} = \frac{1}{3 \sqrt{2}}$

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Three Dimensional Geometry

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