The line l_1 passes through the point (2,6,2) and is perpendicular to the plane 2x+y-2z=10…

Question

The line $l_1$ passes through the point $(2,6,2)$ and is perpendicular to the plane $2x+y-2z=10$. Then the shortest distance between the line $l_1$ and the line $\dfrac{x+1}{2}=\dfrac{y+4}{-3}=\dfrac{z}{2}$ is:

Accepted Answer

Equation of $${l_1} = {{x - 2} \over 2} = {{y - 6} \over 1} = {{z - 2} \over { - 2}}$$

Shortest distance with $${{x + 1} \over 2} = {{y + 4} \over { - 3}} = {z \over 2}$$ is

S.d $$ = \left| {{{\begin{pmatrix} 3 & {10} & 2  \
   2 & 1 & { - 2}  \
   2 & { - 3} & 2  \

\end{pmatrix} } \over {\left| { - 4\widehat i - 8\widehat j - 8\widehat k} \right|}}} \right| = \left| {{{( - 12) - 10(8) + 2( - 8)} \over {12}}} \right|$$

= 9 units

Solutions · 1