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 $\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$ is :
Solution
<p>Equation of ${l_1} = {{x - 2} \over 2} = {{y - 6} \over 1} = {{z - 2} \over { - 2}}$</p>
<p>Shortest distance with ${{x + 1} \over 2} = {{y + 4} \over { - 3}} = {z \over 2}$ is</p>
<p>S.d $$ = \left| {{{\matrix{
3 & {10} & 2 \cr
2 & 1 & { - 2} \cr
2 & { - 3} & 2 \cr
} } \over {\left| { - 4\widehat i - 8\widehat j - 8\widehat k} \right|}}} \right| = \left| {{{( - 12) - 10(8) + 2( - 8)} \over {12}}} \right|$$</p>
<p>= 9 units</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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