The shortest distance between the lines
${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$
and x + y + z + 1 = 0, 2x – y + z
+ 3 = 0 is :
Solution
Plane through line of intersection is
<br><br>x + y + z + 1 + $\lambda$ (2x –y + z + 3) = 0
<br><br>It should be parallel to given line ${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$
<br><br>$\therefore$ 0(1 + 2$\lambda$) - 1(1 - $\lambda$) + 1(1 + $\lambda$) = 0 $\Rightarrow$ $\lambda$ = 0
<br><br>$\therefore$ Required Plane : x + y + z + 1 = 0
<br><br>Shortest distance of (1, –1, 0) from this plane
<br><br>= ${{\left| {1 - 1 + 0 + 1} \right|} \over {\sqrt {{1^2} + {1^2} + {1^2}} }}$ = ${1 \over {\sqrt 3 }}$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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