If the distance between the plane,
23x – 10y – 2z + 48 = 0 and the plane
containing the lines
${{x + 1} \over 2} = {{y - 3} \over 4} = {{z + 1} \over 3}$
and
$${{x + 3} \over 2} = {{y + 2} \over 6} = {{z - 1} \over \lambda }\left( {\lambda \in R} \right)$$
is equal to
${k \over {\sqrt {633} }}$, then k is equal to ______.
Answer (integer)
3
Solution
Required distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)
<br><br>$\Rightarrow$ $$\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$$ = ${k \over {\sqrt {633} }}$
<br><br>$\Rightarrow$ $\left| {{3 \over {\sqrt {633} }}} \right|$ = ${k \over {\sqrt {633} }}$
<br><br>$\Rightarrow$ k = 3
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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