The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines $$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$$ and $$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$$ is :
JEE · Math · previous-year question
- A.$${1 \over 3}$$
- B.$${1 \over {\sqrt 3 }}$$
- C.3
- D.$${\sqrt 3 }$$correct
Answer
D. $${\sqrt 3 }$$
Explanation
Vector of the plane is $$\left| {\begin{matrix} {\hat i} & {\hat j} & {\hat k} \\ 1 & 2 & { - 1} \\ { - 1} & 1 & { - 2} \\ \end{matrix} } \right| = - 3\hat i + 3\hat j + 3\hat k$$ Now equation of plane is $$ - 3x + 3y + 3z = c$$ (1, 1, 0) will satisfy the plane $$ \Rightarrow - 3 + 3 + 0 = c$$ $$ \Rightarrow $$ c = 0 $$ - 3x + 3y + 3z = 0$$ distance from (2, 1, 4) is $$ \Rightarrow \left| {{{ - 6 + 3 + 12} \over {\sqrt {27} }}} \right| = \left| {{9 \over {3\sqrt 3 }}} \right| = \sqrt 3 \,\,units$$
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