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Let the foot of perpendicular from a point \mathrm{P}(1,2,-1) to the straight line L: \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}$ be $N. Let a line be drawn from P parallel to the plane \mathrm{x}+\mathrm{y}+2 \mathrm{z}=0 which meets \mathrm{L}  at point \mathrm{Q}. If \alpha is the acute angle between the lines P N$ and $P Q, then \cos \alpha is equal to__________.
Option: 1 \frac{1}{\sqrt{5}}
Option: 2 \frac{\sqrt{3}}{2}
Option: 3 \frac{1}{\sqrt{3}}
Option: 4 \frac{1}{2 \sqrt{3}}

Answers (1)

best_answer


Let \, N\, be\, \frac{x}{1}= \frac{y}{0}= \frac{z}{-1}= \lambda \Rightarrow x= \lambda ,y= 0,z= -\lambda
N\left ( \lambda ,0,- \lambda \right )
Now\, PN\perp L\Rightarrow \overrightarrow{PN}\cdot \left ( i-k \right )= 0
\Rightarrow \left ( \left ( \lambda -1 \right )i-2j+\left ( -\lambda+1 \right )k \right )\cdot \left ( i-k \right )= 0
\Rightarrow \left ( \lambda -1 \right )+\lambda -1= 0\Rightarrow \lambda = 1
\Rightarrow N\left ( 1,0,-1 \right )



Let\, Q\, be\, \left ( p,0,-p \right )
\therefore \overrightarrow{PQ}\perp \left ( normal\, to \, plane \right )
\Rightarrow \overrightarrow{PQ}\cdot \vec{n}= 0
\Rightarrow \left [ \left ( p-1 \right ) i+\left ( -2 \right )j+\left ( -p+1 \right )k\right ]\cdot \left ( i+j+2k \right )= 0
\Rightarrow p= -1
\Rightarrow Q\left ( -1,0,1 \right )


\therefore \overrightarrow{PN}= 2j\, and\, \overrightarrow{PQ}= -2i-2j+2k
Angle\, between\, these= \cos^{-1}\left | \frac{\overrightarrow{PN}\cdot \overrightarrow{PQ}}{\left | \overrightarrow{PN} \right |\left | \overrightarrow{PQ} \right |} \right |
                                           = \cos^{-1}\left | \frac{4}{2\cdot 2\sqrt{3}} \right |= \cos^{-1}\left ( \frac{1}{\sqrt{3}} \right )

Posted by

Kuldeep Maurya

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