Let L be the line of intersection of the planes r( i- j+2 k)=2 and r(2 i+ j- k)=2.

Question

Let $L$ be the line of intersection of the planes $\vec r\cdot(\hat i-\hat j+2\hat k)=2$ and $\vec r\cdot(2\hat i+\hat j-\hat k)=2$. If $P(\alpha,\beta,\gamma)$ is the foot of perpendicular on $L$ from the point $(1,2,0)$, then the value of $35(\alpha+\beta+\gamma)$ is equal to:

Accepted Answer

${P_1}:x - y + 2z = 2$

${P_2}:2x + y - 3 = 2$

Let line of Intersection of planes P1 and P2 cuts xy plane in point Q.

$$\Rightarrow$$ z-coordinate of point Q is zero

$$ \Rightarrow \left. {\begin{pmatrix} {x - y = 2}  \
   {and\,2x + y = 2}  \

\end{pmatrix} } \right\} \Rightarrow x = {4 \over 3},y = {{ - 2} \over 3}$$

$$ \Rightarrow Q\left( {{4 \over 3},{{ - 2} \over 3},0} \right)$$

Vector parallel to the line of intersection

$$\overrightarrow a  = \left| {\begin{pmatrix} {\widehat i} & {\widehat j} & {\widehat k}  \
   1 & { - 1} & 2  \
   2 & 1 & { - 1}  \

\end{pmatrix} } \right| =  - \widehat i + 5\widehat j + 3\widehat k$$

Equation of Line of intersection

$${{x - {4 \over 3}} \over { - 1}} = {{y + {2 \over 3}} \over 5} = {{z - 0} \over 3} = \lambda $$ (say)

Let coordinates of foot of perpendicular be

$$F\left( { - \lambda  + {4 \over 3},5\lambda  - {2 \over 3},3\lambda } \right)$$

$$\overrightarrow {PF}  = \left( { - \lambda  + {1 \over 3}} \right)\widehat i + \left( {5\lambda  - {8 \over 3}} \right)\widehat j + (3\lambda )\widehat k$$

$$\overrightarrow {PF} .\overrightarrow a  = 0$$

$$ \Rightarrow \lambda  - {1 \over 3} + 25\lambda {{ - 40} \over 3} + 9\lambda  = 0$$

$$ \Rightarrow 35\lambda  = {{41} \over 3} \Rightarrow \lambda  = {{41} \over {105}}$$

Now, $$\alpha  =  - \lambda  + {4 \over 3},\beta  = 5\lambda  - {2 \over 3},\gamma  = 3\lambda $$

$$ \Rightarrow \alpha  + \beta  + \gamma  = 7\lambda  + {2 \over 3}$$

$$ = 7\left( {{{41} \over {105}}} \right) + {2 \over 3}$$

$$ = {{51} \over {15}}$$

$$ \Rightarrow 35(\alpha  + \beta  + \gamma ) = {{51} \over {15}} \times 35 = 119$$

Solutions · 1