Question

Let $L$ be the line of intersection of planes $\overrightarrow{r}\cdot (\hat{i}-\hat{j}+2\hat{k})=2$ and $\overrightarrow{r}\cdot (2\hat{i}+\hat{j}-\hat{k})=2$. 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

• A. $143$
• B. $101$
• C. $134$
• D. $119$

Answer 1

The correct option is D $119$

Given planes:
$\overrightarrow{r}\cdot (\hat{i}-\hat{j}+2\hat{k})=2$
$\overrightarrow{r}\cdot (2\hat{i}+\hat{j}-\hat{k})=2$
Putting
$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$
We get the equation of plane as
$x-y+2z=22x+y-z=2$
Direction of the line of interesection is
$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ 2& 1& -1\end{array}\right|=-\hat{i}+5\hat{j}+3\hat{k}$
Finding point on both of the plane,
Assuming $z=0$
$x-y=22x+y=2\Rightarrow x=\frac{4}{3},y=-\frac{2}{3}$
So, the line of intersection becomes
$\frac{x-\frac{4}{3}}{-1}=\frac{y+\frac{2}{3}}{5}=\frac{z-0}{3}=\lambda$
Let the foot of perpendicular be
$(\alpha ,\beta ,\gamma )=(-\lambda +\frac{4}{3},5\lambda -\frac{2}{3},3\lambda )$
Point from where foot of perpendicular is drawn is $(1,2,0)$
Direction ratios of the line joining the point and foot of perpendicular are
$(-\lambda +\frac{1}{3},5\lambda -\frac{8}{3},3\lambda )$
This is perpendicular to $-\hat{i}+5\hat{j}+3\hat{k}$, so
$-1(-\lambda +\frac{1}{3})+5(5\lambda -\frac{8}{3})+3\left(3\lambda \right)=0\Rightarrow 35\lambda =\frac{41}{3}$
Now,
$35(\alpha +\beta +\gamma )=35(-\lambda +\frac{4}{3}+5\lambda -\frac{2}{3}+3\lambda )=35(7\lambda +\frac{2}{3})=\frac{287}{3}+\frac{70}{3}=119$