Mathematics - Vector Question with Solution | TestHub

Given: $P\left(\alpha ,\beta ,\gamma \right)$ be the foot of the perpendicular from the point $Q\left(1,2,3\right)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ 

Now, let $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda$

So, any point on the line is given by $\left(5\lambda -3, 2\lambda +1, 3\lambda -4\right)$, so let the point be $P\left(5\lambda -3, 2\lambda +1, 3\lambda -4\right)$

Now, direction ratio of $PQ=\left(5\lambda -4, 2\lambda -1, 3\lambda -7\right)$

Now, the line and $PQ$ will be perpendicular,

So, $5\left(5\lambda -4\right)+2\left(2\lambda -1\right)+3\left(3\lambda -7\right)=0$

$\Rightarrow 38\lambda -43=0$

$\Rightarrow \lambda =\frac{43}{38}$

So, the point $P\left(\alpha ,\beta ,\gamma \right)=P\left(5\times \frac{43}{38}-3, 2\times \frac{43}{38}+1, 3\times \frac{43}{38}-4\right)$

Then $19\left(\alpha +\beta +\gamma \right)=19\left[5\times \frac{43}{38}-3+2\times \frac{43}{38}+1+ 3\times \frac{43}{38}-4\right]$

 $\Rightarrow 19\left(\alpha +\beta +\gamma \right)=19\left[\frac{101+124-23}{38}\right]$

$\Rightarrow 19\left(\alpha +\beta +\gamma \right)=101$