Mathematics - Vector Question with Solution | TestHub

Let the equation of $AB$ is $$\frac{x-1}{2}=\frac{y-(-1)}{-3}=\frac{z-(-10)}{8}=k$$ Let $L$ be the foot of the perpendicular drawn form $P(1,0,0)$.
$\therefore L=(2k+1,-3\mathrm{k}-1,8k-10)$. Now, direction ratio of $PL=(2k,-3k-1,8$ $-10)$ and direction ratio of $AB=(2,-3,8)$ Since, $PL$ is perpendicular to $AB$ $$\therefore 2(2k)-3(-3k-1)+8(8k-10)=0$$ Now, $k=\frac{2(1-1)+(-3)(0+1)+8(0+10)}{(2{)}^2+(-3{)}^2+(8{)}^2}$ $$=\frac{0-3+80}{4+9+64}=\frac{77}{77}=1$$ $\therefore$ Required co-ordinate $$=L=(2+1,-3-1,8-10)=(3,-4,-2)\text{. }$$