Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Equation of line joining points (0,-11,4) and (2,-3,1) is $$\frac{x-2}{2}=\frac{y+3}{8}=\frac{z-1}{-3}=\lambda$$ Let $P$ is any point of the above line then coordinate of $P$ is $(2\lambda +2,8\lambda -3,-3\lambda +1)$ $\therefore {\mathrm{DR}}^{\prime }\mathrm{s}$ of $PQ$ is $(2\lambda +1,8\lambda -11,-3\lambda -3$ Now, $(2\lambda +1)(4+(8\lambda -11)(8)+(-3\lambda -3)(-3)=0$ $\Rightarrow 4\lambda +2+64\lambda -88+9\lambda +9=0$ $77\lambda -77=0$ $\therefore$ Required foot of perpendicular. $$P(4,5,-2)$$