Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Equation of st. line joining $Q(2,3,5)$ and $R(1,-1,4)$ is $\frac{x-2}{-1}=\frac{y-3}{-4}=\frac{z-5}{1}=\lambda$ Let $P(-\lambda +2,-4\lambda +3,-\lambda +5)$ Since $P$ also lies on $5x-4y-z=1$ $\therefore -5\lambda +10+16\lambda -12+\lambda -5=1$ $\Rightarrow 12\lambda =8\Rightarrow \lambda =\frac{2}{3}\therefore P=\left(\frac{4}{3},\frac{1}{3},\frac{13}{3}\right)$ Now let another point $S$ on $QR$ be $(-\mu +2,-4\mu +3,-\mu +5)$ Since $S$ is the foot of perpendicular drawn from $T(2,1,4)$ to $QR$, where dr's of $ST$ are $\mu ,4\mu -2,\mu -1$ and dr's of $QR$ are $-1,-4,-1$ $\begin{array}{ll}\therefore & -\mu -16\mu +8-\mu +1=0\Rightarrow 18\mu =9\Rightarrow \mu =\frac{1}{2}\\ \therefore & S=\left(\frac{3}{2},1,\frac{9}{2}\right)\end{array}$ $\therefore$ Distance between $P$ and $S$ $\begin{array}{l}=\sqrt{{\left(\frac{4}{3}-\frac{3}{2}\right)}^2+{\left(\frac{1}{3}-1\right)}^2+{\left(\frac{13}{3}-\frac{9}{2}\right)}^2}\\ =\sqrt{\frac{1}{36}+\frac{4}{9}+\frac{1}{36}}=\frac{1}{\sqrt{2}}\end{array}$