The point $P$ is the intersection of the straight line joining the points $Q(2,3,5)$ and $R(1,-1,4)$ with…
The point $P$ is the intersection of the straight line joining the points $Q(2,3,5)$ and $R(1,-1,4)$ with the plane $5 x-4 y-$ $z=1$. If $S$ is the foot of the perpendicular drawn from the point $T(2,1,4)$ to $Q R$, then the length of the line segment $P S$ is
$\frac{1}{\sqrt{2}}$
$\sqrt{2}$
2
$2 \sqrt{2}$
Solution
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 $5 x-4 y-z=1$
$\therefore-5 \lambda+10+16 \lambda-12+\lambda-5=1$
$\Rightarrow 12 \lambda=8 \Rightarrow \lambda=\frac{2}{3} \quad \therefore P=\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$
Now let another point $S$ on $Q R$ be
$(-\mu+2,-4 \mu+3,-\mu+5)$
Since $S$ is the foot of perpendicular drawn from
$T(2,1,4)$ to $Q R$, where dr's of $S T$ are $\mu, 4 \mu-2, \mu-1$ and dr's of $Q R$ are $-1,-4,-1$
$\begin{array}{ll}
\therefore & -\mu-16 \mu+8-\mu+1=0 \Rightarrow 18 \mu=9 \Rightarrow \mu=\frac{1}{2} \\
\therefore & S=\left(\frac{3}{2}, 1, \frac{9}{2}\right)
\end{array}$
$\therefore \quad$ Distance between $P$ and $S$
$\begin{array}{l}
=\sqrt{\left(\frac{4}{3}-\frac{3}{2}\right)^{2}+\left(\frac{1}{3}-1\right)^{2}+\left(\frac{13}{3}-\frac{9}{2}\right)^{2}} \\
=\sqrt{\frac{1}{36}+\frac{4}{9}+\frac{1}{36}}=\frac{1}{\sqrt{2}}
\end{array}$
