Question

$P(0,3,-2),Q(3,7,-1)$ and $R(1,-3,-1)$ are $3$ given points. Let $L_1$ be the line passing through $P$ and $Q$ and $L_2$ be the line through $R$ and parallel to the vector $V=i+k$.

Answer 1

Given that $P(0,3,-2),Q(3,7,-1),R(1,-3,-1$
$L_1:r=P+lQP=3j-2k+l(3i+4j+k)$
$L_2:r=R+mV=i-3j-k+m(i+k)$
A) Since, $P$ lies on $L_1$
therefore, perpendicular distance from P to $L_1$ is $0$
B) Shortest distance between $L_1$ and $L_2$ $=\left|\frac{[(P-R).QP\times V]}{|QP\times V|}\right|=\left|\frac{-12}{6}\right|=2$
C) $\Delta PQR=\frac{|PQ\times PR|}{2}=\frac{|-10i+2j+22k|}{2}=\frac{|(-3i-4j-k)\times (-i+6j-k)|}{2}=\sqrt{147}$
D) Equation of plane in three point form is $\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
$\Rightarrow 5x-y-11z=19$
Therefore, perpendicular distance from origin to plane $=\frac{19}{\sqrt{5^2+1^2+{11}^2}}=\frac{19}{\sqrt{147}}.$