Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\\ & L_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\end{array}$
$\begin{array}{rl}& P(2\lambda +1,3\lambda +2,4\lambda +3)\\ & Q(3\mu +2,4\mu +4,5\mu +5)\end{array}$ Dr's of $PQ<2\lambda -3\mu -1,3\lambda -4\mu -2$, $\begin{array}{rl}& 4\lambda -5\mu -2>\\ & PQ=\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=-\hat{i}+2\hat{j}-\hat{k}\\ & \Rightarrow \frac{2\lambda -3\mu -1}{-1}=\frac{3\lambda -4\mu -2}{2}=\frac{4\lambda -5\mu -2}{-1}\\ & \Rightarrow \lambda =\frac{1}{3}\mu =\frac{-1}{6}\\ & \Rightarrow P\left(\frac{5}{3},3,\frac{13}{3}\right)Q\left(\frac{3}{2},\frac{10}{3},\frac{25}{6}\right)\end{array}$ Dr's $PQ⟨1,-2,1⟩$ $\therefore \text{ Line }$ $\frac{y-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{y-\frac{13}{3}}{1}$