Question

Let $A(2\hat{i}+3\hat{j}+5\hat{k}),B(-\hat{i}+3\hat{j}+2\hat{k})$ , and $C(\lambda \hat{i}+5\hat{j}+\mu \hat{k})$ are the vertices of a $\Delta ABC$ and its median through $A$ is equally inclined to the positive directions of axes. Then find the value $2\lambda -\mu$.

Answer 1

Position vector of $D=\left(\frac{\lambda -1}{2}\right)\hat{i}+4\hat{j}+\left(\frac{\mu +2}{2}\right)\hat{k}$
Direction ratios of $AD$ will be,
$\left(\frac{\lambda -5}{2}\right),1,\left(\frac{\mu -8}{2}\right)$
But direction cosines of $AD$ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{\lambda -5}{2}=1=\frac{\mu -8}{2}$
$\Rightarrow \lambda =7,\mu =10$
$\therefore 2\lambda -\mu =14-10=4$