Question

$OABC$ is a tetrahedron in with $O$ as the origin and position vectors of points $A,B,C$ as $\hat{i}+2\hat{j}+3\hat{k},2\hat{i}+\alpha \hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}+2\hat{k}$ respectively, then the integral value of $\alpha$ to have shortest distance between $\overrightarrow{OA}\&\overrightarrow{BC}$ as $\sqrt{\frac{3}{2}}$, is

Answer 1

Answer: 3

Unit vector $\left(\hat{n}\right)$ perpendicular to $\overrightarrow{OA}$ and $\overrightarrow{CB}$
$\hat{n}=\frac{\overrightarrow{CB}\times \overrightarrow{OA}}{\left|\overrightarrow{CB}\times \overrightarrow{OA}\right|}=\frac{\left(3\alpha -7\right)\hat{i}-4\hat{j}+\left(5-\alpha \right)\hat{k}}{\sqrt{{\left(3\alpha -7\right)}^2+16+{\left(5-\alpha \right)}^2}}$
As $\overrightarrow{CB}\times \overrightarrow{OA}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& \alpha -3& -1\\ 1& 2& 3\end{array}\right|=\left(3\alpha -7\right)\hat{i}-4\hat{j}+\left(5-\alpha \right)\hat{k}$
$\overrightarrow{BA}=-\hat{i}+\left(2-\alpha \right)\hat{j}+2\hat{k}$
$\Rightarrow \left|\overrightarrow{BA}.\hat{n}\right|=\left|\frac{\left(7-3\alpha \right)-4\left(2-\alpha \right)+2\left(5-\alpha \right)}{\sqrt{{\left(3\alpha -7\right)}^2+16+{\left(5-\alpha \right)}^2}}\right|$
Shortest distance $=\left|\overrightarrow{}.\hat{n}\right|=\sqrt{\frac{3}{2}}$
$\Rightarrow \left|\frac{\left(9-\alpha \right)}{\sqrt{{\left(3\alpha -7\right)}^2+16+{\left(5-\alpha \right)}^2}}\right|=\sqrt{\frac{3}{2}}$
$\Rightarrow \frac{{\left(9-\alpha \right)}^2}{{\left(3\alpha -7\right)}^2+16+{\left(5-\alpha \right)}^2}=\frac{3}{2}$
$\Rightarrow 2{\left(9-\alpha \right)}^2=3{\left(3\alpha -7\right)}^2+48+3{\left(5-\alpha \right)}^2$
$\Rightarrow 2{\alpha }^2-36\alpha +162=30{\alpha }^2-156\alpha +270$
$\Rightarrow 7{\alpha }^2-30\alpha +27=0$
$\Rightarrow \alpha =3,\frac{9}{7}$
So, integral value of $\alpha =3$