Question

$O A B C$ 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 $\overrightarrow{O A} \& \overrightarrow{B C}$ as $\sqrt{\frac{3}{2}}$, is

Answer 1

Unit vector $\left(\hat{n}\right)$ perpendicular to $\overrightarrow{O A}$ and $\overrightarrow{C B}$
$\hat{n}=\frac{\overrightarrow{C B} \times \overrightarrow{O A}}{\left|\overrightarrow{C B} \times \overrightarrow{O A}\right|}=\frac{\left(3 \alpha - 7\right) \hat{i} - 4 \hat{j} + \left(5 - \alpha \right) \hat{k}}{\sqrt{\left(3 \alpha - 7\right)^{2} + 16 + \left(5 - \alpha \right)^{2}}}$
As $\overrightarrow{C B}\times \overrightarrow{O A}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha -3 & -1 \\ 1 & 2 & 3 \end{vmatrix}=\left(3 \alpha - 7\right)\hat{i}-4\hat{j}+\left(5 - \alpha \right)\hat{k}$
$\overrightarrow{B A}=-\hat{i}+\left(2 - \alpha \right)\hat{j}+2\hat{k}$
$\Rightarrow \left|\overrightarrow{B A} \, . \hat{n}\right|=\left|\frac{\left(7 - 3 \alpha \right) - 4 \left(2 - \alpha \right) + 2 \left(5 - \alpha \right)}{\sqrt{\left(3 \alpha - 7\right)^{2} + 16 + \left(5 - \alpha \right)^{2}}}\right|$
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 $\alpha =3$