Question

Let $\overrightarrow{OB}=\hat{i}+2\hat{j}+2\hat{k}$ and $\overrightarrow{OA}=4\hat{i}+2\hat{j}+2\hat{k}$. The distance of the point $B$ from the straight line passing through $A$ and parallel to the vector $2\hat{i}+3\hat{j}+6\hat{k}$ is

• A. $\frac{7\sqrt{5}}{9}$
• B. $\frac{5\sqrt{7}}{9}$
• C. $\frac{3\sqrt{5}}{7}$
• D. $\frac{9\sqrt{5}}{7}$
• E. $\frac{9\sqrt{7}}{5}$

Answer 1

The correct option is D $\frac{9\sqrt{5}}{7}$

Let the equation of line be $r=a+\lambda b$.

Since, line passes through $A$, so $a=4\hat{i}+2\hat{j}+2\hat{k}$ and line is parallel to the vector $2\hat{i}+3\hat{j}+6\hat{k}$, so $b=2\hat{i}+3\hat{j}+6\hat{k}$
Hence, equation of the line is
$r=4\hat{i}+2\hat{j}+2\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$
Distance of point $B$ from the line $=\left|\frac{\left|(a_2-a_1)\right|\times b}{\left|b\right|}\right|$
$a_2-a_1=3\hat{i}+0\hat{j}+0\hat{k}$
$(a_2-a_1)\times b=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 0& 0\\ 2& 3& 6\end{array}\right|=-18\hat{j}+9\hat{k}$
$|(a_2-a_1)\times b|=\sqrt{{(-18)}^2+{\left(9\right)}^2}$
$=\sqrt{405}=9\sqrt{5}$
$\left|b\right|=\sqrt{4+9+36}=7$
Therefore, required distance is $\frac{9\sqrt{5}}{7}$.