Question

Let \[OB=\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+2\overset{\hat{\ }}{\mathop{k}}\,\ and\ OA=4\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+2\overset{\hat{\ }}{\mathop{k}}\,\] . The distance of the point B from the straight line passing through A and parallel to the vector \[2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\,\] is,
A. $\dfrac{7\sqrt{5}}{9}$
B. $\dfrac{5\sqrt{7}}{9}$
C. $\dfrac{3\sqrt{5}}{7}$
D. $\dfrac{9\sqrt{5}}{7}$
E. $\dfrac{9\sqrt{7}}{5}$

Answer 1

Hint: At first write the equation of line passing through A and parallel to \[2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\,\]. Equation of line passing through a point $\overset{\to }{\mathop{a}}\,$ and parallel to $\overset{\to }{\mathop{b}}\,$ can be written as $\overset{\to }{\mathop{a}}\,+\lambda \overset{\to }{\mathop{b}}\,$ . Now, find the distance of point B from this line by using the formula $\left| \dfrac{\left( {{a}_{2}}-{{a}_{1}} \right)\times b}{\left| b \right|} \right|$ , where $\overset{\to }{\mathop{{{a}_{2}}}}\,$ is the point from which distance is to be find, $\overset{\to }{\mathop{{{a}_{1}}}}\,$ is the point through which the line is passing and $\overset{\to }{\mathop{b}}\,$ is vector parallel to the line.

Complete step-by-step answer:
We have to find the distance of B from the straight line passing through A and parallel to vector \[2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\,\].
For this let us find the equation of straight line passing through A and parallel to \[2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\,\].
We know equation of a line passing through point A and parallel to a vector $\overset{\to }{\mathop{b}}\,$ can be written as $\overset{\to }{\mathop{a}}\,+\lambda \overset{\to }{\mathop{b}}\,$ where $'\lambda '$ is an arbitrary constant.
Here,
 \[\begin{align}
  & \overset{\to }{\mathop{a}}\,=\ \overset{\to }{\mathop{OA}}\,=4\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+2\overset{\hat{\ }}{\mathop{k}}\, \\
 & and\ \overset{\to }{\mathop{b}}\,=2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\, \\
\end{align}\]
So, equation of the line \[\left( 4\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+2\overset{\hat{\ }}{\mathop{k}}\, \right)+\lambda \left( 2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\, \right)\] .
Now, we have to find the distance of point B from the above obtained line.
We know the formula for distance of a point B $\left( \overset{\to }{\mathop{{{a}_{2}}}}\, \right)$ from a line $\overset{\to }{\mathop{a}}\,+\lambda \overset{\to }{\mathop{b}}\,$is \[\left| \dfrac{\left( \overset{\to }{\mathop{{{a}_{2}}}}\,-\overset{\to }{\mathop{a}}\, \right)\times \overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{b}}\, \right|} \right|\].
Here,
\[\begin{align}
  & \overset{\to }{\mathop{{{a}_{2}}}}\,=\ \overset{\to }{\mathop{OB}}\,=\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+2\overset{\hat{\ }}{\mathop{k}}\, \\
 & and\ \overset{\to }{\mathop{a}}\,=4\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+2\overset{\hat{\ }}{\mathop{k}}\, \\
 & and\ \overset{\to }{\mathop{b}}\,=2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\, \\
\end{align}\]
Hence, the required distance,
 \[\begin{align}
  & =\left| \dfrac{\left[ \left( \overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+2\overset{\hat{\ }}{\mathop{k}}\, \right)-\left( 4\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,+2\overset{\hat{\ }}{\mathop{k}}\, \right) \right]\times \overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{b}}\, \right|} \right| \\
 & =\left| \left[ \dfrac{\left( -3\overset{\hat{\ }}{\mathop{i}}\,+0\overset{\hat{\ }}{\mathop{j}}\,+0\overset{\hat{\ }}{\mathop{k}}\, \right)\times \left( 2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\, \right)}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( 6 \right)}^{2}}}} \right] \right| \\
\end{align}\]
We know, $\left| \overset{\to }{\mathop{x}}\, \right|$ of a vector \[\overset{\to }{\mathop{x}}\,={{x}_{1}}\overset{\hat{\ }}{\mathop{i}}\,+{{x}_{2}}\overset{\hat{\ }}{\mathop{j}}\,+{{x}_{3}}\overset{\hat{\ }}{\mathop{k}}\,\] is given by $\sqrt{{{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}}$ . so, we have replaced $\left| \overset{\to }{\mathop{b}}\, \right|$ with \[\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( 6 \right)}^{2}}}\].
The required distance,
\[\begin{align}
  & =\left| \dfrac{\left( -3\overset{\hat{\ }}{\mathop{i}}\, \right)\times \left( 2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\, \right)}{\sqrt{4+9+36}} \right| \\
 & =\left| \dfrac{\left( -3\overset{\hat{\ }}{\mathop{i}}\, \right)\times \left( 2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\, \right)}{\sqrt{49}} \right| \\
 & =\left| \dfrac{\left( -3\overset{\hat{\ }}{\mathop{i}}\, \right)\times \left( 2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\, \right)}{7} \right| \\
\end{align}\]
We know cross product of \[\left( {{a}_{1}}\overset{\hat{\ }}{\mathop{i}}\,+{{a}_{2}}\overset{\hat{\ }}{\mathop{j}}\,+{{a}_{3}}\overset{\hat{\ }}{\mathop{k}}\, \right)\ and\ \left( {{b}_{1}}\overset{\hat{\ }}{\mathop{i}}\,+{{b}_{2}}\overset{\hat{\ }}{\mathop{j}}\,+{{b}_{3}}\overset{\hat{\ }}{\mathop{k}}\, \right)\] is given by $\left| \begin{matrix}
   i & j & k \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right|$ .
So, cross product of \[\left( -3\overset{\hat{\ }}{\mathop{i}}\, \right)\ and\ \left( 2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\, \right)\] will be,
$\begin{align}
  & =\left| \begin{matrix}
   i & j & k \\
   -3 & 0 & 0 \\
   2 & 3 & 6 \\
\end{matrix} \right| \\
 & =-j\left( -18 \right)+k\left( -9 \right) \\
 & =18\overset{\hat{\ }}{\mathop{j}}\,-9\overset{\hat{\ }}{\mathop{k}}\, \\
\end{align}$
So, the required distance,
$\begin{align}
  & =\left| \dfrac{18\overset{\hat{\ }}{\mathop{j}}\,-9\overset{\hat{\ }}{\mathop{k}}\,}{7} \right| \\
 & =\dfrac{\sqrt{{{\left( 18 \right)}^{2}}+{{\left( -9 \right)}^{2}}}}{7} \\
 & =\dfrac{\sqrt{324+81}}{7} \\
 & =\dfrac{\sqrt{2105}}{7} \\
 & =\dfrac{9\sqrt{5}}{7} \\
\end{align}$
Therefore, the required distance is $\dfrac{9\sqrt{5}}{7}$ and option (D) is the correct answer.

Note: We have used $\overset{\to }{\mathop{r}}\,=\overset{\to }{\mathop{a}}\,+\lambda \overset{\to }{\mathop{b}}\,$ form of equation of line. We can also use the form $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ for writing the equation of line passing through A and parallel to given vector. And then find the foot of perpendicular from point B to the obtained line and finally calculate the distance between points B and foot of perpendicular to get the answer.