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,

& \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{aligned}$$ So, equation of the line $$\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{aligned} & \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{aligned}$$ Hence, the required distance, $$\begin{aligned} & =\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{aligned}$$ We know, $\left| \overset{\to }{\mathop{x}}\, \right|$ of a vector $$\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{aligned} & =\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{aligned}$$ We know cross product of $$\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, $\begin{aligned} & =\left| \begin{matrix} i & j & k \\\ -3 & 0 & 0 \\\ 2 & 3 & 6 \\\ \end{matrix} \right| \\\ & =-j\left( -18 \right)+k\left( -9 \right) \\\ & =18\overset{\hat{\ }}{\mathop{j}}\,-9\overset{\hat{\ }}{\mathop{k}}\, \\\ \end{aligned}$ So, the required distance, $\begin{aligned} & =\left| \dfrac{18\overset{\hat{\ }}{\mathop{j}}\,-9\overset{\hat{\ }}{\mathop{k}}\,}{7} \right| \\\ & =\dfrac{\sqrt{{{\left( 18 \right)}^{2}}+{{\left( -9 \right)}^{2}}}}{7} \\\ & =\dfrac{\sqrt{324+81}}{7} \\\ & =\dfrac{\sqrt{2105}}{7} \\\ & =\dfrac{9\sqrt{5}}{7} \\\ \end{aligned}$ 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.