Question

If the vector $\overrightarrow{b}=3\hat{j}+4\hat{k}$ is written as the sum of a vector $\overrightarrow{{{b}_{1}}}$ , parallel to $\overrightarrow{a}=\hat{i}+\hat{j}$ and a vector $\overrightarrow{{{b}_{2}}}$ ,perpendicular to $\vec{a}$ ,then ${{\vec{b}}_{1}}\times {{\vec{b}}_{2}}$ is equal to.
(A) $3\hat{i}-3\hat{j}+9\hat{k}$
(B) $-6\hat{i}+6\hat{j}-\dfrac{9}{2}\hat{k}$
(C) $6\hat{i}-6\hat{j}+\dfrac{9}{2}\hat{k}$
(D) $-3\hat{i}+3\hat{j}-9\hat{k}$

Answer 1

In this question we have been asked to find the value of ${{\vec{b}}_{1}}\times {{\vec{b}}_{2}}$, when the given information is stated as “the vector $\overrightarrow{b}=3\hat{j}+4\hat{k}$ is written as the sum of a vector $\overrightarrow{{{b}_{1}}}$ , parallel to $\overrightarrow{a}=\hat{i}+\hat{j}$ and a vector $\overrightarrow{{{b}_{2}}}$ ,perpendicular to $\vec{a}$”. We know that the dot product of two perpendicular vectors is zero and the cross product of two vectors $A={{A}_{1}}\hat{i}+{{A}_{2}}\hat{j}+{{A}_{3}}\hat{k}$ and $B={{B}_{1}}\hat{i}+{{B}_{2}}\hat{j}+{{B}_{3}}\hat{k}$ is given as $A\times B=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{A}_{1}} & {{A}_{2}} & {{A}_{3}} \\\ {{B}_{1}} & {{B}_{2}} & {{B}_{3}} \\\ \end{matrix} \right|$ .

Complete step by step solution:
Now considering from the question we have been asked to find the value of ${{\vec{b}}_{1}}\times {{\vec{b}}_{2}}$, when the given information is stated as “the vector $\overrightarrow{b}=3\hat{j}+4\hat{k}$ is written as the sum of a vector $\overrightarrow{{{b}_{1}}}$ , parallel to $\overrightarrow{a}=\hat{i}+\hat{j}$ and a vector $\overrightarrow{{{b}_{2}}}$, perpendicular to $\vec{a}$”.
Now we can say that $\vec{b}={{\vec{b}}_{1}}+{{\vec{b}}_{2}}\Rightarrow 3\hat{j}+4\hat{k}$.
As the vector $\overrightarrow{{{b}_{1}}}$ is parallel to $\overrightarrow{a}=\hat{i}+\hat{j}$ , we can assume that $\overrightarrow{{{b}_{1}}}=p\left( \hat{i}+\hat{j} \right)$.
As the vector $\overrightarrow{{{b}_{2}}}$ is perpendicular to $\overrightarrow{a}=\hat{i}+\hat{j}$ , we can assume that $\overrightarrow{{{b}_{2}}}=s\hat{i}+r\hat{j}+q\hat{k}$.
Now we can say that the dot product of $\vec{a}$ and ${{\vec{b}}_{2}}$ will be zero as they are perpendicular to each other. Hence
$\begin{aligned} & \overrightarrow{a}.\overrightarrow{{{b}_{2}}}\Rightarrow \left( \hat{i}+\hat{j} \right).\left( s\hat{i}+r\hat{j}+q\hat{k} \right)=0 \\\ & \Rightarrow s+r=0 \\\ \end{aligned}$
We can conclude that
$\begin{aligned} & \vec{b}={{{\vec{b}}}_{1}}+{{{\vec{b}}}_{2}}\Rightarrow 3\hat{j}+4\hat{k} \\\ & \Rightarrow 3\hat{j}+4\hat{k}=p\left( \hat{i}+\hat{j} \right)+\left( s\hat{i}+r\hat{j}+q\hat{k} \right) \\\ & \Rightarrow s+p=0 \\\ & r+p=3 \\\ & q=4 \\\ \end{aligned}$
Hence we have $s=-r$ so we can say that
$\begin{aligned} & s+p=0 \\\ & \Rightarrow -r+p=0 \\\ \end{aligned}$
From $r+p=3$ and $p-r=0$ we will have $p=r=\dfrac{3}{2}$ and $s=\dfrac{-3}{2}$ .
Hence we can say that
$\begin{aligned} & {{{\vec{b}}}_{1}}=\dfrac{3}{2}\left( \hat{i}+\hat{j} \right) \\\ & {{{\vec{b}}}_{2}}=\left( \dfrac{-3}{2}\hat{i}+\dfrac{3}{2}\hat{j}+4\hat{k} \right) \\\ \end{aligned}$ .
From the basic concept we know that the cross product of two vectors $A={{A}_{1}}\hat{i}+{{A}_{2}}\hat{j}+{{A}_{3}}\hat{k}$ and $B={{B}_{1}}\hat{i}+{{B}_{2}}\hat{j}+{{B}_{3}}\hat{k}$ is given as $A\times B=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{A}_{1}} & {{A}_{2}} & {{A}_{3}} \\\ {{B}_{1}} & {{B}_{2}} & {{B}_{3}} \\\ \end{matrix} \right|$ .
Hence
$\begin{aligned} & {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ \dfrac{3}{2} & \dfrac{3}{2} & 0 \\\ \dfrac{-3}{2} & \dfrac{3}{2} & 4 \\\ \end{matrix} \right| \\\ & \Rightarrow {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}=\hat{i}\left| \begin{matrix} \dfrac{3}{2} & 0 \\\ \dfrac{3}{2} & 4 \\\ \end{matrix} \right|-\hat{j}\left| \begin{matrix} \dfrac{3}{2} & 0 \\\ \dfrac{-3}{2} & 4 \\\ \end{matrix} \right|+\hat{k}\left| \begin{matrix} \dfrac{3}{2} & \dfrac{3}{2} \\\ \dfrac{-3}{2} & \dfrac{3}{2} \\\ \end{matrix} \right| \\\ & \Rightarrow {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}=\left( \dfrac{3}{2}\times 4-0 \right)\hat{i}-\left( \dfrac{3}{2}\times 4-0 \right)\hat{j}+\left( \dfrac{3}{2}\times \dfrac{3}{2}-\dfrac{3}{2}\times \dfrac{-3}{2} \right)\hat{k} \\\ & \Rightarrow {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}=6\hat{i}-6\hat{j}+\dfrac{9}{2}\hat{k} \\\ \end{aligned}$
Therefore we can conclude that ${{\vec{b}}_{1}}\times {{\vec{b}}_{2}}=6\hat{i}-6\hat{j}+\dfrac{9}{2}\hat{k}$ .

So, the correct answer is “Option C”.

Note: While answering questions of this type we should be sure with our concept. Someone may confuse and assume that the cross product of two perpendicular vectors is zero and end up having the answer zero which is clearly a wrong answer.