Question

Let \[\vec{a}=3\hat{i}+\hat{j}\] and $\vec{\beta }=2\hat{i}-\hat{j}+3\hat{k}$. If $\vec{\beta }={{\vec{\beta }}_{1}}-{{\vec{\beta }}_{2}}$ , where ${{\vec{\beta }}_{1}}$ is parallel to $\vec{\alpha }$ and ${{\vec{\beta }}_{2}}$ is perpendicular to $\vec{\alpha }$ then ${{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}$ is equal to?
$\begin{align}
  & a)-3\hat{i}+9\hat{j}+5\hat{k} \\
 & b)3\hat{i}-9\hat{j}-5\hat{k} \\
 & c)\dfrac{1}{2}\left( -3\hat{i}+9\hat{j}+5\hat{k} \right) \\
 & d)\dfrac{1}{2}\left( 3\hat{i}-9\hat{j}+5\hat{k} \right) \\
\end{align}$

Answer 1

Hint: We are given that ${{\vec{\beta }}_{1}}$ is parallel to $\vec{\alpha }$. Hence we have ${{\vec{\beta }}_{1}}=\lambda \vec{\alpha }$ substituting this in $\vec{\beta }={{\vec{\beta }}_{1}}-{{\vec{\beta }}_{2}}$ we get ${{\vec{\beta }}_{2}}$ in terms of $\lambda $ . Now are given that ${{\vec{\beta }}_{2}}$ is perpendicular to $\vec{\alpha }$ and we know that dot product of perpendicular vectors is 0. Hence using this we will find the value of $\lambda $ and hence ${{\vec{\beta }}_{1}}$ and ${{\vec{\beta }}_{2}}$ . Now we can find ${{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}$

Complete step by step answer:
Now we have two vectors \[\vec{a}=3\hat{i}+\hat{j}\] and $\vec{\beta }=2\hat{i}-\hat{j}+3\hat{k}$
And we know that ${{\vec{\beta }}_{1}}$ is parallel to $\vec{\alpha }$.
Now we know if $\vec{a}$ is parallel to $\vec{b}$ then $\vec{a}=\lambda \vec{b}$
Hence we have ${{\vec{\beta }}_{1}}=\lambda \left( 3\hat{i}+\hat{j} \right)$
\[{{\vec{\beta }}_{1}}=3\lambda \hat{i}+\lambda \hat{j}....................\left( 1 \right)\]
Now $\vec{\beta }={{\vec{\beta }}_{1}}-{{\vec{\beta }}_{2}}$ and $\vec{\beta }=2\hat{i}-\hat{j}+3\hat{k}$
Hence we have
\[2\hat{i}-\hat{j}+3\hat{k}=3\lambda \hat{i}+\lambda \hat{j}-\left( {{{\vec{\beta }}}_{2}} \right)\]
Now rearranging the terms of above equation we get
\[\begin{align}
  & {{{\vec{\beta }}}_{2}}=3\lambda \hat{i}+\lambda \hat{j}-2\hat{i}+\hat{j}-3\hat{k} \\
 & {{{\vec{\beta }}}_{2}}=\left( 3\lambda -2 \right)\hat{i}+\left( \lambda +1 \right)\hat{j}-3\hat{k}......................\left( 2 \right) \\
\end{align}\]
Now we have that ${{\vec{\beta }}_{2}}$ is perpendicular to $\vec{\alpha }$ , and we know if two vectors are perpendicular then the dot product is 0. That is \[{{a}_{1}}\hat{i}+{{b}_{1}}\hat{j}+{{c}_{1}}\hat{k}\] and ${{a}_{2}}\hat{i}+{{b}_{2}}\hat{j}+{{c}_{2}}\hat{k}$ are perpendicular if ${{a}_{1}}.{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}.{{c}_{2}}=0$
Now \[\vec{a}=3\hat{i}+\hat{j}\]
And from equation (2) we have \[{{\vec{\beta }}_{2}}=\left( 3\lambda -2 \right)\hat{i}+\left( \lambda +1 \right)\hat{j}-3\hat{k}\]

Now using this condition on $\vec{\alpha }$ and ${{\vec{\beta }}_{2}}$ we get.
$\begin{align}
  & 3\left( 3\lambda -2 \right)+1\left( \lambda +1 \right)-3\left( 0 \right)=0 \\
 & \Rightarrow 9\lambda -6+\lambda +1=0 \\
 & \Rightarrow 10\lambda -5=0 \\
 & \Rightarrow 10\lambda =5 \\
\end{align}$
Now dividing the equation by 10 we get \[\lambda =\dfrac{5}{10}=\dfrac{1}{2}\]
Substituting the value in equation (2) we get
\[\begin{align}
  & {{{\vec{\beta }}}_{2}}=\left( 3\times \dfrac{1}{2}-2 \right)\hat{i}+\left( \dfrac{1}{2}+1 \right)\hat{j}-3\hat{k} \\
 & \Rightarrow {{{\vec{\beta }}}_{2}}=\left( \dfrac{3}{2}-2 \right)\hat{i}+\left( \dfrac{1+2}{2} \right)\hat{j}-3\hat{k} \\
 & \Rightarrow {{{\vec{\beta }}}_{2}}=\left( \dfrac{3-4}{2} \right)\hat{i}+\left( \dfrac{3}{2} \right)\hat{j}-3\hat{k} \\
 & \therefore {{{\vec{\beta }}}_{2}}=\dfrac{-1}{2}\hat{i}+\dfrac{3}{2}\hat{j}-3\hat{k} \\
\end{align}\]
And substituting the value in equation (1) we get
\[\begin{align}
  & {{{\vec{\beta }}}_{1}}=\dfrac{1}{2}\left( 3\hat{i}+\hat{j} \right) \\
 & \therefore {{{\vec{\beta }}}_{1}}=\dfrac{3}{2}\hat{i}+\dfrac{1}{2}\hat{j} \\
\end{align}\]
Now we need to calculate ${{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}$
Now we know that cross product of two vectors \[{{a}_{1}}\hat{i}+{{b}_{1}}\hat{j}+{{c}_{1}}\hat{k}\] and ${{a}_{2}}\hat{i}+{{b}_{2}}\hat{j}+{{c}_{2}}\hat{k}$ is given by $\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right|$
Hence we get ${{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}$
\[\begin{align}
  & =\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   \dfrac{3}{2} & \dfrac{1}{2} & 0 \\
   \dfrac{-1}{2} & \dfrac{3}{2} & -3 \\
\end{matrix} \right| \\
 & =\hat{i}\left( -3\times \dfrac{1}{2}-0\times \left( \dfrac{3}{2} \right) \right)-\hat{j}\left( \dfrac{-3\times 3}{2}-\dfrac{-1}{2}\times 0 \right)+\hat{k}\left( \dfrac{3}{2}\times \dfrac{3}{2}-\dfrac{-1}{2}\times \dfrac{1}{2} \right) \\
 & =\vec{i}\left( \dfrac{-3}{2} \right)-\hat{j}\left( \dfrac{-9}{2} \right)+\hat{k}\left( \dfrac{9}{4}-\dfrac{-1}{4} \right) \\
 & =-\left( \dfrac{3}{2} \right)\vec{i}+\left( \dfrac{9}{2} \right)\hat{j}+\hat{k}\left( \dfrac{10}{4} \right) \\
 & =-\dfrac{3}{2}\hat{i}+\dfrac{9}{2}\hat{j}+\dfrac{5}{2}\hat{k} \\
\end{align}\]
Hence the value of ${{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}=\dfrac{1}{2}\left( -3\hat{i}+9\hat{j}+5\hat{k} \right)$
Option c is the correct option

Note:
Now note that in vector we have two different products. Dot product and cross product. When we have the vectors are perpendicular dot product is 0. This is because dot profucto of vectors \[\vec{a}\] and $\vec{b}$ is given by $|\vec{a}||\vec{b}|\cos \theta $ and we know that cos 90 = 0. Hence we get dot product of two perpendicular vectors as 0.