Question

Find a vector $\overrightarrow{c}$, such that $\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{b}$ and $\overrightarrow{a}.\overrightarrow{c}=3$ if $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{j}-\hat{k}$.

Answer 1

We start solving this problem by going through the concept of dot product, cross product and vector triple product. First, we apply the cross product to $\overrightarrow{a}\times \overrightarrow{c}$ and $\overrightarrow{b}$, and we use the vector triple product formula $\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c}$. Then, we find the cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$, and equate them. Then we solve the obtained equation to find the vector $\overrightarrow{c}$.

Complete step by step answer:
For solving this problem, we need to go through the concept of dot product and the cross product of two vectors and also vector triple product.
Dot product of vectors $\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and $\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ can be given as,
$~\overrightarrow{a}.\overrightarrow{b}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}$
$\left( \overrightarrow{a}\times \overrightarrow{b} \right)$ is the cross product of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, and can be given as,
$\left( \overrightarrow{a}\times \overrightarrow{b} \right)=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ \end{matrix} \right|$
The vector triple product of three vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ is given by
$\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c}$
We are given that $\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{b}$ and $\overrightarrow{a}.\overrightarrow{c}=3$.
Now let us consider, $\overrightarrow{a}\times \left( \overrightarrow{a}\times \overrightarrow{c} \right)$. Then we get,
$\overrightarrow{a}\times \left( \overrightarrow{a}\times \overrightarrow{c} \right)=\overrightarrow{a}\times \overrightarrow{b}............\left( 1 \right)$
Using the triple vector product, we get that
$\overrightarrow{a}\times \left( \overrightarrow{a}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{a}-\left( \overrightarrow{a}.\overrightarrow{a} \right)\overrightarrow{c}$
Now using the formula for dot product and as $\overrightarrow{a}.\overrightarrow{c}=3$, we get
$\begin{aligned} & \Rightarrow \overrightarrow{a}\times \left( \overrightarrow{a}\times \overrightarrow{c} \right)=3\overrightarrow{a}-\left( 1\left( 1 \right)+1\left( 1 \right)+1\left( 1 \right) \right)\overrightarrow{c} \\\ & \Rightarrow \overrightarrow{a}\times \left( \overrightarrow{a}\times \overrightarrow{c} \right)=3\overrightarrow{a}-3\overrightarrow{c}..............\left( 2 \right) \\\ \end{aligned}$
Now let us find the value of $\overrightarrow{a}\times \overrightarrow{b}$.
Using the formula for cross product we get,
$\begin{aligned} & \Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 1 & 1 & 1 \\\ 0 & 1 & -1 \\\ \end{matrix} \right| \\\ & \Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)=\left| \begin{matrix} 1 & 1 \\\ 1 & -1 \\\ \end{matrix} \right|\hat{i}-\left| \begin{matrix} 1 & 1 \\\ 0 & -1 \\\ \end{matrix} \right|\hat{j}+\left| \begin{matrix} 1 & 1 \\\ 0 & 1 \\\ \end{matrix} \right|\hat{k} \\\ & \Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)=\left( -1-1 \right)\hat{i}-\left( -1-0 \right)\hat{j}+\left( 1-0 \right)\hat{k} \\\ & \Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)=-2\hat{i}+\hat{j}+\hat{k}..............\left( 3 \right) \\\ \end{aligned}$
Using equations (2) and (3) and substituting them in equation (1), we get
$\Rightarrow 3\overrightarrow{a}-3\overrightarrow{c}=-2\hat{i}+\hat{j}+\hat{k}$
Now, let us substitute the vector $\overrightarrow{a}$ in it, then we get
$\begin{aligned} & \Rightarrow 3\left( \hat{i}+\hat{j}+\hat{k} \right)-3\overrightarrow{c}=-2\hat{i}+\hat{j}+\hat{k} \\\ & \Rightarrow 3\hat{i}+3\hat{j}+3\hat{k}-3\overrightarrow{c}=-2\hat{i}+\hat{j}+\hat{k} \\\ & \Rightarrow 3\overrightarrow{c}=3\hat{i}+3\hat{j}+3\hat{k}-\left( -2\hat{i}+\hat{j}+\hat{k} \right) \\\ & \Rightarrow 3\overrightarrow{c}=3\hat{i}+3\hat{j}+3\hat{k}+2\hat{i}-\hat{j}-\hat{k} \\\ & \Rightarrow 3\overrightarrow{c}=5\hat{i}+2\hat{j}+2\hat{k} \\\ & \Rightarrow \overrightarrow{c}=\dfrac{5}{3}\hat{i}+\dfrac{2}{3}\hat{j}+\dfrac{2}{3}\hat{k} \\\ \end{aligned}$
So, we get that the vector $\overrightarrow{c}$ is $\dfrac{5}{3}\hat{i}+\dfrac{2}{3}\hat{j}+\dfrac{2}{3}\hat{k}$.

Hence, answer is $\dfrac{5}{3}\hat{i}+\dfrac{2}{3}\hat{j}+\dfrac{2}{3}\hat{k}$.

Note: There is possibility of making a mistake while taking the vector triple product as $\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a}$ which is actually the formula for $\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}$. So, one needs to be careful while applying the vector triple product while solving the problem.