Question

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

Answer 1

Given,

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{j}-\hat{k}$

Let $\overrightarrow{c}=x\hat{i}+y\hat{j}+z\hat{k}$

Then, $\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{b}$

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ x& y& z\end{array}\right|=\hat{j}-\hat{k}$

$\hat{i}(z-y)-\hat{j}(z-x)+\hat{k}(y-x)=\hat{j}-\hat{k}$

On comparing the coefficients of $\hat{i},\hat{j}$ and $\hat{k}$ from both sides, we get,

$z-y=0$ .......(1)

$x-z=1$ .......(2)

And, $x-y=1$ .......(3)

Also, $\overrightarrow{a}.\overrightarrow{c}=3$

$(\hat{i}+\hat{j}+\hat{k}).(x\hat{i}+y\hat{j}+z\hat{k})=3$

$x+y+z=3$ ......(4)

On adding equation 2 and 3, we get,

$2x-y-z=2$ ........(5)

On adding equation 4 and 5, we get,

$x=\frac{5}{3}$

Then, $y=\frac{2}{3}$ and $z=\frac{2}{3}$

Hence, $\overrightarrow{c}=\frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k})$