Question

If $\overrightarrow{a}=-\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}\text{and}\overrightarrow{b}=2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k},$ then the vector $\overrightarrow{c}$ satisfying the conditions
$1)\text{coplanar with}\overrightarrow{a}\text{and}\overrightarrow{b}$
$2)\text{perpendicular to}\overrightarrow{b}$
$3)\overrightarrow{a}\cdot \overrightarrow{c}=7$ is

• A. $-\frac{3}{2}\hat{i}+\frac{5}{2}\hat{j}+3\hat{k}$
• B. $-\frac{7}{3}\hat{i}+\frac{7}{3}\hat{j}+\frac{7}{3}\hat{k}$
• C. $-6\hat{i}+\hat{k}$
• D. $-\hat{k}$

Answer 1

The correct option is B $-\frac{7}{3}\hat{i}+\frac{7}{3}\hat{j}+\frac{7}{3}\hat{k}$

Let $\overrightarrow{c}=p\hat{i}+q\hat{j}+r\hat{k}$
Given $\overrightarrow{a}\cdot \overrightarrow{c}=7\Rightarrow -p+q+r=7$
Also$\overrightarrow{c}$ is perpendicular to $\overrightarrow{b}\Rightarrow 2p+q+r=0$
Since all three vectors are coplanar, $\left|\begin{array}{ccc}2& 1& 1\\ -1& 1& 1\\ p& q& r\end{array}\right|=0$
$\Rightarrow 2(r-q)-1(-r-p)+1(-q-p)=0\Rightarrow 2r-2q+r+p-q-p=3r-3q=0\Rightarrow q=r$
Using the previously obtained results, we have $p=-q$ and so, $q=r=-p=\frac{7}{3}$
$\therefore \overrightarrow{c}=-\frac{7}{3}\hat{i}+\frac{7}{3}\hat{j}+\frac{7}{3}\hat{k}$