Question

$a=-\hat{i}+\hat{j}+\hat{k}$ and $b=2\hat{i}+\hat{j}+\hat{k}$, then find the vector $c$ satisfying the conditions

$\left(i\right)$ that it is coplanar with $a$ and $b$

$\left(ii\right)$ that it is perpendicular to $b$

$(iii$) $a.c=7$

• A. $-\frac{3}{2}\hat{i}+\frac{5}{2}\hat{j}+3\hat{k}$
• B. $\frac{7(-\hat{i}+\hat{j}+\hat{k})}{3}$
• C. $-6\hat{i}+0\hat{j}+\hat{k}$
• D. $-\hat{i}+2\hat{j}+2\hat{k}$

Answer 1

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

Let $\overrightarrow{c}=p\hat{i}+q\hat{j}+r\hat{k}$
$\overrightarrow{a}\cdot \overrightarrow{c}=7\Rightarrow -p+q+r=7$
$\overrightarrow{c}\text{is perpendicular to}\overrightarrow{b}\Rightarrow 2p+q+r=0$
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, $q=r=-p=\frac{7}{3}$
$\therefore \overrightarrow{c}=-\frac{7}{3}\hat{i}+\frac{7}{3}\hat{j}+\frac{7}{3}\hat{k}$