Question

If $\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+2\hat{j}-3\hat{k},\overrightarrow{c}=3\hat{i}+\mu \hat{j}+5\hat{k}$ are coplanar, then $\mu$ is a root of the equation

• A. $x^2+3x=4$
• B. $x^2+2x=6$
• C. $x^2+3x=6$
• D. None of these

Answer 1

Answer: (A)

$x^2+3x=4$

lf $\overline{a}=2\overline{i}-\overline{j}+\overline{k}$, $\overline{b}=\overline{i}+2\overline{j}-3\overline{k},$ $\overline{c}=3\overline{i}+\mu \overline{j}+5\overline{k}$ are coplanar then,

$\left[\overline{a}\overline{b}\overline{c}\right]=0$

$\left|\begin{array}{ccc}2& -1& 1\\ 1& 2& -3\\ 3& \mu & 5\end{array}\right|=0$

$\therefore 0=2(10+3\mu )+1(5+9)+1(\mu -6)$
$\Rightarrow 0=20+6\mu +14+\mu -6$
$\Rightarrow 0=28+7\mu$
$\Rightarrow \mu =-4$

Which is the root of $x^2+3x-4=0$