Question

$\text{If}\overrightarrow{a}=2\hat{i}+6\hat{j}+27\hat{k};\overrightarrow{b}=\hat{i}+\lambda \hat{j}+\mu \hat{k}$ and $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{0},$ then $(\lambda ,\mu )=$

• A. $(6,27)$
• B. $(27,6)$
• C. $(3,\frac{27}{2})$
• D. $(\frac{27}{2},3)$

Answer 1

The correct option is C $(3,\frac{27}{2})$

$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{0}$

$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 6& 27\\ 1& \lambda & \mu \end{array}\right|=\overrightarrow{0}$
$\Rightarrow (6\mu -27\lambda )\hat{i}-(2\mu -27)\hat{j}+(2\lambda -6)\hat{k}=\overrightarrow{0}$

Equating the coefficients, we get
$\mu =\frac{27}{2},\lambda =3$