Question

If the points $2a+3b-c,a-2b+3c$, $3a+\lambda b-2c$ and $a-6b+6c$ are coplanar, then the direction cosines of the vector $\lambda \hat{i}-2\lambda \hat{j}+\hat{k}$ are

• A. $\frac{2}{\sqrt{21}},\frac{-4}{\sqrt{21}},\frac{1}{\sqrt{21}}$
• B. $-\frac{2}{\sqrt{78}},\frac{5}{\sqrt{78}},\frac{7}{\sqrt{78}}$
• C. $\frac{4}{9},\frac{8}{9},\frac{1}{9}$
• D. $\frac{2}{3},-\frac{2}{3},-\frac{1}{3}$

Answer 1

Answer: (C)

We have,
$2a+3b-c,a-2b+3c,3a+\lambda b-2c$ and
$a-6b+6c$ are coplanar.
$\therefore \left|\begin{array}{ccc}1-2& -2-3& 3+1\\ 3-2& \lambda -3& -2+1\\ 1-2& -6-3& 6+1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}-1& -5& 4\\ 1& \lambda -3& -1\\ -1& -9& 7\end{array}\right|=0$
$\Rightarrow -1(7\lambda -21-9)+5(7-1)+4(-9+\lambda -3)=0$
$\Rightarrow -7\lambda +30+30-48+4\lambda =0$
$\Rightarrow 3\lambda =12$
$\Rightarrow \lambda =4$
Direction cosine of $\lambda \hat{i}-2\lambda \hat{j}+\hat{k}$
is $\frac{\lambda }{\sqrt{5{\lambda }^2+1}},\frac{-2\lambda }{\sqrt{5{\lambda }^2+1}},\frac{1}{\sqrt{5{\lambda }^2+1}}$
Put $\lambda =4$
$\therefore$ Direction cosine is $\frac{4}{9},\frac{8}{9},\frac{1}{9}$