Question

If the points $2 a +3 b - c , a -2 b +3 c$, $3 a +\lambda b -2 c$ and $a -6 b +6 c$ 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,
$2 a +3 b - c , a -2 b +3 c , 3 a +\lambda b -2 c$ and
$a -6 b +6 c$ are coplanar.
$\therefore \begin{vmatrix}1-2 & -2-3 & 3+1 \\ 3-2 & \lambda-3 & -2+1 \\ 1-2 & -6-3 & 6+1\end{vmatrix}=0$
$\Rightarrow \begin{vmatrix}-1 & -5 & 4 \\ 1 & \lambda-3 & -1 \\ -1 & -9 & 7\end{vmatrix}=0$
$\Rightarrow -1(7 \lambda-21-9)+5(7-1)+4(-9+\lambda-3)=0$
$\Rightarrow \,-7 \lambda+30+30-48+4 \lambda=0$
$\Rightarrow \, 3 \lambda=12$
$\Rightarrow \,\lambda=4$
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 $\frac{4}{9}, \frac{8}{9}, \frac{1}{9}$