Question

If the vectors $3\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$, $\overrightarrow{i}+2\overrightarrow{j}-3\overrightarrow{k}$, $3\overrightarrow{i}+\lambda \overrightarrow{j}+5\overrightarrow{k}$ are co-planar, the value of $\lambda$ is

• A. $-4$
• B. $4$
• C. $8$
• D. $-8$

Answer 1

Answer: (A)

$-4$

When three vectors are co-planar, their scalar triple product is always zero.

For $3i+j-2k,i+2j-3k,3i+\lambda j+5k$, triple product is

$\left|\begin{array}{ccc}3& 1& -2\\ 1& 2& -3\\ 3& \lambda & 5\end{array}\right|=0\Rightarrow 3(2\times 5-(-3\left)\lambda \right)-1(5-(-3\left)3\right)+(-2)(\lambda -2\times 3)=0\Rightarrow 3(10+3\lambda )-(5+9)-2(\lambda -6)=0\Rightarrow 30+9\lambda -14-2\lambda +12=0\Rightarrow 28+7\lambda =0\Rightarrow \lambda =-4$