Question

If the points whose position vectors are $3\overline{i}-2\overline{j}-\overline{k},2\overline{i}+3\overline{j}-4\overline{k},-\overline{i}+\overline{j}+2\overline{k}$ and $4\overline{i}+5\overline{j}+\lambda \overline{k}$ are coplanar then show that $\lambda =\frac{-146}{m}$.Find $m$

Answer 1

$\begin{array}{c}\overrightarrow{OA}=3\hat{i}-2\hat{j}-\hat{k}\\ \overrightarrow{OB}=2\hat{i}+3\hat{j}-4\hat{k}\\ \overrightarrow{OC}=-\hat{i}+\hat{j}+2\hat{k}\\ \overrightarrow{OD}=4\hat{i}+5\hat{j}+\lambda \hat{k}\end{array}$

$\begin{array}{c}\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-\hat{i}+5\hat{j}-3\hat{k}\\ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=-4\hat{i}+3\hat{j}+3\hat{k}\\ \overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}=\hat{i}+7\hat{j}+\left(\lambda +1\right)\hat{k}\end{array}$

Since $A,B,C,D$ are co-planar

$\left|\begin{array}{ccc}-1& 5& -3\\ -4& 3& 3\\ 1& 7& \lambda +1\end{array}\right|=0$

$C_2\to C_2+5C_1$

$C_3\to C_3-3C_1$

$\left|\begin{array}{ccc}-1& 0& 0\\ -4& -17& 15\\ 1& 12& \lambda -2\end{array}\right|=0$

Expanding along $R_1$, we get

$\lambda =-\frac{-146}{17}$

On comparing $\lambda$ with $\frac{-146}{m}$

we get, $m=17$