Question

The position vectors of the points $A,B,C$ and $D$ are $3\hat{i}-2\hat{j}-\hat{k},2\hat{i}+3\hat{j}-4\hat{k},-\hat{i}+\hat{j}+2\hat{k}$ and $4\hat{i}+5\hat{j}+\lambda \hat{k}$ respectively. If the points $A,B,C$ and $D$ lie on a plane, find the value of $\lambda$.

Answer 1

$\begin{array}{c}\overrightarrow{OA}=3\hat{i}-2\hat{j}-\hat{k}\\ \\ \overrightarrow{OB}=2\hat{i}+3\hat{j}-4\hat{k}\\ \\ \overrightarrow{OC}=-\hat{i}+\hat{j}+2\hat{k}\\ \\ \overrightarrow{OD}=4\hat{i}+4\hat{j}+\lambda \hat{k}\\ \\ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-\hat{i}+5\hat{j}-3\hat{k}\\ \\ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=-4\hat{i}+3\hat{j}+3\hat{k}\\ \\ \overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}=\hat{i}+7\hat{j}+\left(\lambda +1\right)\hat{k}\end{array}$

Given points will be coplanar, If

$\left[\overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD}\right]=0$

$\begin{array}{c}\left|\begin{array}{ccc}-1& 5& -3\\ -4& 3& 3\\ 1& 7& \lambda +1\end{array}\right|=0\\ \\ -1\left(3\lambda +3-21\right)-5\left(-4\lambda -4-3\right)-3\left(-28-3\right)=0\\ \\ -3\lambda +18+35+20\lambda +93=0\\ \\ 17\lambda +146=0\\ \\ \lambda =-\frac{146}{17}\end{array}$