Question

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

• A. $0$
• B. $\frac{37}{4}$
• C. $\frac{-37}{4}$
• D. $1$

Answer 1

The correct option is C $\frac{-37}{4}$

Let $O$ be the fixed point.

$\overrightarrow{OA}=$ position vector of $A=3\hat{i}–2\hat{j}–\hat{k}=(3,-2,-1)$

$\overrightarrow{OB}=$ position vector of $B=2\hat{i}+3\hat{j}–4\hat{k}=(2,3,-4)$

$\overrightarrow{OC}=$ position vector of $C=-\hat{i}+2\hat{j}+2\hat{k}=(-1,2,2)$

$\overrightarrow{OD}=$ position vector of $D=4\hat{i}+5\hat{j}+\lambda \hat{k}=(4,5,\lambda )$

$\overrightarrow{AB}=\overrightarrow{OB}–\overrightarrow{OA}=(-1,5,-3)$

$\overrightarrow{AC}=\overrightarrow{OC}–\overrightarrow{OA}=(-4,4,3)$

$\overrightarrow{AD}=\overrightarrow{OD}–\overrightarrow{OA}=(1,7,\lambda +1)$

Since $A,B,C,D$ are coplanar , $\left[\overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD}\right]=0$

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

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

$\Rightarrow 1(-12–15)–7(12+3)+(\lambda +1)(-20+4)=0$

$\Rightarrow -27-105-16\lambda –16=0$

$\Rightarrow -148–16\lambda =0$

$\Rightarrow 16\lambda =-148$

$\therefore \lambda =-\frac{148}{16}=-\frac{37}{4}$