Question

Find the value of $\lambda$, if four points with position vectors $3\hat{i}+6\hat{j}+9\hat{k},\hat{i}+2\hat{j}+3\hat{k},2\hat{i}+3\hat{j}+\hat{k}$ and $4\hat{i}+6\hat{j}+\lambda \hat{k}$ are coplanar.

Answer 1

Given $4$ vectors $3\hat{i}+6\hat{j}+9\hat{k}=OA$

$\hat{i}+2\hat{j}+3\hat{k}=OB$

$2\hat{i}+3\hat{j}+\hat{k}=OC$

$4\hat{i}+6\hat{j}+\lambda \hat{k}=OD$

$\overline{AB}=OB-OA$

$=(1-3)\hat{i}+(2-6)\hat{j}+(3-9)\hat{k}$

$=-2\hat{i}-4\hat{j}-6\hat{k}$

$\overline{AC}=OC-OA$

$=(2-3)\hat{i}+(3-6)\hat{j}+(1-9)\hat{k}$

$=-\hat{i}+(-3)\hat{j}-8\hat{k}$

$\overline{AD}=OD-OA$

$=(4-3)\hat{i}+(6-6)\hat{j}+(\lambda -9)\hat{k}$

$=\hat{i}+o\hat{j}+(\lambda -9)\hat{k}$

$\left|\begin{array}{ccc}-2& -4& -6\\ -1& -3& -8\\ 1& 0& \lambda -9\end{array}\right|=0$

$-2[-3(\lambda -9)+0]+4[-(\lambda -9)+8]-6[0+3]=0$

$-2[-3x+27]+4[-\lambda +9+8]-6\left(3\right)=0$

$6\lambda +(-54)-4\lambda +68-18=0$

$+2\lambda -4=0$

$2\lambda =4$

$\lambda =2$