Question

Find the value of $x$ for which the points $A(3,2,1),B(4,x,5),C(4,2,-2)$ and D(6,5,-1) are coplanar.

Answer 1

We have A($3,2,1$); B($4,x,5$); C($4,2,—2$) and D($0,5,—1$) as the given points. AD now,

$\overrightarrow{AB}$= position vector of B — position vector of A

=$\left(4\hat{i}+x\hat{j}+5\hat{k}\right)-\left(3\hat{i}+2\hat{j}+\hat{k}\right)=\hat{i}+(x-2)\hat{j}+4\hat{k}$

$\overrightarrow{AC}$ = position vector of C — position vector of A

= $\left(4\hat{i}+2\hat{j}-2\hat{k}\right)-\left(3\hat{i}+2\hat{j}+\hat{k}\right)=\hat{i}+0\hat{j}-3\hat{k}$

$\overrightarrow{AD}$ = position vector of D — position vector of A

=$\left(6\hat{i}+5\hat{j}-\hat{k}\right)-\left(3\hat{i}+2\hat{j}+\hat{k}\right)=3\hat{i}+3\hat{j}-2\hat{k}$

Since, A. B. C and D arc coplanar. then

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

$\left|\begin{array}{ccc}1& (x-2)& 4\\ 1& 0& -3\\ 3& 3& -2\end{array}\right|=0$

$\Rightarrow 1\left[0+9\right]-(x-2)\left[-2+9\right]+4\left[3-0\right]=0$

$\Rightarrow 9-7(x-2)+4\times 3=0$

$\Rightarrow 7x=35$

$\Rightarrow x=5$