Question

The vector equation of the plane passing through the points $\hat{i}+\hat{j}-2\hat{k},2\hat{i}-\hat{j}-\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$ is

• A. $\overline{r}.(9\hat{i}+3\hat{j}+\hat{k})=14$
• B. $\overline{r}.(9\hat{i}+3\hat{j}+\hat{k})+14=0$
• C. $\overline{r}.(9\hat{i}+3\hat{j}-\hat{k})=14$
• D. $\overline{r}.(9\hat{i}+3\hat{j}-\hat{k})+14=0$

Answer 1

The correct option is C $\overline{r}.(9\hat{i}+3\hat{j}-\hat{k})=14$

The plane passes through three given points
Let $A(\hat{i}+\hat{j}-2\hat{k}),B(2\hat{i}-\hat{j}-\hat{k}),C(\hat{i}+2\hat{j}+\hat{k})$
then $\overline{AB}=P.VofB-P.VofA=\hat{i}-2\hat{j}+3\hat{k}$
and $\overline{BC}=P.VofC-P.VofB=-\hat{i}+3\hat{j}+0\hat{k}$
Now any vector normal to the plane containing the points
$A,B,Cis\overline{n}=\overline{AB}\times \overline{BC}=\left|\begin{array}{ccc}i& j& k\\ 1& -2& 3\\ -1& 3& 0\end{array}\right|=-9\hat{i}-3\hat{j}+\hat{k}$
$\therefore (\overline{r}-\overline{a})\cdot \overline{n}=0$
$\Rightarrow \overline{r}\cdot \overline{n}=\overline{a}\cdot \overline{n}$
$\Rightarrow \overline{r}\cdot (-9\hat{i}-3\hat{j}+\hat{k})=(\hat{i}+\hat{j}-2\hat{k})\cdot (-9\hat{i}-3\hat{j}+\hat{k})$
$\Rightarrow \overline{r}\cdot (9\hat{i}+3\hat{j}-\hat{k})=14$
Hence choice (c) is the correct one.