Question

Given 3 points given by position vectors $\overline{a}=\hat{i}+\hat{j},\overline{b}=\hat{j}+\hat{k},\overline{c}=-\hat{i}-\hat{j}-\hat{k}$. Find the plane passing through these 3 points.

• A. $\overline{r}.(\overline{i}-3\hat{j}+\hat{k})=-1$
• B. $\overline{r}.(-\overline{i}+3\hat{j}-\hat{k})=1$
• C. $\overline{r}.(\overline{i}-3\hat{j}+\hat{k})=2$
• D. $\overline{r}.(-\overline{i}+3\hat{j}-\hat{k})=-2$

Answer 1

Answer: (A), (B)

The correct options are
A $\overline{r}.(\overline{i}-3\hat{j}+\hat{k})=-1$
B $\overline{r}.(-\overline{i}+3\hat{j}-\hat{k})=1$

First we have to find the vector perpendicular to the plane. It will be along $(\overline{a}-\overline{b})\times (\overline{b}-\overline{c})$
$(\overline{a}-\overline{b})\times (\overline{b}-\overline{c})=\left(\right(\hat{i}+\hat{j})-(\hat{j}+\hat{k})\times (\hat{j}+\hat{k})-(-\hat{i}-\hat{j}-\hat{k}\left)\right)=(\hat{i}-\hat{k})\times (\hat{i}+2\hat{j}+2\hat{k})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& -1\\ 1& 2& 2\end{array}\right|=\hat{i}(0+2)-\hat{j}(2+1)+\hat{k}(2-0)=2\hat{i}-3\hat{j}+2\hat{k}\Rightarrow \hat{n}ordirectionperpendiculartotheplaneneededis\frac{2\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{17}}$
Without loss of generality we will choose $\overline{a}$ to be the point on the plane. Therefore equation of the plane will be $(\overline{r}-\overline{a}).\hat{n}=0$
$(\overline{r}-(\hat{i}+\hat{j}\left)\right).\hat{n}=0(\overline{r}-(\hat{i}+\hat{j}\left)\right).\frac{(2\hat{i}-3\hat{j}+2\hat{k})}{\sqrt{17}}=0\overline{r}.(\overline{2i}-3\hat{j}+2\hat{k})=(\hat{i}+\hat{j}).(2\hat{i}-3\hat{j}+2\hat{k})\overline{r}.(2\hat{i}-3\hat{j}+2\hat{k})=2-3=-1\therefore \overline{r}.(\overline{i}-3\hat{j}+\hat{k})=-1$
This can also be written as $\overline{r}.(-\overline{i}+3\hat{j}-\hat{k})=1$
Therefore both a and b are correct.