Question

Equation of the plane passing through three points $A$, $B$, $C$ with position vectors $-6\hat{i}+3\hat{j}+2\hat{k}$, $3\hat{i}-2\hat{j}+4\hat{k}$, $5\hat{i}+7\hat{j}+3\hat{k}$ is

• A. $\vec{r}\cdot\left(\hat{i}-\hat{j}-7\hat{k}\right)+23=0$
• B. $\vec{r}\cdot\left(\hat{i}+\hat{j}+7\hat{k}\right)=23$
• C. $\vec{r}\cdot\left(\hat{i}+\hat{j}-7\hat{k}\right)+23=0$
• D. $\vec{r}\cdot\left(\hat{i}-\hat{j}-7\hat{k}\right)=23$

Answer 1

Answer: (A)

Equation of the plane passing through three points $A$, $B$, $C$ with position vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ is
$\left(\vec{r}-\vec{a}\right)\cdot\left[\left(\vec{b}-\vec{a}\right)\times\left({\vec{c}-\vec{a}}\right)\right]=0$
We have, $\left(\vec{b}-\vec{a}\right)=\left(3\hat{i}-2\hat{j}+4\hat{k}\right)-\left(-6\hat{i}+3\hat{j}+2\hat{k}\right)$
$=9\hat{i}-5\hat{j}+2\hat{k}$
$\left(\vec{c}-\vec{a}\right)=\left(5\hat{i}+7\hat{j}+3\hat{k}\right)-\left(6\hat{j}+3\hat{j}+2\hat{k}\right)$
$=11\hat{i}+4\hat{j}+\hat{k}$
Now, $\left(\vec{b}-\vec{a}\right)\times\left(\vec{c}-\vec{a}\right)=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 9&-5&2\\ 11&4&1\end{vmatrix}=-13\hat{i}+13\hat{j}+91\hat{k}$
$\therefore $ Equation of plane is
$\left[\vec{r}-\left(-6\hat{i}+3\hat{j}+2\hat{k}\right)\right]\cdot\left[-13\hat{i}+13\hat{j}+91\hat{k}\right]=0$
$\Rightarrow \vec{r}\cdot\left(-13\hat{i}+13\hat{j}+91\hat{k}\right)-\left(78+39+182\right)=0$
$\Rightarrow \vec{r}\cdot\left(-13\hat{i}+13\hat{j}+91\hat{k}\right)=299$
$\Rightarrow \vec{r}\cdot\left(\hat{i}-\hat{j}-7\hat{k}\right)=-23$
$\Rightarrow \vec{r}\cdot\left(\hat{i}-\hat{j}-7\hat{k}\right)+23=0$