Question

Vector equation of the plane $\overrightarrow{r}=\hat{i}-\hat{j}+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})$ in the scalar dot product form is

• A. $\overrightarrow{r}.(5\hat{i}+2\hat{j}-3\hat{k})=7$
• B. $\overrightarrow{r}.(5\hat{i}-2\hat{j}+3\hat{k})=7$
• C. $\overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7$
• D. $\overrightarrow{r}.(5\hat{i}+2\hat{j}+3\hat{k})=17$

Answer 1

The correct option is C $\overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7$

We have been given a plane which passes through three points namely, $P=(1,-1,0),Q=(2,0,1),R=(2,-3,3)$ substituting $0,1$ for $\lambda ,\mu .$

$\overrightarrow{PQ}=\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{PR}=\hat{i}-2\hat{j}+3\hat{k}$

Normal vector $=\overrightarrow{PQ}\times \overrightarrow{PR}$

$=(\overline{i}-\overline{j}+\overline{k})\times (\overline{i}-2\overline{j}+3\overline{k})$

$=-2\overline{k}-3\overline{j}-\overline{k}+3\overline{i}+\overline{j}+2\overline{i}$

$=5\overline{i}-2\overline{j}-3\overline{k}$

Now, $5(x-1)-2(y+1)-3(z-0)=0$ represents the equation of plane.
$\Rightarrow 5x-2y-3z=7$ or $\overline{r}.(5\overline{i}-2\overline{j}-3\overline{k})=0$