Question

The scalar product form of equation of plane $\overline{r}=(s-2t)\hat{i}+(3-t)\hat{j}+(2s-t)\hat{k}$ is

• A. $\overrightarrow{r}\cdot (2\hat{i}-5\hat{j}-\hat{k})+15=0$
• B. $\overrightarrow{r}\cdot (2\hat{i}-5\hat{j}-\hat{k})=15$
• C. $\overrightarrow{r}\cdot (2\hat{i}-5\hat{j}-\hat{k})=3$
• D. $\overrightarrow{r}\cdot (2\hat{i}-5\hat{j}-\hat{k})=-3$

Answer 1

Answer: (A)

$\overrightarrow{r}\cdot (2\hat{i}-5\hat{j}-\hat{k})+15=0$

Given $\overrightarrow{r}=(s-2t)\hat{i}+(3-t)\hat{j}+(2s-t)\hat{k}$
$\Rightarrow \overrightarrow{r}=3\hat{j}+s(\hat{i}+2\hat{k})+t(-2\hat{i}+\hat{j}-\hat{k})$
This equation represent the equation ofthe plane through a point having position vector $\overrightarrow{a}=3\hat{j}$ parallel to the vectors
$\overrightarrow{b}=\hat{i}+2\hat{k}$ and $\overrightarrow{c}=-2\hat{i}-\hat{j}+2\hat{k}$.
Normal Vector to the plane is given by
$\overrightarrow{n}=|\overrightarrow{b}\times \overrightarrow{c}|=2\hat{i}-5\hat{j}-\hat{k}$
Now using scalar product form of equation of plane
$\overrightarrow{r}\cdot \overrightarrow{n}=\overrightarrow{a}\cdot \overrightarrow{n}$
$\Rightarrow \overrightarrow{r}\cdot (2\hat{i}-5\hat{j}-\hat{k})=\left(3\hat{j}\right)(2\hat{i}-5\hat{j}-\hat{k})$
$\Rightarrow \overrightarrow{r}\cdot (2\hat{i}-5\hat{j}-\hat{k})=-15$
Hence choice (a) is correct.