Question

Find the equation of the plane passing through the line of intersection of the planes ${\displaystyle \overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})}$ = 1 and ${\displaystyle \overrightarrow{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})+4}$ = 0 and parallel to x-axis

Options

• ${\displaystyle \overrightarrow{r}\cdot (\hat{j}-3\hat{k})+6}$ = 0

• ${\displaystyle \overrightarrow{r}\cdot (\hat{i}-2\hat{j}-3\hat{k})}$ = 14

• ${\displaystyle \overrightarrow{r}\cdot (2\hat{i}+\hat{j}-\hat{k})}$ = – 5

• ${\displaystyle \overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})-1}$ = 0

Answer 1

${\displaystyle \overrightarrow{r}\cdot (\hat{j}-3\hat{k})+6}$ = 0

Explanation:

The two given plane are

${\displaystyle \overrightarrow{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})+4}$ = 0 and ${\displaystyle \overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})-1}$ = 0

The equation of the plane that passes through the intersection of these planes.

${\displaystyle \overrightarrow{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})+4+\lambda [\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})-1]}$ = 0

or ${\displaystyle \overrightarrow{r}\cdot [(2+\lambda )\hat{i}+(3+\lambda )\hat{j}+(-1+\lambda )\hat{k}]+4-\lambda }$ = 0

⇒ The x-axis is perpendicular to the plane (1)

⇒ Its normal is parallel to the x-axis.

Direction ratios of normal are ${\displaystyle 2+\lambda ,3+\lambda ,-1+\lambda }$

Direction ratios of x-axis are 1, 0, 0

The plane's normal and the x-axis are perpendiculars.

${\displaystyle (2+\lambda )\times 1+(3+\lambda )\times 0+(-1+\lambda )\times 0}$ = 0

or ${\displaystyle 2+\lambda }$ = 0

∴ ${\displaystyle \lambda =-2}$

Putting ${\displaystyle \lambda =-2}$ in (1)

${\displaystyle \overrightarrow{r}\cdot [(2-2)\hat{i}+(3-2)\hat{j}+(-1-2)\hat{k}]+4+2}$ = 0

∴ The required plane equation is ${\displaystyle \overrightarrow{r}\cdot (\hat{j}-3\hat{k})+6}$ = 0.