Question

For given vector equation of the plane $\overrightarrow{r}=\hat{i}-\hat{j}+\lambda (\hat{i}-\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k}),$ the cartesian equation of the plane is

• A. $x-2y+z+1=0$
• B. $x+2y+z+1=0$
• C. $x+2y+2z-3=0$
• D. $x-2y+3z-4=0$

Answer 1

The correct option is B $x+2y+z+1=0$

Given plane :
$\overrightarrow{r}=\overrightarrow{i}-\overrightarrow{j}+\lambda (\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})$
Plane passes through $(1,-1,0)$
Plane is parallel to $\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-2\hat{j}+3\hat{k}$
Normal vector to plane is:
$(\hat{i}-\hat{j}+\hat{k})\times (\hat{i}-2\hat{j}+3\hat{k})$
$=-\hat{i}-2\hat{j}-\hat{k}$
So, $D.r^{\prime }s$ of normal to the plane is $(-1,-2,-1)$
So, equation of plane is :
$-1(x-1)-2(y+1)-1(z-0)=0\Rightarrow x+2y+z+1=0$