Question

Consider three vectors $\overrightarrow{P}=\hat{i}+\hat{j}+\hat{k};\overrightarrow{q}=2\hat{i}+4\hat{j}-\hat{k}$ and $\overrightarrow{r}=2\hat{i}+4\hat{j}+3\hat{k}$. If $\overrightarrow{p},\overrightarrow{q}$ and $\overrightarrow{r}$ denotes the position vector of three non-collinear points, then the equation of the plane containing these points is

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

Answer 1

The correct option is D $3x-y-2=0$

Equation of plane passing through $\overrightarrow{p}=\hat{i}+\hat{j}+\hat{k}$ is given by,
$a(x-1)+b(y-1)+c(z-1)=0$
Given plane containing $\overrightarrow{q}=2\hat{i}+4\hat{j}-1\hat{k}$
$\Rightarrow a+3b-2c=\mathrm{0......}\left(1\right)$
it also containing $\overrightarrow{r}=2\hat{i}+4\hat{j}+3\hat{k}$
$\Rightarrow a+3b+3c=\mathrm{0.....}\left(2\right)$
Solving (1) and (2), we get $c=0,a=-3b$
Hence, required plane is
$3x-y-2=0$