Question

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

• A. $8x-3y-z-5=0$
• B. $x-3y+2z=0$
• C. $3x-y+z=0$
• D. $3x-y-2=0$

Answer 1

The correct option is A $8x-3y-z-5=0$

Equation of plane passing through $\overrightarrow{p}=\hat{i}+\hat{j}$ is given by,
$a(x-1)+b(y-1)+cz=0$
Given plane containing $\overrightarrow{q}=2\hat{i}+4\hat{j}-1\hat{k}$
$\Rightarrow a+3b-c=\mathrm{0......}\left(1\right)$
It also has $\overrightarrow{r}=\hat{i}+3\hat{k}$
$\Rightarrow -b+3c=\mathrm{0.....}\left(2\right)$
Solving (1) and (2), we get $a=-8c,b=3c$
Hence, required plane is

$-8c(x-1)+3c(y-1)+cz=0$

$\Rightarrow -8x+8+3y-3+z=0$

$\Rightarrow 8x-3y-z-5=0$