The distance of the point with position vector $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $\vec{r} . (-2 \hat{i}+\hat{j}+3 \hat{k})=8$ is :
Answer & explanation
Correct answer: option 3
Point → $(2 \hat{i}+\hat{j}-\hat{k}) ≡ P(2, 1, -1)$
Plane → $\vec{r}(-2 \hat{i} + \hat{j} + 3\hat{k}) = 8$
⇒ $(x \hat{i}+y \hat{i}+z \hat{k})(-2 \hat{i}+\hat{j}+3 \hat{k})-8=0$
$\Rightarrow -2 x+y+3 z-8=0$ → equation of plane
as for point (x0, y0, z0) and plane
Ax + By + Cz + D = 0
$d=\frac{\left|A x_0+B y_0+C z_0+D\right|}{\sqrt{A^2+B^2+C^2}}$
so distance
$=\frac{|-2(2)+(1)+3(-1)-8|}{\sqrt{(-2)^2+1^2+3^2}}$
$=\frac{14}{\sqrt{14}}$
$=\sqrt{14}$