Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the given line is $\bar{r}=3\hat{i}-2\hat{j}+\hat{k}+\lambda (\hat{i}-2\hat{j})$.
The direction vector of the line is $\vec{b}=\hat{i}-2\hat{j}$.

The equation of the given plane is $\bar{r}\cdot (2\hat{i}+\hat{j}+\hat{k})=4$.
The normal vector to the plane is $\vec{n}=2\hat{i}+\hat{j}+\hat{k}$.

Taking the dot product of $\vec{b}$ and $\vec{n}$:
$\vec{b}\cdot \vec{n}=(1)(2)+(-2)(1)+(0)(1)=2-2=0$
Since $\vec{b}\cdot \vec{n}=0$, the line is parallel to the plane.

The distance between the line and the plane is the perpendicular distance from any point on the line to the plane. A point on the line is given by its position vector $\vec{a}=3\hat{i}-2\hat{j}+\hat{k}$, which corresponds to the coordinates $(3,-2,1)$.

The Cartesian equation of the plane is $2x+y+z-4=0$.

The perpendicular distance $d$ from the point $(3,-2,1)$ to the plane is:
$d=\frac{|2(3)+1(-2)+1(1)-4|}{\sqrt{2^2+1^2+1^2}}$

$d=\frac{|6-2+1-4|}{\sqrt{4+1+1}}$

$d=\frac{|1|}{\sqrt{6}}=\frac{1}{\sqrt{6}}\text{ units}$

Answer: $\frac{1}{\sqrt{6}}$ units