Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The position vector of the point $A(-1,2,0)$ is $\vec{a}=-\hat{i}+2\hat{j}$.

The plane passes through $A(-1,2,0)$ and $B(2,2,-1)$. A vector parallel to the plane is $\overrightarrow{AB}=(2-(-1))\hat{i}+(2-2)\hat{j}+(-1-0)\hat{k}=3\hat{i}-\hat{k}$.

The given line is $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$. Rewriting it in standard form, we get $\frac{x-1}{1}=\frac{y+1/2}{1}=\frac{z+1}{-1}$.

The direction vector of the line is $\vec{v}=\hat{i}+\hat{j}-\hat{k}$. Since the plane is parallel to this line, $\vec{v}$ is also parallel to the plane.

The parametric vector equation of a plane passing through a point with position vector $\vec{a}$ and parallel to two non-collinear vectors $\vec{b}$ and $\vec{c}$ is given by $\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$.

Substituting $\vec{a}=-\hat{i}+2\hat{j}$, $\vec{b}=\overrightarrow{AB}=3\hat{i}-\hat{k}$, and $\vec{c}=\vec{v}=\hat{i}+\hat{j}-\hat{k}$, we get:

$\vec{r}=(-\hat{i}+2\hat{j})+\lambda (3\hat{i}-\hat{k})+\mu (\hat{i}+\hat{j}-\hat{k})$

Answer: $\bar{r}=(-\hat{i}+2\hat{j})+\lambda (3\hat{i}-\hat{k})+\mu (\hat{i}+\hat{j}-\hat{k})$