Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The coordinates of the midpoint $M$ of the line segment joining $A(4,5,-10)$ and $B(-1,2,1)$ are given by:

$M=\left(\frac{4-1}{2},\frac{5+2}{2},\frac{-10+1}{2}\right)=\left(\frac{3}{2},\frac{7}{2},-\frac{9}{2}\right)$

The normal vector $\vec{n}$ to the plane is parallel to the vector $\overrightarrow{AB}$ since the plane is perpendicular to $AB$.

$\vec{n}=\overrightarrow{AB}=(-1-4)\hat{i}+(2-5)\hat{j}+(1-(-10))\hat{k}=-5\hat{i}-3\hat{j}+11\hat{k}$

The equation of the plane passing through $M$ with normal vector $\vec{n}$ is:

$(\bar{r}-\vec{m})\cdot \vec{n}=0$

$\bar{r}\cdot \vec{n}=\vec{m}\cdot \vec{n}$

Substituting the values of $\vec{m}$ and $\vec{n}$:

$\vec{m}\cdot \vec{n}=\left(\frac{3}{2}\hat{i}+\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k}\right)\cdot (-5\hat{i}-3\hat{j}+11\hat{k})$

$\vec{m}\cdot \vec{n}=\frac{3}{2}(-5)+\frac{7}{2}(-3)+\left(-\frac{9}{2}\right)(11)$

$\vec{m}\cdot \vec{n}=-\frac{15}{2}-\frac{21}{2}-\frac{99}{2}=-\frac{135}{2}$

Thus, the equation of the plane is:

$\bar{r}\cdot (-5\hat{i}-3\hat{j}+11\hat{k})=-\frac{135}{2}$

$\bar{r}\cdot (-5\hat{i}-3\hat{j}+11\hat{k})+\frac{135}{2}=0$

Answer: $\bar{r}\cdot \left(-5\hat{i}-3\hat{j}+11\hat{k}\right)+\frac{135}{2}=0$