Question

The equation of the plane which bisects the line joining $(3,0,5)$ and $(1,2,-1)$ at right angles is

• A. $2x+y+2z=7$
• B. $-2x+2y-6z=7$
• C. $x-y+2z=87$
• D. $x-y+3z=7$

Answer 1

The correct option is D $x-y+3z=7$

The given points are $A(3,0,5)$ and $B(1,2,-1)$

The line segment $AB$is given by $(x_2-x_1,y_2-y_1,z_2-z_1)=(-2,2,-6)=(-1,1,-3)$

Since the plane bisects $AB$ at right angles, $\overrightarrow{AB}$ is the normal to the plane which is $\overrightarrow{n}$ .

Therefore $\overrightarrow{n}=-\overrightarrow{i}+\overrightarrow{j}-3\overrightarrow{k}$

Let $C$ be the midpoint of $AB$.

$C=(\frac{3+1}{2},\frac{0+2}{2},\frac{5-1}{2})$

$\Rightarrow C=(2,1,2)$

Let $\overrightarrow{a}=2\overrightarrow{i}+\overrightarrow{j}+2\overrightarrow{k}$

Hence the equation of the plane is given by $(\overrightarrow{r}-\overrightarrow{a})\cdot \overrightarrow{n}=0$

We know that $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$

$\Rightarrow \left(\right(x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k})-(2\overrightarrow{i}+\overrightarrow{j}+2\overrightarrow{k}\left)\right)\cdot (-\overrightarrow{i}+\overrightarrow{j}-3\overrightarrow{k})=0$

$\Rightarrow \left(\right(x-2)\overrightarrow{i}+(y-1)\overrightarrow{j}+(z-2\left)\overrightarrow{k}\right)\cdot (-\overrightarrow{i}+\overrightarrow{j}-3\overrightarrow{k})$=0

$\Rightarrow -(x-2)+(y-1)-3(z-2)=0$

$\Rightarrow -x+2+y-1-3z+6=0$

$\Rightarrow x-y+3x=7$

Hence the required equation of plane is $x-y+3z=7$