Question

Equation of the plane through the mid-point of the line segment joining the points $P(4,5,-10)$ and $Q(-1,2,1)$ and perpendicular to $PQ$ is

• A. $r\cdot \left(\frac{3}{2}\hat{i}+\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k}\right)=45$
• B. $r\cdot \left(-\hat{i}+2\hat{j}+k\right)=\frac{135}{2}$
• C. $r\cdot \left(5\hat{i}+3\hat{j}-11k\right)+\frac{135}{2}=0$
• D. $r\cdot \left(4\hat{i}+5\hat{j}-10k\right)=85$
• E. $r\cdot \left(5\hat{i}+3\hat{j}-11k\right)=\frac{135}{2}$

Answer 1

Answer: (E)

Coordinate of mid-point of $P(4,5,-10)$ and $Q(-1,2,1)$ is
$\left(\frac{4-1}{2},\frac{5+2}{2},\frac{-10+1}{2}\right)\text{ i.e., }\left(\frac{3}{2},\frac{7}{2},\frac{-9}{2}\right)$
Now, $DR$'s of $PQ$ is $(-1-4,2-5,1+10)$
i.e.,$(-5,-3,11)$ or $(5,3,-11)$
$\therefore$ Equation of plane passing through
$\left(\frac{3}{2},\frac{7}{2},\frac{-9}{2}\right)$ and having $D{R}^{\prime }s(5,3,-11)$ is
$5\left(x-\frac{3}{2}\right)+3\left(y-\frac{7}{2}\right)-11\left(z+\frac{9}{2}\right)=0$
$\Rightarrow 5x+3y-11z=\frac{15}{2}+\frac{21}{2}+\frac{99}{2}$
$\Rightarrow 5x+3y-11z=\frac{135}{2}$
It is written in vector form
$r\cdot (5\hat{i}+3\hat{j}-11\hat{k})=\frac{135}{2}$