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 $P Q$ 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 $DR's \,(5,3,-11)$ is
$ 5\left(x-\frac{3}{2}\right)+3\left(y-\frac{7}{2}\right)-11\left(z+\frac{9}{2}\right)=0 $
$\Rightarrow 5 x+3 y-11 z=\frac{15}{2}+\frac{21}{2}+\frac{99}{2} $
$\Rightarrow 5 x+3 y-11 z=\frac{135}{2}$
It is written in vector form
$r \cdot(5 \hat{ i }+3 \hat{ j }-11 \hat{ k })=\frac{135}{2}$