Question
Equation of the plane through the mid–point of the line segment joining the points $$P\left( {4,\,5,\, - 10} \right)$$ and $$Q\left( { - 1,\,2,\,1} \right)$$ and perpendicular to $$PQ$$ is :
A.
$$\overrightarrow r .\left( {\frac{3}{2}\hat i + \frac{7}{2}\hat j - \frac{9}{2}\hat k} \right) = 45$$
B.
$$\overrightarrow r .\left( { - \hat i + 2\hat j - \hat k} \right) = \frac{{135}}{2}$$
C.
$$\overrightarrow r .\left( {5\hat i + 3\hat j - 11\hat k} \right) + \frac{{135}}{2} = 0$$
D.
$$\overrightarrow r .\left( {5\hat i + 3\hat j - 11\hat k} \right) = \frac{{135}}{2}$$
Answer :
$$\overrightarrow r .\left( {5\hat i + 3\hat j - 11\hat k} \right) = \frac{{135}}{2}$$
Solution :
$$\eqalign{
& {\text{Mid point of }}PQ{\text{ is}} = \left( {\frac{3}{2},\,\frac{7}{2},\,\frac{{ - 9}}{2}} \right) \cr
& {\text{DR of the normal is}} \cr
& = \left( {4 - \left( { - 1} \right)} \right),\,\left( {5 - 2} \right),\,\left( { - 10 - 1} \right) \cr
& = 5,\,3,\, - 11 \cr
& \therefore {\text{ Equation of plane is}} \cr
& 5\left( {x - \frac{3}{2}} \right) + 3\left( {y - \frac{7}{2}} \right) - 11\left( {z + \frac{9}{2}} \right) = 0 \cr
& \Rightarrow 5x + 3y - 11z = \frac{{135}}{2} \cr
& \Rightarrow \overrightarrow r .\left( {5\hat i + 3\hat j - 11\hat k} \right) = \frac{{135}}{2} \cr} $$