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Question: Equation of the plane through the midpoint of the line segment joining the points \(P(4,5, - 10),Q...

Equation of the plane through the midpoint of the line segment joining the points
P(4,5,−10),Q(−1,2,1)P(4,5, - 10),Q( - 1,2,1) and perpendicular to PQPQ is
A. r.(23i^+72j^−92k^)=45r.\left( {\dfrac{2}{3}\hat i + \dfrac{7}{2}\hat j - \dfrac{9}{2}\hat k} \right) = 45
B. r.(−i^+2j^+k^)=1352r.\left( { - \hat i + 2\hat j + \hat k} \right) = \dfrac{{135}}{2}
C. r.(5i^+3j^−11k^)+1352=0r.\left( {5\hat i + 3\hat j - 11\hat k} \right) + \dfrac{{135}}{2} = 0
D. r.(4i^+5j^−10k^)=85r.\left( {4\hat i + 5\hat j - 10\hat k} \right) = 85
E. r.(5i^+3j^−11k^)=1352r.\left( {5\hat i + 3\hat j - 11\hat k} \right) = \dfrac{{135}}{2}

Explanation

Solution

Hint:- find midpoint and direction ratio between two points. Use midpoint and direction ratio to solve this question.

Complete step-by-step answer:
Here the given two points are P(4,5,−10),Q(−1,2,1)P(4,5, - 10),Q( - 1,2,1)
Therefore coordinate of the midpoint of PPand QQis
As we know the midpoint formula between two points is (x1+x22,y1+y22,z1+z22)\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)
(4−12,5+22,−10+12)\left( {\dfrac{{4 - 1}}{2},\dfrac{{5 + 2}}{2},\dfrac{{ - 10 + 1}}{2}} \right) i.e. (32,72,−92)\left( {\dfrac{3}{2},\dfrac{7}{2},\dfrac{{ - 9}}{2}} \right)
And we know that the direction ratio between two points are (x2−x1,y2−y1,z2−z1)\left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)
Now direction ratio of PQPQ is (−1−4,2−5,1+10)\left( { - 1 - 4,2 - 5,1 + 10} \right) i.e. is (−5,−3,11)\left( { - 5, - 3,11} \right) or (5,3,−11)\left( {5,3, - 11} \right)
Therefore, equation of plane passing through (32,72,−92)\left( {\dfrac{3}{2},\dfrac{7}{2},\dfrac{{ - 9}}{2}} \right) and having direction ratio (5,3,−11)\left( {5,3, - 11} \right) is
5(x−32)+3(y−72)−11(z+92)=05\left( {x - \dfrac{3}{2}} \right) + 3\left( {y - \dfrac{7}{2}} \right) - 11\left( {z + \dfrac{9}{2}} \right) = 0
as we know when the point (x1,y1,z1)\left( {{x_1},{y_1},{z_1}} \right) and the direction ratio (x2,y2,z2)\left( {{x_2},{y_2},{z_2}} \right) are given the equation of line is x2(x−x1)+y2(y−y1)+z2(z−z1)=0{x_2}\left( {x - {x_1}} \right) + {y_2}\left( {y - {y_1}} \right) + {z_2}\left( {z - {z_1}} \right) = 0
⇒5x+3y−11z=152+212+992 ⇒5x+3y−11z=1352  \Rightarrow 5x + 3y - 11z = \dfrac{{15}}{2} + \dfrac{{21}}{2} + \dfrac{{99}}{2} \\\ \Rightarrow 5x + 3y - 11z = \dfrac{{135}}{2} \\\
It is written in vector form
r.(5i^+3j^−11k^)=1352r.\left( {5\hat i + 3\hat j - 11\hat k} \right) = \dfrac{{135}}{2}
Hence, option E is the correct answer.

Note:- first of all you have to find the midpoint as the plane must pass through the midpoint given in question then for the equation of plane we must have to find the direction ratio between two points then apply the formula of writing the equation of plane. As we know the direction ratio of planes are always perpendicular to the plane.