Question

Write the Cartesian equation of a plane, bisecting the line segment joining the points A (2,3,5) and
B(4,5,7) at right angles.

Answer 1

The mid points of the line joining P(3,−2,1)P(3,−2,1) and a(1,4,−3)a(1,4,−3) is

M(3+12,−2+42,1−32)M(3+12,−2+42,1−32)

M=(2,1,−1)M=(2,1,−1)

The point lies on the plane

Equation of the line PQPQ is

r→=3i^+2j^+k^+μ(2i^+j^−k^)r→=3i^+2j^+k^+μ(2i^+j^−k^)

The required plane is r→(2i^+j^−k^)(3i^−2j^+k^).(2i^+j^−k^)r→(2i^+j^−k^)(3i^−2j^+k^).(2i^+j^−k^)

=r→.(2i^+j^−k^)=3=r→.(2i^+j^−k^)=3

The cartesian equation is 2x+y−k=3


Plzzzzz mark me brainlist plzz.......
HOPE THIS ANSWER IS HELPFUL FOR U

Answer 2

Hey !!

One point of required plane = Mid-point of given line segment

            = $\frac{2+4}{2} , \frac{3+5}{2} , \frac{5+7}{2} = (3,4,5)$

Also Dr's of normal to the plane = ( 4 , -2) , (5 , -3) , (7 , -5)

       = 2 , 2 , 2

Therefore, the required equation of plane is

           $2(x-3) + 2(y-4) + 2(z-6) = 0$

               $2x + 2y + 2z = 26\\\\(OR)\\\\x + y + z = 13$

GOOD LUCK !!