Find the equation of the plane that bisects the line segment joining points (1, 2, 3) and (3, 4, 5) and is at right angle to it.

The given plane bisects the line segment joining points A(1, 2, 3) and B(3, 4, 5) and is at right angle to it.

This means the plane passes through the midpoint of the line AB


Therefore,






And it is also given the plane is normal to the line joining the points A(1, 2, 3) and B(3, 4, 5)


Then


Position vector of - position vector of




We know that vector equation of a plane passing through point and perpendicular/normal to the vector is given by



Substituting the values from eqn(i) and eqn(ii) in the above equation, we get




(by multiplying the two vectors using the formula )




is the vector equation of a required plane.


Let


Then, the above vector equation of the plane becomes,



Now multiplying the two vectors using the formula, we get




This is the Cartesian form of equation of the required plane.


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