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.