Find the equation of a plane which bisects perpendicularly the line joining the points A (2, 3, 4) and B (4, 5, 8) at right angles.

Question

Philoid · Accepted Answer

Given that,

There is a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.

We need to find the equation of such plane.

For this, let us find the mid-point of line AB.

Since, mid-point is the halfway between two end-points. So,

⇒ Midpoint of AB = (3, 4, 6)

This can be represented into position vector,

Also, we need to find normal of the plane, .

And we know that, equation of the plane which bisects perpendicularly the line joining two points, here, A and B is

Where,

Substitute the values of , and in the above equation, we get

⇒ 2(x – 3) + 2(y – 4) + 4(z – 6) = 0

Further simplifying it,

⇒ 2x – 6 + 2y – 8 + 4z – 24 = 0

⇒ 2x + 2y + 4z – 6 – 8 – 24 = 0

⇒ 2x + 2y + 4z – 38 = 0

⇒ 2(x + y + 2z – 19) = 0

⇒ x + y + 2z – 19 = 0

⇒ x + y + 2z = 19

Thus, required equation of the plane is x + y + 2z = 19.