Show that the points and are equidistant from the plane and lies on opposite side of it.

Question

Philoid · Accepted Answer

We are given with points,

Also, .

Where,

So,

⇒ 5x + 2y – 7z + 9 = 0

We need to show that the points A and B are equidistant from the plane .

Also that, the points lie on the opposite side of the plane.

Normal of the plane 5x + 2y – 7z + 9 = 0 is,

We know that,

Perpendicular distance of the position vector of a point, to the the plane, P: ax + by + cz + d = 0 is given as

Where,

So, perpendicular distance of the point to the plane 5x + 2y – 7z + 9 = 0 having normal is,

∴, |D₁| = |D₂|

But, since D₁ and D₂ have different signs.

⇒ The points A and B lie on the opposite sides of the plane.

Thus, we have shown that the given points are equidistant from the plane and lies on the opposite side of the plane.