Find the distance of the point P (–1, –5, –10) from the point of intersection of the line joining the points A (2, –1, 2) and B (5, 3, 4) with the plane x – y + z = 5?

Answer 1

If you plug in point A into the formula that describes the plane, you'll see that it's a solution, 5 - (-1) + 2 = 5, so that IS the point where the line intersects the plane (phew, they already gave it to you!).

So the distance from the point of intersection of the line and the plane (point A) to the point P is just the distance formula, or

√((x₂-x₁)² + (y₂ - y₁)² + (z₂ - z₁)²) = √((-1 - 2)² + (-5 - -1)² + (-10 - 2)²) = √(9 + 16 + 144) = √169 = 13 units