Find the plane perpendicular to the XY plane.

Question

Say I have a right triangle with the hypotenuse in quadrant 1 lying on the XY plane. How would I go about finding the vertical plane that intersects the XY plane along the hypotenuse?

Answer 1

Simply find the equation of the hypotenuse $f(x,y)$ and that's the equation of your plane, since $z$ can be anything.

Answer 2

Let the vector $v \in \mathbb{R}^3$ be the hypotenuse. Then the normal vector to the plane that you described is $n = v \times (0, 0, 1).$ So one can describe the plane you want as $\{x \in \mathbb{R}^3 \colon x \cdot n = 0 \}$