What is the the Cartesian equation of the plane having x-, y-, and z-intercepts of 2, 5, and 3 respectively?

Question

What is the the Cartesian equation of the plane having x-, y-, and z-intercepts of 2, 5, and 3 respectively? Im not sure how to do this

Answer 1

Should be known by heart: if the $x$, $y$ and $z$_intercepts are $a,\ b,\ c$ ($\neq 0$) respectively, an equation is $$\frac xa+\frac yb+\frac zc=1.$$

Answer 2

Generally, the equation of the plane can be written as

$$ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0, $$

where $\vec n = (a, b, c)$ is a normal vector, and $\vec v_0 = (x_0, y_0, z_0)$ is a point on the plane. So to write an equation of a plane, a stand method is to find a point on the plane and find its normal vector.

Here we have $3$ points on the plane, namely $(2, 0, 0), (0, 5, 0), (0, 0, 3)$. To find a normal vector, usually we use cross product of two vectors in the plane.

$$ \vec{n} = (2, -5, 0) \times (2, 0, -3) = (15, 6, 10). $$

Then the equation of the plane is:

$$ 15(x - 2) + 6y + 10z = 0. $$