Answer to Question #96502 in Analytic Geometry for Olajide Olaitan

Question

Find an equation for the plane perpendicular to the vector A=2i+3j+6k and passing through the terminal point of the vector B=1+5j+3k

Answer 1

The equation of the plane with normal $\bold n = (A, B, C)$ that goes through point $\bold r_0 = (x_0, y_0, z_0)$ is $\bold n \cdot (\bold r - \bold r_0) = A (x- x_0) + B(y - y_0) + C (z- z_0) = 0$.

In our case, vector $\bold A = (2, 3, 6)$ can be used as normal vector, and the terminal point of vector $\bold B$ can be used as $\bold r_0$. Hence, the equation of the plane is $2 (x-1) + 3 (y-5) + 6 (z - 3) = 0$, or simplified $2 x + 3 y + 6z - 35 = 0$.