I have a question regarding vectors:
Find the equation of the plane perpendicular to the vector $\vec{n}\space=(2,3,6)$ and which goes through the point $ A(1,5,3)$. (A cartesian and parametric equation).
Also find the distance between the beginning of axis and this plane.
I'm not really sure where to start. Any help would be appreciated.
Suppose you take any point on the plane, say, $\vec{OR}$, where $O$ is the origin. Now, $\vec{AR}$ would be some sort of a line segment that lies on the plane.
But we know that this line segment must be perpendicular to $\space(2,3,6)$. This means that the dot product of $\vec{AR}$ (which is $\vec{OR} - \vec{OA}$) with $\space(2,3,6)$ must be equal to $0$. So we have:
$$(\vec{OR} - \vec{OA})\cdot \space(2,3,6) = 0,$$ $$\vec{OR}\cdot\space(2,3,6) = \vec{OA}\cdot(2,3,6).$$
But you know that $\vec{OA} = \space(1,5,3)$. Hence, the equation of the plane is:
$$\vec{OR}\cdot\space(2,3,6) = \space(1,5,3)\cdot\space(2,3,6),$$
$$\vec{OR}\cdot\space(2,3,6) = 35.$$
In Cartesian form, this is just
$$2x + 3y + 6z = 35.$$