Find a vector parallel to the plane $z=2x+3y$.

Question

The suggested solution is $3i-2j$. Why? The first part of the question was to find a perpendicular vector.

Answer 1

You probably found that a vector perpendicular to the plane is $2i+3j-k$. Or, in notation that I personally prefer, $(2,3,-1)$.

We need to find a vector that is perpendicular to the vector $(2,3,-1)$. So we want a non-zero vector $(a,b,c)$ such that the inner product (dot product) of $(a,b,c)$ and $(2,3,1)$ is $0$.

There are many choices. The vector $(-3,2,0)$ will do the job. So will the vector $(1,0,-2)$. So will any linear combination of these.

Answer 2

This is not a explicit answer to your question, but remember that a plane can be thought of a set of vectors which are perpendicular (dot product is zero) to a "normal vector" which is used to describe the orientation of the plane. It is the components of the coefficients. The question can also be phrased as trying to find a vector which is orthogonal to the direction vector.

Answer 3

The plane can be written as: $-2x - 3y + z = 0$. So it has a normal vector: $N = -2i - 3j + k$. So if $v$ is a vector that is parallel to this plane, then $v \perp N$. So let $v = ai + bj + ck$, then $v\cdot N = 0 \iff -2a - 3b + c = 0$. So $c = 2a + 3b$, and $v = ai + bj + (2a + 3b)k$. For example we can take $a = 1$, and $b = 2$. Thus: $v = (1,2,8)$

Answer 4

To find a vector parallel to the plane we need only find two points which lie on the plane.

That is, which solve the equation:

$$z = 2x + 3y$$

Clearly, $(x,y,z) = (0,0,0)$ is a solution, as is $(x,y,z)=(3,-2,0)$ since

$$0 = 2(3) + 3(-2)$$

The vector with initial point $(0,0,0)$ and terminal point $(3,-2,0)$ is

$$\vec{v} = (3,-2,0) - (0,0,0) = (3,-2,0) = 3i - 2j$$

As these two points lie on the plane, $\vec{v}$ lies on the plane, and is therefore parallel to it.