vector question assistance

Question

let there be 2 lines: $(2,-3,1) + s(3,-2,1)$ and $(2,-1,-3) +t(3,-2,1)$ which are parallel to each other.

find the formula of the plane determined by them.

my try: a vector perpendicular to $(3,-2,1)$ is for instance $(-1,-1,1)$ since their dot product is $0$. by taking cross product of the 2 we can find the normal to the desired plane which would yield $(-1 ,-4, -5)$ . later we will combine that info with a known point on the said plane like $(2,-3,1)$ to give $-2+12-5+d=0$ and hence $d = -5$. from all this we can say the formula of the plane is $-x-4y-5z-5=0$ . the answer says otherwise so I would like to ask for help here.

Answer 1

There are many vectors perpendicular to $(3, -2, 1)$; there's no reason to suppose the one you picked is related to the plane normal.

On the other hand, the point $(2, -3, 1)$ is on one line (hence in the plane); the point $(2, -1, -3)$ is on the other line (hence in the plane). So the vector between them is also in the plane, i.e., $v = (2, -3, 1) - (2, -1, -3)$ is a vector in the plane. So is the direction vector of the lines, $w = (3, -2, 1)$. From $v$ and $w$, you can find the normal.

Does that help?

Answer 2

The problem lies therein that while $(-1, -1, 1)$ is perpendicular to $(3, -2, 1)$, it isn't necessarily in the direction of any line in the plane requested and therefore the cross product $(3, -2, 1) \times (-1, -1, 1)$ doesn't yield a normal of the requested plane. We're working in the three dimensions: there are many lines perpendicular to $(3, -2, 1)$ (which I'll denote by $\mathbf{v}$); only one of them is in the plane.

(The following may or may not help with visualizing that: Pick up, say, a pencil and imagine that is your vector $\mathbf{v}$. Point e.g. your first finger to it so that the finger and the pencil are in a straight angle relative to each other. You can move the finger around the pencil, yet the angle between them remains straight.)

Hint To define a plane, we need at least two different vectors. The normal to the plane ought be perpendicular to both of them. As you noticed, one vector we can use is $\mathbf{v} =(3, -2, 1)$. The other? Luckily, we know two points on the plane, namely $(2, -3, 1)$ and $(2, -1, -3)$: by definition, the vector of the line going through them is not parallel to $\mathbf{v}$.