1. ## Help with vectors

Hello I need help with the following:

u = 2i +3j v= -i + 2j

Find a cartesian equation for the plane P containing the point Q(1,2,3) and parallel to u and w.

Find the perpendicular distance between the plane P and the origin

2. hi

Do you know what a cross product is?

You could use it to get a vector that is orthogonal to both u and w.

However in this case, itīs clear that both u and v are in the xy plane so to speak. So you will get a normal pointing "straight up" from the xy-plane.

You will need your normal to satisfy $(0,0,1) \cdot (1-x,2-y,3-z)=0$ . So you get $z = 3$ . Note here that the $\cdot$ means dot-product.

So you have a plane parallell to the xy-plane, with perpendicular distance of 3 to the origin.

3. Hello, gva0324!

Are you sure of the wording of this problem?
As given, it is too simple . . .

Given: . $\begin{array}{ccc}\vec u &=& 2i +3j \\ \vec v&=& \text{-}i + 2j \end{array}$

(a) Find a cartesian equation for the plane $P$ containing the point $Q(1,2,3)$
. . . and parallel to $\vec u$ and $\vec v$

Since $\vec u$ and $\vec v$ are in the $xy$-plane, then plane $P$ is parallel to the $xy$-plane.

Since plane $P$ contains $Q(1,2,3)$, its equation is: . $z \:=\:3$

(b) Find the perpendicular distance between the plane $P$ and the origin

The point on plane $P$ nearest the origin is $(0,0,3).$

Its distance from the origin is $3.$