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
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 $\displaystyle (0,0,1) \cdot (1-x,2-y,3-z)=0$ . So you get $\displaystyle z = 3 $ . Note here that the $\displaystyle \cdot $ means dot-product.
So you have a plane parallell to the xy-plane, with perpendicular distance of 3 to the origin.
Hello, gva0324!
Are you sure of the wording of this problem?
As given, it is too simple . . .
Given: .$\displaystyle \begin{array}{ccc}\vec u &=& 2i +3j \\ \vec v&=& \text{-}i + 2j \end{array}$
(a) Find a cartesian equation for the plane $\displaystyle P$ containing the point $\displaystyle Q(1,2,3)$
. . . and parallel to $\displaystyle \vec u$ and $\displaystyle \vec v$
Since $\displaystyle \vec u$ and $\displaystyle \vec v$ are in the $\displaystyle xy$-plane, then plane $\displaystyle P$ is parallel to the $\displaystyle xy$-plane.
Since plane $\displaystyle P$ contains $\displaystyle Q(1,2,3)$, its equation is: .$\displaystyle z \:=\:3$
(b) Find the perpendicular distance between the plane $\displaystyle P$ and the origin
The point on plane $\displaystyle P$ nearest the origin is $\displaystyle (0,0,3).$
Its distance from the origin is $\displaystyle 3.$