Answer to Question #206627 in Analytic Geometry for Tshego

Question

We assume given a plane u passing by the tip of the vectors u=<-1,1,2, v=<2, - 1,0> and w=<1, 1, 3.

a.) Find the dot products u.v and w.v

b.) Determine whether or not there is a vector n that is perpendicular to u. If yes, then find the vector n. Otherwise explain why such a vector does not exist?

Answer 1

a)

$$\vec u\cdot \vec v=-1(2)+1(-1)+2(0)=-3$$

$$\vec w\cdot \vec v=1(2)+1(-1)+3(0)=1$$

b)

$$\vec u\times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ -1 & 1 & 2 \\ 2 & -1 & 0 \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 1 & 2 \\ -1 & 0 \end{vmatrix}-\vec j\begin{vmatrix} -1 & 2 \\ 2 & 0 \end{vmatrix}+\vec k\begin{vmatrix} -1 & 1 \\ 2 & -1 \end{vmatrix}$$

$$=-2\vec i+4\vec j-\vec k$$

The vector $\vec n=-2\vec i+4\vec j-\vec k$ is perpendicular to the vector $\vec u.$

Find the equation of the plane $U$

$$\begin{vmatrix} x-(-1) & y-1 & z-2 \\ 2-(-1) & -1-1 & 0-2 \\ 1-(-1) & 1-1 & 3-2 \\ \end{vmatrix}=0$$

$$(x+1)\begin{vmatrix} -2 & -2 \\ 0 & 1 \end{vmatrix}-(y-1)\begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix}+(z-2)\begin{vmatrix} 3 & -2 \\ 2 & 0 \end{vmatrix}=0$$

$$-2x-2-7y+7+4z-8=0$$

$$2x+7y-4z+3=0$$

The vector $\vec n=2\vec i+7\vec j-4\vec k$ is perpendicular to the plane $U.$