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 $U.$