Answer to Question #200630 in Analytic Geometry for vina

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 >.

(6.1) Find the dot products ~u · ~v and w~ · ~v

(6.2) 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

(6.1)

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

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

(6.2)

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

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

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

$$\vec u\times\vec w=\begin{vmatrix} \vec i & \vec j & \vec k \\ -1 & 1 & 2 \\ 1 & 1 & 3 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}-\vec j\begin{vmatrix} -1 & 2 \\ 1 & 3 \end{vmatrix}+\vec k\begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix}$$

$$=\vec i+5\vec j-2\vec k$$

Two vectors $2\vec i+4\vec j-\vec k$ and $\vec i+5\vec j-2\vec k$ are not collinear.

Three vectors are not in the single plane. Then we cannot find the vector $\vec n.$ Such vector $\vec n$ does not exist.