Answer to Question #204513 in Linear Algebra for Felicia

Question

Assume that U is a plane. Find out whether or not the following vectors lie in U:

(10.1) ~u =< 3.8, 1 >, ~v =< −4, 1, 1 > and w~ = −~v

(10.2) ~u =< 3.8, 1 >, ~v =< −4, 1, 1 > and w~ = ~u − ~v

Answer 1

1) The three vectors are coplanar if their scalar triple product is zero.

2) The three vectors are coplanar if they are linearly dependent.

(10.1)

$$\vec u\cdot(\vec v\times \vec w)=\begin{vmatrix} 3 & 8 & 1 \\ -4 & 1 & 1 \\ 4 & -1 & -1 \end{vmatrix}$$

$$=3\begin{vmatrix} 1 & 1 \\ -1 & -1 \end{vmatrix}-8\begin{vmatrix} -4 & 1 \\ 4 & -1 \end{vmatrix}+1\begin{vmatrix} -4 &1 \\ 4 & -1 \end{vmatrix}$$

$$=3(0)-8(0)+0=0$$

The following vectors lie in U.

 Since the vector $\vec u, \vec v$ and $-\vec v$ are linearly dependent, they lie in U.

(10.2)

$$\vec u\cdot(\vec v\times \vec w)=\begin{vmatrix} 3 & 8 & 1 \\ -4 & 1 & 1 \\ 7 & 7 & 0 \end{vmatrix}$$

$$=3\begin{vmatrix} 1 & 1 \\ 7 & 0 \end{vmatrix}-8\begin{vmatrix} -4 & 1 \\ 7 & 0 \end{vmatrix}+1\begin{vmatrix} -4 &1 \\ 7 & 7 \end{vmatrix}$$

$$=3(-7)-8(-7)+(-35)=0$$

The following vectors lie in U.

 Since the vector $\vec u, \vec v$ and $\vec u-\vec v$ are linearly dependent, they lie in U.