Question #204513

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


1
Expert's answer
2021-06-08T18:34:38-0400

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)


u(v×w)=381411411\vec u\cdot(\vec v\times \vec w)=\begin{vmatrix} 3 & 8 & 1 \\ -4 & 1 & 1 \\ 4 & -1 & -1 \end{vmatrix}

=3111184141+14141=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=3(0)-8(0)+0=0

The following vectors lie in U.


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


(10.2)


u(v×w)=381411770\vec u\cdot(\vec v\times \vec w)=\begin{vmatrix} 3 & 8 & 1 \\ -4 & 1 & 1 \\ 7 & 7 & 0 \end{vmatrix}

=3117084170+14177=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=3(-7)-8(-7)+(-35)=0

The following vectors lie in U.


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



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