Question #209780

Let ~u =< −2, 1, −1, ~v =< −3, 2, −1 > and w~ =< 1, 3, 5 >. Compute:

(6.1) ~u × w~ ,

(6.2) ~u × (w~ × ~v) and (~u × w~ ) × ~v.


1
Expert's answer
2021-06-27T18:38:50-0400

(6.1)


u×w=ijk211135\vec u\times \vec w=\begin{vmatrix} \vec i & \vec j & \vec k \\ -2 & 1 & -1 \\ 1 & 3 & 5 \\ \end{vmatrix}

=i1135j2115+k2113=\vec i\begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix}-\vec j\begin{vmatrix} -2 & -1 \\ 1 & 5 \end{vmatrix}+\vec k\begin{vmatrix} -2 & 1 \\ 1 & 3 \end{vmatrix}

=8i+9j7k=8\vec i+9\vec j-7\vec k

u×w=8,9,7\vec u\times \vec w=\langle8, 9, -7\rangle

(6.2)

w×v=ijk135321\vec w\times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ 1 & 3 & 5 \\ -3 & 2 & -1 \\ \end{vmatrix}

=i3521j1531+k1332=\vec i\begin{vmatrix} 3 & 5 \\ 2 & -1 \end{vmatrix}-\vec j\begin{vmatrix} 1 & 5 \\ -3 & -1 \end{vmatrix}+\vec k\begin{vmatrix} 1 & 3 \\ -3 & 2 \end{vmatrix}

=13i14j+11k=-13\vec i-14\vec j+11\vec k

u×(w×v)=ijk211131411\vec u\times (\vec w\times \vec v)=\begin{vmatrix} \vec i & \vec j & \vec k \\ -2 & 1& -1 \\ -13 & -14 & 11 \\ \end{vmatrix}

=i111411j211311+k211314=\vec i\begin{vmatrix} 1 & -1 \\ -14 & 11 \end{vmatrix}-\vec j\begin{vmatrix} -2 & -1 \\ -13 & 11 \end{vmatrix}+\vec k\begin{vmatrix} -2 & 1 \\ -13 & -14 \end{vmatrix}

=3i+35j+41k=-3\vec i+35\vec j+41\vec k

u×(w×v)=3,35,41\vec u\times (\vec w\times \vec v)=\langle -3, 35, 41 \rangle


(u×w)×v=ijk897321(\vec u\times \vec w)\times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ 8 & 9 & -7 \\ -3 & 2 & -1 \\ \end{vmatrix}

=i9721j8731+k8932=\vec i\begin{vmatrix} 9 & -7 \\ 2 & -1 \end{vmatrix}-\vec j\begin{vmatrix} 8 & -7 \\ -3 & -1 \end{vmatrix}+\vec k\begin{vmatrix} 8 & 9 \\ -3 & 2 \end{vmatrix}

=5i+29j+43k=5\vec i+29\vec j+43\vec k

(u×w)×v=5,29,43(\vec u\times \vec w)\times \vec v=\langle 5, 29, 43 \rangle




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