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# help

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211
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Let a, b, c be vectors such that$$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix}, \mathbf{a} \times \mathbf{c} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}, \text{ and } \mathbf{b} \times \mathbf{c} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}.$$

1. Then evaluate

$$(\mathbf{a} + \mathbf{b})\times \mathbf{c}, \mathbf{a}\times(2\mathbf{b} -3 \mathbf{c}), (\mathbf{a} + 2 \mathbf{c})\times \mathbf{b}$$

2. Then evaluate

$$(\mathbf{b} + \mathbf{c})\times \mathbf{b}, \mathbf{a}\times(\mathbf{b} + 4 \mathbf{a}), (\mathbf{a} + \mathbf{b} + \mathbf{c})\times \mathbf{a}$$

Aug 5, 2019

#1
+24389
+2

Let a, b, c be vectors such that

$$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix},\ \mathbf{a} \times \mathbf{c} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix},\ \text{ and } \mathbf{b} \times \mathbf{c} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}$$.
1.
Then evaluate

$$(\mathbf{a} + \mathbf{b})\times \mathbf{c},\ \mathbf{a}\times(2\mathbf{b} -3 \mathbf{c}),\ (\mathbf{a} + 2 \mathbf{c})\times \mathbf{b}$$

$$\begin{array}{|l|rcll|} \hline \mathbf{1.} & (\mathbf{a} + \mathbf{b})\times \mathbf{c} \\ & &=& \mathbf{a}\times \mathbf{c} + \mathbf{b} \times \mathbf{c} \\ &&=& \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}+\begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \\ &&=& \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix} \\ \hline \mathbf{2.} & \mathbf{a}\times(2\mathbf{b} -3 \mathbf{c}) \\ & &=& \mathbf{a}\times(2\mathbf{b}) - \mathbf{a}\times (3 \mathbf{c}) \\ & &=& 2(\mathbf{a}\times\mathbf{b}) - 3(\mathbf{a}\times \mathbf{c}) \\ & &=& 2 \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix} - 3 \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \\ & &=& \begin{pmatrix} -2 \\- 2 \\ -2 \end{pmatrix} - \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix} \\ & &=& \begin{pmatrix} 1 \\- 8 \\ -5 \end{pmatrix} \\ \hline \mathbf{3.} & (\mathbf{a} + 2 \mathbf{c})\times \mathbf{b} \\ & &=& \mathbf{a}\times \mathbf{b} + (2 \mathbf{c})\times \mathbf{b} \\ & &=& \mathbf{a}\times \mathbf{b} + 2 (\mathbf{c}\times \mathbf{b}) \\ & &=& \mathbf{a}\times \mathbf{b} - 2 (\mathbf{b}\times \mathbf{c}) \\ & &=& \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix} - 2 \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \\ & &=& \begin{pmatrix} -1 \\- 5 \\ -7 \end{pmatrix} \\ \hline \end{array}$$

Aug 6, 2019
#2
+24389
+2

Let a, b, c be vectors such that
$$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix},\ \mathbf{a} \times \mathbf{c} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix},\ \text{ and } \mathbf{b} \times \mathbf{c} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}$$.

2.
Then evaluate

$$(\mathbf{b} + \mathbf{c})\times \mathbf{b},\ \mathbf{a}\times(\mathbf{b} + 4 \mathbf{a}),\ (\mathbf{a} + \mathbf{b} + \mathbf{c})\times \mathbf{a}$$

$$\begin{array}{|l|rcll|} \hline \mathbf{1.} & (\mathbf{b} + \mathbf{c})\times \mathbf{b} \\ & &=& \mathbf{b}\times \mathbf{b} + \mathbf{c}\times \mathbf{b} \\ & &=& 0 + \mathbf{c}\times \mathbf{b} \\ & &=& - (\mathbf{b}\times \mathbf{c}) \\ & &=& - \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \\ & &=& \begin{pmatrix} 0 \\ -2 \\ -3 \end{pmatrix} \\ \hline \mathbf{2.} & \mathbf{a}\times(\mathbf{b} + 4 \mathbf{a}) \\ & &=& \mathbf{a}\times \mathbf{b} + \mathbf{a}\times (4 \mathbf{a}) \\ & &=& \mathbf{a}\times \mathbf{b} + 4(\mathbf{a}\times \mathbf{a}) \\ & &=& \mathbf{a}\times \mathbf{b} + 4 * 0 \\ & &=& \mathbf{a}\times \mathbf{b} \\ & &=& \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix} \\ \hline \mathbf{3.} & (\mathbf{a} + \mathbf{b} + \mathbf{c})\times \mathbf{a} \\ & &=& \mathbf{a}\times \mathbf{a} + \mathbf{b}\times \mathbf{a} + \mathbf{c}\times \mathbf{a} \\ & &=& \mathbf{a}\times \mathbf{a} - \mathbf{a}\times \mathbf{b} - \mathbf{a}\times \mathbf{c} \\ & &=& 0 - (\mathbf{a}\times \mathbf{b}) - (\mathbf{a}\times \mathbf{c}) \\ & &=& - (\mathbf{a}\times \mathbf{b}) - (\mathbf{a}\times \mathbf{c}) \\ & &=& - \begin{pmatrix} -1 \\- 1 \\ -1 \end{pmatrix} - \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \\ & &=& \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \\ & &=& \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} \\ \hline \end{array}$$

Aug 6, 2019