To find a vector \( v \) such that the cross product \( u \times v = (5, 1, -3) \), we can use the formula for the cross product in terms of the components of the vectors.
Let \( u = \begin{pmatrix} -1 \ -1 \ -2 \end{pmatrix} \) and \( v = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix} \). The cross product \( u \times v \) is given by the determinant of the following matrix:
\[ u \times v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & -1 & -2 \ v_1 & v_2 & v_3 \end{vmatrix} \]
Calculating the determinant, we have:
\[ u \times v = \mathbf{i} \begin{vmatrix} -1 & -2 \ v_2 & v_3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -1 & -2 \ v_1 & v_3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -1 & -1 \ v_1 & v_2 \end{vmatrix} \]
This leads us to the components:
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The \( x \)-component: \[ -1 v_3 + 2 v_2 = 5 \]
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The \( y \)-component: \[ -(-1 v_3 + 2 v_1) = 1 \implies v_3 - 2 v_1 = 1 \implies v_3 = 1 + 2 v_1 \]
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The \( z \)-component: \[ -(-1 v_2 + -1 v_1) = -3 \implies v_2 - v_1 = -3 \implies v_2 = v_1 - 3 \]
Now, we can express \( v \) in terms of \( v_1 \):
Let \( v_1 = t \) (a parameter):
- \( v_2 = t - 3 \)
- \( v_3 = 1 + 2t \)
Thus, we have:
\[ v = \begin{pmatrix} t \ t - 3 \ 1 + 2t \end{pmatrix} \]
To find a specific \( v \), let’s choose \( t = 0 \): \[ v = \begin{pmatrix} 0 \ -3 \ 1 \end{pmatrix} \]
So, one vector \( v \) that is consistent with the given expressions is:
\[ v = \begin{pmatrix} 0 \ -3 \ 1 \end{pmatrix} \]
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