We set this equal to 5:

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• \mathbf{v} = egin{bmatrix} a \ b \end{bmatrix}, \quad \mathbf{a} = egin{bmatrix} 3 \ -4 \end{bmatrix}
• The 2D cross product is:
• \mathbf{v} imes \mathbf{a} = a(-4) - b(3) = -4a - 3b
• -4a - 3b = 5
• This is a linear equation in two variables. We can express one variable in terms of the other. Solving for \(b\):
• -3b = 4a + 5 \quad \Rightarrow \quad b = -rac{4a + 5}{3}

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