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Chapter 4: Problem 29

Find \((a) u-v\) (b) \(2(\mathbf{u}+3 \mathbf{v}),\) and \((\mathbf{c}) 2 \mathbf{v}-\mathbf{u}\). $$\mathbf{u}=(0,4,3,4,4), \quad \mathbf{v}=(6,8,-3,3,-5)$$

Short Answer

Expert verified
\((a) u-v = (-6, -4, 6, 1, 9)\), (b) \(2(\mathbf{u}+3 \mathbf{v}) = (36, 56, -12, 26, -22)\), (c) \(2\mathbf{v}-\mathbf{u} = (12, 12, -9, 2, -14)\)

Step by step solution

01

Subtract \(\mathbf{v}\) from \(\mathbf{u}\)

Calculate \(u-v\) by subtracting each corresponding component of \(\mathbf{v}\) from \(\mathbf{u}\). You get (0 - 6, 4 - 8, 3 - (-3), 4 - 3, 4 - (-5)) = (-6, -4, 6, 1, 9)
02

Find \(2(\mathbf{u} + 3\mathbf{v})\)

First, multiply \(\mathbf{v}\) by 3, yielding (6*3, 8*3, -3*3, 3*3, -5*3) = (18, 24, -9, 9, -15). Then add \(\mathbf{u}\) to this product, yielding (0 + 18, 4 + 24, 3 - 9, 4 + 9, 4 - 15) = (18, 28, -6, 13, -11). Finally multiply this result by 2 to get the answer 2*(18, 28, -6, 13, -11) = (36, 56, -12, 26, -22)
03

Calculate \(2\mathbf{v} - \mathbf{u}\)

Multiply the vector \(\mathbf{v}\) by 2 to get (6*2, 8*2, -3*2, 3*2, -5*2) = (12, 16, -6, 6, -10). Then subtract \(\mathbf{u}\) from this product, yielding (12 - 0, 16 - 4, -6 - 3, 6 - 4, -10 - 4) = (12, 12, -9, 2, -14)

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