Chapter 3: Problem 17
Let \(\mathbf{u}=(5,-1,0,3,-3), \mathbf{v}=(-1,-1,7,2,0),\) and \(\mathbf{w}=(-4,2,-3,-5,2) .\) Find the components of (a) \(w-u\) (b) \(2 v+3 u\) (c) \(-w+3(v-u)\) (d) \(5(-v+4 u-w)\) (e) \(-2(3 \mathbf{w}+\mathbf{v})+(2 \mathbf{u}+\mathbf{w})\) (f) \(\frac{1}{2}(w-5 v+2 u)+v\)
Short Answer
Expert verified
The components are (a) (-9, 3, -3, -8, 5), (b) (13, -5, 14, 13, -9), (c) (-14, -2, 24, 2, 7), (d) (125, -25, -20, 75, -70), (e) (32, -10, 1, 27, -16), (f) (9/2, 3/2, -12, -5/2, -2).
Step by step solution
01
Calculate w - u
Subtract each component of \( \mathbf{u} \) from the corresponding component of \( \mathbf{w} \):\[ \mathbf{w} - \mathbf{u} = (-4, 2, -3, -5, 2) - (5, -1, 0, 3, -3) = (-4-5, 2+1, -3-0, -5-3, 2+3) = (-9, 3, -3, -8, 5). \]
02
Calculate 2v + 3u
Multiply each component of \( \mathbf{v} \) by 2 and \( \mathbf{u} \) by 3, then add the resulting vectors:\[ 2 \mathbf{v} = 2(-1, -1, 7, 2, 0) = (-2, -2, 14, 4, 0), \]\[ 3 \mathbf{u} = 3(5, -1, 0, 3, -3) = (15, -3, 0, 9, -9), \]\[ 2 \mathbf{v} + 3 \mathbf{u} = (-2+15, -2-3, 14+0, 4+9, 0-9) = (13, -5, 14, 13, -9). \]
03
Calculate -w + 3(v-u)
First compute \( \mathbf{v} - \mathbf{u} \):\[ \mathbf{v} - \mathbf{u} = (-1, -1, 7, 2, 0) - (5, -1, 0, 3, -3) = (-6, 0, 7, -1, 3). \]Then multiply by 3:\[ 3(\mathbf{v} - \mathbf{u}) = 3(-6, 0, 7, -1, 3) = (-18, 0, 21, -3, 9). \]Now calculate \(-\mathbf{w} + 3(\mathbf{v} - \mathbf{u}) \):\[ -\mathbf{w} = -(-4, 2, -3, -5, 2) = (4, -2, 3, 5, -2), \]\[ -\mathbf{w} + 3(\mathbf{v} - \mathbf{u}) = (4, -2, 3, 5, -2) + (-18, 0, 21, -3, 9) = (-14, -2, 24, 2, 7). \]
04
Calculate 5(-v + 4u - w)
Calculate \(-\mathbf{v} + 4\mathbf{u} - \mathbf{w} \):\[ -\mathbf{v} = (1, 1, -7, -2, 0), \]\[ 4\mathbf{u} = 4(5, -1, 0, 3, -3) = (20, -4, 0, 12, -12), \]\[ -\mathbf{v} + 4\mathbf{u} = (1+20, 1-4, -7+0, -2+12, 0-12) = (21, -3, -7, 10, -12), \]Subtract \(\mathbf{w}\):\[ -\mathbf{v} + 4\mathbf{u} - \mathbf{w} = (21, -3, -7, 10, -12) - (-4, 2, -3, -5, 2) = (21+4, -3-2, -7+3, 10+5, -12-2) = (25, -5, -4, 15, -14). \]Finally, multiply by 5:\[ 5(-\mathbf{v} + 4\mathbf{u} - \mathbf{w}) = 5(25, -5, -4, 15, -14) = (125, -25, -20, 75, -70). \]
05
Calculate -2(3w + v) + (2u + w)
First find \(3\mathbf{w} + \mathbf{v} \):\[ 3\mathbf{w} = 3(-4, 2, -3, -5, 2) = (-12, 6, -9, -15, 6), \]\[ 3\mathbf{w} + \mathbf{v} = (-12+(-1), 6+(-1), -9+7, -15+2, 6+0) = (-13, 5, -2, -13, 6). \]Now multiply by -2:\[ -2(3\mathbf{w} + \mathbf{v}) = -2(-13, 5, -2, -13, 6) = (26, -10, 4, 26, -12). \]Calculate \(2\mathbf{u} + \mathbf{w} \):\[ 2\mathbf{u} = 2(5, -1, 0, 3, -3) = (10, -2, 0, 6, -6), \]\[ 2\mathbf{u} + \mathbf{w} = (10, -2, 0, 6, -6) + (-4, 2, -3, -5, 2) = (6, 0, -3, 1, -4). \]Finally combine results:\[ -2(3\mathbf{w} + \mathbf{v}) + (2\mathbf{u} + \mathbf{w}) = (26, -10, 4, 26, -12) + (6, 0, -3, 1, -4) = (32, -10, 1, 27, -16). \]
06
Calculate \frac{1}{2}(w-5v+2u)+v
First, find \( \mathbf{w} - 5\mathbf{v} + 2\mathbf{u} \):\[ 5\mathbf{v} = 5(-1, -1, 7, 2, 0) = (-5, -5, 35, 10, 0), \]\[ 2\mathbf{u} = 2(5, -1, 0, 3, -3) = (10, -2, 0, 6, -6), \]\[ \mathbf{w} = (-4, 2, -3, -5, 2), \]Therefore,\[ \mathbf{w} - 5\mathbf{v} + 2\mathbf{u} = (-4, 2, -3, -5, 2) - (-5, -5, 35, 10, 0) + (10, -2, 0, 6, -6) = (-4+5-5, 2+5-2, -3-35+0, -5-10+6, 2+0-6), \]\[ = (11, 5, -38, -9, -4). \]Now, multiply by \( \frac{1}{2} \):\[ \frac{1}{2}(11, 5, -38, -9, -4) = \left(\frac{11}{2}, \frac{5}{2}, -19, -\frac{9}{2}, -2\right). \]Finally, add \( \mathbf{v} \):\[ \left( \frac{11}{2}, \frac{5}{2}, -19, -\frac{9}{2}, -2 \right) + (-1, -1, 7, 2, 0) = \left( \frac{11}{2}-1, \frac{5}{2}-1, -19+7, -\frac{9}{2}+2, -2+0 \right), \]\[ = \left( \frac{9}{2}, \frac{3}{2}, -12, -\frac{5}{2}, -2 \right). \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Subtraction
Vector subtraction is a fundamental part of vector arithmetic. It involves taking one vector from another, which means subtracting each corresponding component. To subtract vector \( \mathbf{u} \) from vector \( \mathbf{w} \), each component of \( \mathbf{u} \) is subtracted from the corresponding component of \( \mathbf{w} \).
For example, if \( \mathbf{u} = (5, -1, 0, 3, -3) \) and \( \mathbf{w} = (-4, 2, -3, -5, 2) \), the vector subtraction \( \mathbf{w} - \mathbf{u} \) results in:
For example, if \( \mathbf{u} = (5, -1, 0, 3, -3) \) and \( \mathbf{w} = (-4, 2, -3, -5, 2) \), the vector subtraction \( \mathbf{w} - \mathbf{u} \) results in:
- First component: \(-4 - 5 = -9\)
- Second component: \(2 + 1 = 3\)
- Third component: \(-3 - 0 = -3\)
- Fourth component: \(-5 - 3 = -8\)
- Fifth component: \(2 + 3 = 5\)
Vector Addition
Vector addition involves combining two vectors by adding their corresponding components. This operation results in a new vector.
To add two vectors, such as \( \mathbf{v} = (-1, -1, 7, 2, 0) \) and \( 3 \mathbf{u} = (15, -3, 0, 9, -9) \), each component of \( \mathbf{v} \) is added to the corresponding component of \( 3\mathbf{u} \). The process is straightforward:
To add two vectors, such as \( \mathbf{v} = (-1, -1, 7, 2, 0) \) and \( 3 \mathbf{u} = (15, -3, 0, 9, -9) \), each component of \( \mathbf{v} \) is added to the corresponding component of \( 3\mathbf{u} \). The process is straightforward:
- First component: \(-1 + 15 = 14\)
- Second component: \(-1 - 3 = -4\)
- Third component: \(7 + 0 = 7\)
- Fourth component: \(2 + 9 = 11\)
- Fifth component: \(0 - 9 = -9\)
Scalar Multiplication
Scalar multiplication in vector arithmetic means multiplying a vector by a number, or scalar, and this influences every component of the vector individually. If we look into multiplying the vector \( \mathbf{u} = (5, -1, 0, 3, -3) \) by a scalar 3, each component of \( \mathbf{u} \) is multiplied by 3:
- First component: \(3 \times 5 = 15\)
- Second component: \(3 \times -1 = -3\)
- Third component: \(3 \times 0 = 0\)
- Fourth component: \(3 \times 3 = 9\)
- Fifth component: \(3 \times -3 = -9\)
Linear Combinations
Linear combinations allow for scalable and weighted combinations of vectors via addition and scalar multiplication. A linear combination of vectors involves taking each vector multiplied by a constant, then adding these scaled vectors together.
Let's consider constructing a linear combination \( 2\mathbf{v} + 3\mathbf{u} \), where:
Let's consider constructing a linear combination \( 2\mathbf{v} + 3\mathbf{u} \), where:
- \( 2\mathbf{v} \) means each component of \( \mathbf{v} \) is multiplied by 2, results in \((-2, -2, 14, 4, 0)\).
- \( 3\mathbf{u} \) scales \( \mathbf{u} \) by 3, resulting in \((15, -3, 0, 9, -9)\).