Chapter 12: Problem 39
In Problems 39 through 42, give a componentwise proof of the indicated property of vector algebra. Take \(\mathbf{a}=\left\langle a_{1}, a_{2}\right\rangle\), \(\mathbf{b}=\left\langle b_{1}, b_{2}\right\rangle\), and \(\mathbf{c}=\left\langle c_{1}, c_{2}\right\rangle\) throughout. \(\mathbf{a}+(\mathbf{b}+\mathbf{c})=(\mathbf{a}+\mathbf{b})+\mathbf{c}\)
Short Answer
Step by step solution
Understand the Problem
Express Vectors Componentwise
Compute \(\mathbf{b} + \mathbf{c}\)
Calculate \(\mathbf{a} + (\mathbf{b} + \mathbf{c})\)
Compute \(\mathbf{a} + \mathbf{b}\)
Calculate \((\mathbf{a} + \mathbf{b}) + \mathbf{c}\)
Compare the Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Associative Property
- \( \mathbf{a} + (\mathbf{b} + \mathbf{c}) = (\mathbf{a} + \mathbf{b}) + \mathbf{c} \)
Componentwise Proof
- \( \mathbf{a} = \langle a_{1}, a_{2} \rangle \)
- \( \mathbf{b} = \langle b_{1}, b_{2} \rangle \)
- \( \mathbf{c} = \langle c_{1}, c_{2} \rangle \)
Vector Addition
- \( \mathbf{b} + \mathbf{c} = \langle b_{1} + c_{1}, b_{2} + c_{2} \rangle \)
- \( \mathbf{a} + (\mathbf{b} + \mathbf{c}) = \langle a_{1} + (b_{1} + c_{1}), a_{2} + (b_{2} + c_{2}) \rangle \)
- \( (\mathbf{a} + \mathbf{b}) + \mathbf{c} = \langle (a_{1} + b_{1}) + c_{1}, (a_{2} + b_{2}) + c_{2} \rangle \)
Mathematical Proof
- Begin by stating known facts or axioms. In this case, the vector definitions and math operations.
- Perform logical operations step by step, such as calculating the individual components of vector sums.
- Reach a conclusion where the initial hypothesis is shown to hold true, as done here by equating both expressions of vector addition.