Chapter 11: Problem 6
Let \(\mathbf{a}=\langle\sqrt{2}, \sqrt{2}, 0\rangle, \mathbf{b}=\langle 1,-1,1\rangle\), and \(\mathbf{c}=\langle-2,2,1\rangle\). Find each of the following: (a) \(a^{\cdot} \mathbf{c}\) (b) \((a-c) \cdot b\) (c) \(\mathbf{a} /\|\mathbf{a}\|\) (d) \((\mathbf{b}-\mathbf{c}) \cdot \mathbf{a}\) (e) \(\frac{\mathbf{b} \cdot \mathbf{c}}{\|\mathbf{b}\| \mathbf{c} \|}\) (f) \(\mathbf{a} \cdot \mathbf{a}-\|\mathbf{a}\|^{2}\)
Short Answer
Step by step solution
Calculate \( \mathbf{a} \cdot \mathbf{c} \)
Calculate \((\mathbf{a} - \mathbf{c}) \cdot \mathbf{b} \)
Calculate \( \mathbf{a} / \|\mathbf{a}\| \)
Calculate \((\mathbf{b} - \mathbf{c}) \cdot \mathbf{a} \)
Calculate \( \frac{\mathbf{b} \cdot \mathbf{c}}{\|\mathbf{b}\| \mathbf{c} \|} \)
Calculate \( \mathbf{a} \cdot \mathbf{a} - \|\mathbf{a}\|^2 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
- \( \mathbf{a} \cdot \mathbf{c} = a_1c_1 + a_2c_2 + a_3c_3 \)
- If the dot product is zero, it implies that the vectors are orthogonal, meaning they are at right angles to each other. This was the case in Step 1 and Step 4.
Vector Magnitude
- \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)
Understanding magnitude is crucial for normalizing a vector, a process that involves dividing each component by the vector's magnitude to yield a unit vector, which has a magnitude of 1. This was shown in Step 3 of the solution.
Vector Operations
- **Addition and Subtraction:** Adding or subtracting corresponding components of vectors. For example, \( \mathbf{a} - \mathbf{c} \) was calculated to determine a new vector.
- **Scalar Multiplication:** Involves multiplying each component of a vector by a scalar, which scales the vector but does not change its direction.
- **Dot Product:** Already discussed but foundational; it provides a way to multiply vectors to find useful scalar values like work or projections.
Orthogonal Vectors
- In vector notation, if \( \mathbf{a} \cdot \mathbf{b} = 0 \), then \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal.
In Steps 1 and 4 of the solution, demonstrating that certain computed dot products equal zero confirmed orthogonality, showing that these vectors stand at right angles in vector space.