Chapter 11: Problem 1
Let \(\mathbf{a}=-2 \mathbf{i}+3 \mathbf{j}, \mathbf{b}=2 \mathbf{i}-3 \mathbf{j}\), and \(\mathbf{c}=-5 \mathbf{j}\). Find each of the following: (a) \(2 \mathbf{a}-4 \mathbf{b}\) (b) \(\mathbf{a} \cdot \mathbf{b}\) (c) \(\mathbf{a} \cdot(\mathbf{b}+\mathbf{c})\) (d) \((-2 \mathbf{a}+3 \mathbf{b}) \cdot 5 \mathbf{c}\) (e) \(\|\mathbf{a}\| \mathbf{c} \cdot \mathbf{a}\) (f) \(\mathbf{b} \cdot \mathbf{b}-\|\mathbf{b}\|\)
Short Answer
Step by step solution
Calculate 2a - 4b
Find dot product a · b
Find a · (b + c)
Calculate (-2a + 3b) · 5c
Compute ||a|| c · a
Evaluate b · b - ||b||
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]
This operation is particularly useful because it has several properties, such as:
- It is commutative, meaning \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \).
- If the dot product is zero, it indicates that the vectors are perpendicular.
Magnitude of a Vector
\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \]
This scalar value helps us understand how "big" or "long" a vector is. Some key uses for calculating the magnitude include:
- Normalization, where a vector is scaled down to have a magnitude of 1.
- Finding distances in vector spaces.
Vector Addition
\[ \mathbf{r} = (a_1 + b_1)\mathbf{i} + (a_2 + b_2)\mathbf{j} \]
Key points to remember about vector addition include:
- It is commutative, meaning \( \mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a} \).
- It allows for the graphical addition using the tip-to-tail method, visually demonstrating how vectors add to one another.
Vector Scaling
\[ k\mathbf{v} = (kv_1)\mathbf{i} + (kv_2)\mathbf{j} \]
The important aspects of vector scaling include:
- Scaling by a positive scalar increases or decreases the vector's magnitude, maintaining the direction.
- Scaling by a negative scalar inverts the direction of the vector.