
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![**Problem Statement: Compute the Dot Product**
Given two vectors, calculate their dot product.
Vectors:
\[ (\mathbf{i} + \mathbf{j}) \cdot (\mathbf{j} + \mathbf{k}) \]
*Please enter your answer in the provided input box.*
**Explanation:**
To compute the dot product, follow these steps:
1. Multiply the corresponding components of the vectors.
2. Add up all the products obtained.
The dot product formula for two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\) is calculated as follows:
\[ \mathbf{a} \cdot \mathbf{b} = (a_1 \cdot b_1) + (a_2 \cdot b_2) + (a_3 \cdot b_3) \]
In this specific problem, expand the vectors and compute the products accordingly:
\[ (\mathbf{i} + \mathbf{j}) \cdot (\mathbf{j} + \mathbf{k}) \]
Now, apply distributive property:
\[ \mathbf{i} \cdot \mathbf{j} + \mathbf{i} \cdot \mathbf{k} + \mathbf{j} \cdot \mathbf{j} + \mathbf{j} \cdot \mathbf{k} \]
Remember:
- \(\mathbf{i} \cdot \mathbf{i} = 1\)
- \(\mathbf{j} \cdot \mathbf{j} = 1\)
- \(\mathbf{k} \cdot \mathbf{k} = 1\)
- \(\mathbf{i} \cdot \mathbf{j} = 0\)
- \(\mathbf{i} \cdot \mathbf{k} = 0\)
- \(\mathbf{j} \cdot \mathbf{k} = 0\)
Combine all the results to find the final dot product value.](https://content.bartleby.com/qna-images/question/c3d3bee5-0c39-4b7b-a553-642953afa184/81797eca-dade-4515-a364-ec8d6005da9b/71aqmd6_thumbnail.png)
Transcribed Image Text:**Problem Statement: Compute the Dot Product**
Given two vectors, calculate their dot product.
Vectors:
\[ (\mathbf{i} + \mathbf{j}) \cdot (\mathbf{j} + \mathbf{k}) \]
*Please enter your answer in the provided input box.*
**Explanation:**
To compute the dot product, follow these steps:
1. Multiply the corresponding components of the vectors.
2. Add up all the products obtained.
The dot product formula for two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\) is calculated as follows:
\[ \mathbf{a} \cdot \mathbf{b} = (a_1 \cdot b_1) + (a_2 \cdot b_2) + (a_3 \cdot b_3) \]
In this specific problem, expand the vectors and compute the products accordingly:
\[ (\mathbf{i} + \mathbf{j}) \cdot (\mathbf{j} + \mathbf{k}) \]
Now, apply distributive property:
\[ \mathbf{i} \cdot \mathbf{j} + \mathbf{i} \cdot \mathbf{k} + \mathbf{j} \cdot \mathbf{j} + \mathbf{j} \cdot \mathbf{k} \]
Remember:
- \(\mathbf{i} \cdot \mathbf{i} = 1\)
- \(\mathbf{j} \cdot \mathbf{j} = 1\)
- \(\mathbf{k} \cdot \mathbf{k} = 1\)
- \(\mathbf{i} \cdot \mathbf{j} = 0\)
- \(\mathbf{i} \cdot \mathbf{k} = 0\)
- \(\mathbf{j} \cdot \mathbf{k} = 0\)
Combine all the results to find the final dot product value.
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