Chapter 11: Problem 33
Find each of the given projections if \(\mathbf{u}=3 \mathbf{i}+2 \mathbf{j}+\mathbf{k}, \mathbf{v}=2 \mathbf{i}-\mathbf{k}\), and \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}-3 \mathbf{k}\). \(\operatorname{proj}_{\mathbf{k}} \mathbf{u}\)
Short Answer
Expert verified
The projection \( \operatorname{proj}_{\mathbf{k}} \mathbf{u} \) is \( \mathbf{k} \).
Step by step solution
01
Understand Projection Formula
The projection of a vector \( \mathbf{u} \) onto another vector \( \mathbf{a} \) is given by the formula \( \operatorname{proj}_{\mathbf{a}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{a}}{\|\mathbf{a}\|^2} \mathbf{a} \). To find \( \operatorname{proj}_{\mathbf{k}} \mathbf{u} \), you need to identify the direction vector, which in this case is \( \mathbf{k} = 0\mathbf{i} + 0\mathbf{j} + 1\mathbf{k} \).
02
Find Dot Product
Calculate the dot product \( \mathbf{u} \cdot \mathbf{k} \). This is calculated as follows: \( (3\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \cdot (0\mathbf{i} + 0\mathbf{j} + 1\mathbf{k}) = 3 \cdot 0 + 2 \cdot 0 + 1 \cdot 1 = 1 \).
03
Calculate Magnitude Squared
Find the magnitude squared \( \|\mathbf{k}\|^2 \). Since \( \mathbf{k} \) is the unit vector in the \( k \) direction, \( \|\mathbf{k}\| = 1 \). Therefore, \( \|\mathbf{k}\|^2 = 1^2 = 1 \).
04
Compute Projection
Now, use the projection formula: \( \operatorname{proj}_{\mathbf{k}} \mathbf{u} = \frac{1}{1} (0\mathbf{i} + 0\mathbf{j} + 1\mathbf{k}) = \mathbf{k} = 0\mathbf{i} + 0\mathbf{j} + 1\mathbf{k} \). Therefore, the projection of \( \mathbf{u} \) onto \( \mathbf{k} \) is simply \( \mathbf{k} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
The dot product is a fundamental operation in vector algebra that allows you to measure the interaction between two vectors. It is represented by the symbol \( \cdot \) and involves multiplying corresponding components of two vectors and summing those products.
For vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is given by:
In vector projection, we use the dot product to find how much of one vector goes in the direction of another, which is crucial for calculating projections correctly.
For vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is given by:
- \( a_1b_1 + a_2b_2 + a_3b_3 \)
In vector projection, we use the dot product to find how much of one vector goes in the direction of another, which is crucial for calculating projections correctly.
Magnitude of a Vector
The magnitude of a vector, often represented by \( \| \mathbf{a} \| \), is its length or size. It is calculated using the Pythagorean theorem for the components of the vector.
For a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \), the magnitude is calculated as:
This tells us how large or small the resulting projection should be based on the direction vector's size.
Moreover, in this particular exercise, \( \mathbf{k} \) is a unit vector, so its magnitude is 1, simplifying the projection calculations.
For a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \), the magnitude is calculated as:
- \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)
This tells us how large or small the resulting projection should be based on the direction vector's size.
Moreover, in this particular exercise, \( \mathbf{k} \) is a unit vector, so its magnitude is 1, simplifying the projection calculations.
Unit Vector
A unit vector is a vector with a magnitude of 1. It provides a direction without any influence of scale. To convert any non-zero vector into a unit vector, you divide it by its magnitude.
Mathematically, if \( \mathbf{a} \) is a vector, its unit vector \( \hat{\mathbf{a}} \) is given by:
In our exercise, the vector \( \mathbf{k} \) is already a unit vector, which means that when we project onto it, the direction is preserved without needing additional calculations for normalization.
Mathematically, if \( \mathbf{a} \) is a vector, its unit vector \( \hat{\mathbf{a}} \) is given by:
- \( \hat{\mathbf{a}} = \frac{\mathbf{a}}{\| \mathbf{a} \|} \)
In our exercise, the vector \( \mathbf{k} \) is already a unit vector, which means that when we project onto it, the direction is preserved without needing additional calculations for normalization.
Projection Formula
The projection formula is used to project a vector \( \mathbf{u} \) onto another vector \( \mathbf{a} \). This is visualized as the shadow that is cast by one vector onto the line defined by another.
The formula for the projection of \( \mathbf{u} \) onto \( \mathbf{a} \) is:
By understanding these components, the concept of projection becomes clear and approachable.
The formula for the projection of \( \mathbf{u} \) onto \( \mathbf{a} \) is:
- \( \operatorname{proj}_{\mathbf{a}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{a}}{\| \mathbf{a} \|^2} \mathbf{a} \)
By understanding these components, the concept of projection becomes clear and approachable.