Chapter 10: Problem 20
Find \(\operatorname{com} p_{b}\) a and projo a. \(\mathbf{a}=3 \mathbf{i}+\mathbf{j}, \mathbf{b}=4 \mathbf{i}-3 \mathbf{j}\)
Short Answer
Expert verified
The projection of vector a onto vector b is 0.36 and the component of vector a along b is \(1.44\mathbf{i} - 1.08\mathbf{j}\)
Step by step solution
01
Calculate the Dot Product of a and b
The dot product of vectors \(\mathbf{a}\) and \(\mathbf{b}\) is calculated using their components. In this case, \(\mathbf{a} . \mathbf{b}=a_{i} \cdot b_{i}+a_{j} \cdot b_{j}=3\cdot4+1\cdot(-3)=12-3=9\)
02
Calculate the Magnitude of b
The magnitude of vector \(\mathbf{b}\) is given by \(\sqrt{b_{i}^{2}+b_{j}^{2}}=\sqrt{4^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5\)
03
Compute the projection
The projection of \(\mathbf{a}\) onto \(\mathbf{b}\) uses the dot product of \(\mathbf{a}\) and \(\mathbf{b}\) and the square of the magnitude of \(\mathbf{b}\). Thus \(\operatorname{projo}_{\mathbf{b}} \mathbf{a}= (\mathbf{a} . \mathbf{b}) / |\mathbf{b}|^2=(9)/(5^2)=(9)/(25)=0.36\)
04
Compute the component of a
\(\operatorname{com}_{\mathbf{b}} \mathbf{a} = \operatorname{projo}_{\mathbf{b}} \mathbf{a} \cdot \mathbf{b} =0.36 \cdot( 4\mathbf{i}-3 \mathbf{j})=1.44 \mathbf{i}- 1.08 \mathbf{j}\)
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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 when working with vectors. It combines two vectors and results in a scalar. To find the dot product of two vectors, such as \(\mathbf{a} = 3\mathbf{i} + \mathbf{j}\) and \(\mathbf{b} = 4\mathbf{i} - 3\mathbf{j}\), you multiply their corresponding components and sum the results.
For these vectors, the dot product \(\mathbf{a} \cdot \mathbf{b}\) is calculated as follows:
For these vectors, the dot product \(\mathbf{a} \cdot \mathbf{b}\) is calculated as follows:
- Multiply the \(\mathbf{i}\)-components: \(3 \cdot 4 = 12\).
- Multiply the \(\mathbf{j}\)-components: \(1 \cdot (-3) = -3\).
- Add these two results together: \(12 - 3 = 9\).
Magnitude of a Vector
Magnitude, or the length of a vector, is a measurement of how long the vector is. It's calculated using the Pythagorean theorem in the context of vector components. For a vector \(\mathbf{b} = 4\mathbf{i} - 3\mathbf{j}\), the magnitude is derived from its components.
The formula to find the magnitude \(|\mathbf{b}|\) is:
The formula to find the magnitude \(|\mathbf{b}|\) is:
- Square each component: \(4^2 = 16\) and \((-3)^2 = 9\).
- Sum these squared values: \(16 + 9 = 25\).
- Take the square root of this sum: \(\sqrt{25} = 5\).
Vector Components
Vector components allow us to break down a vector into its base parts. This breakdown is essential for calculations, such as the dot product and finding projections. A vector like \(\mathbf{a}=3 \mathbf{i}+ \mathbf{j}\) has components:
- In the \(\mathbf{i}\) direction: it has a component of \(3\).
- In the \(\mathbf{j}\) direction: it has a component of \(1\).
Projection of Vectors
The projection of one vector onto another serves to "project" a vector in the direction of another. This provides a vector parallel to the second vector. In the case of vectors \(\mathbf{a}\) onto \(\mathbf{b}\), the projection is computed using the dot product and the magnitude of the vector \(\mathbf{b}\).
The formula used here is:
The formula used here is:
- Calculate the dot product of \(\mathbf{a}\) and \(\mathbf{b}\): \(9\).
- Determine the square of the magnitude of \(\mathbf{b}\): \(5^2 = 25\).
- Divide the dot product by this square: \(\frac{9}{25} = 0.36\).
- \(0.36 \cdot (4\mathbf{i}-3\mathbf{j}) = 1.44 \mathbf{i} - 1.08 \mathbf{j}\).