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Magazine Chapter 12: Problem 5
Find a. \(\mathbf{v} \cdot \mathbf{u},|\mathbf{v}|,|\mathbf{u}|\) b. the cosine of the angle between \(\mathbf{v}\) and \(\mathbf{u}\) c. the scalar component of \(u\) in the direction of \(v\) d. the vector proj, u. $$ \mathbf{v}=5 \mathbf{j}-3 \mathbf{k}, \quad \mathbf{u}=\mathbf{i}+\mathbf{j}+\mathbf{k} $$
Short Answer
Expert verified
a. 2, \(\sqrt{34}\), \(\sqrt{3}\); b. \(\frac{2}{\sqrt{102}}\); c. \(\frac{2}{\sqrt{34}}\); d. \(0, \frac{5}{17}, \frac{-3}{17}\).
Step by step solution
01
Calculate the Dot Product
The dot product of two vectors \( \mathbf{v} \) and \( \mathbf{u} \) is given by \( \mathbf{v} \cdot \mathbf{u} = v_i u_i + v_j u_j + v_k u_k \). First, write the vectors with their components: \( \mathbf{v} = (0, 5, -3) \) and \( \mathbf{u} = (1, 1, 1) \). Now, compute: \[ \mathbf{v} \cdot \mathbf{u} = 0 \times 1 + 5 \times 1 - 3 \times 1 = 0 + 5 - 3 = 2. \]
02
Calculate the Magnitudes
The magnitude of a vector \( \mathbf{v} = (v_i, v_j, v_k) \) is given by the formula: \[ |\mathbf{v}| = \sqrt{v_i^2 + v_j^2 + v_k^2}. \]For \( \mathbf{v} = (0, 5, -3) \), the magnitude is:\[ |\mathbf{v}| = \sqrt{0^2 + 5^2 + (-3)^2} = \sqrt{0 + 25 + 9} = \sqrt{34}. \]For \( \mathbf{u} = (1, 1, 1) \), the magnitude is:\[ |\mathbf{u}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}. \]
03
Find Cosine of the Angle Between Vectors
The cosine of the angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{u} \) is given by \[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| \, |\mathbf{u}|}. \]Substitute the values obtained:\[ \cos \theta = \frac{2}{\sqrt{34} \, \sqrt{3}} = \frac{2}{\sqrt{102}}. \]
04
Calculate the Scalar Component of u in the Direction of v
The scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by:\[ \text{Scalar component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|}. \]Using the previously calculated values:\[ \text{Scalar component} = \frac{2}{\sqrt{34}}. \]
05
Calculate the Projection of u onto v
The vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is given by:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \right) \mathbf{v}. \]Substituting the known values:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{2}{34} \right) (0, 5, -3) = (0, \frac{10}{34}, \frac{-6}{34}) = \left( 0, \frac{5}{17}, \frac{-3}{17} \right). \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
In vector calculus, the dot product, also known as the scalar product, is a way to multiply two vectors resulting in a scalar. It is calculated by multiplying corresponding components of two vectors and summing the results. For example, if you have vectors \( \mathbf{v} = (v_i, v_j, v_k) \) and \( \mathbf{u} = (u_i, u_j, u_k) \), the dot product is computed as:
- \( \mathbf{v} \cdot \mathbf{u} = v_i u_i + v_j u_j + v_k u_k \)
Magnitude of a Vector
The magnitude of a vector, also called the vector's length, is a measure of how long the vector is. For a three-dimensional vector \( \mathbf{v} = (v_i, v_j, v_k) \), it is found using the formula:
- \( |\mathbf{v}| = \sqrt{v_i^2 + v_j^2 + v_k^2} \)
- \( |\mathbf{v}| = \sqrt{0^2 + 5^2 + (-3)^2} = \sqrt{34} \)
Cosine of the Angle
The cosine of the angle between two vectors \( \mathbf{v} \) and \( \mathbf{u} \) can be used to measure the orientation between them. The formula to find it is:
- \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \)
- \( \cos \theta = \frac{2}{\sqrt{34} \sqrt{3}} = \frac{2}{\sqrt{102}} \)
Scalar Component
A scalar component represents how much of one vector lies in the direction of another vector. It is essentially a projection of one vector on a line parallel to another. The formula used is:
- \( \text{Scalar component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \)
- \( \frac{2}{\sqrt{34}} \)
Vector Projection
The vector projection of one vector onto another shows the shadow or footprint of a vector on another vector. It can be visualized as how much of one vector aligns with the direction of a second vector. Here's the formula to calculate it:
- \[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \right) \mathbf{v} \]
- \( \left( 0, \frac{5}{17}, \frac{-3}{17} \right) \)
