Problem 13
Question
(a) find the dot product v \(\cdot \mathbf{w} ;\) (b) find the angle between \(\mathbf{v}\) and \(\mathbf{w} ;\) (c) state whether the vectors are parallel, orthogonal, or neither. $$ \mathbf{v}=\sqrt{3} \mathbf{i}-\mathbf{j}, \quad \mathbf{w}=\mathbf{i}+\mathbf{j} $$
Step-by-Step Solution
Verified Answer
\(\mathbf{v} \cdot \mathbf{w} = \sqrt{3} - 1\), \(\theta = \cos^{-1}\left(\frac{\sqrt{3} - 1}{2 \sqrt{2}}\right)\), They are neither parallel nor orthogonal.
1Step 1: Identify Vector Components
Extract the components of vectors \(\mathbf{v}\) and \(\mathbf{w}\). \(\mathbf{v} = \sqrt{3} \mathbf{i} - \mathbf{j}\) means \(\mathbf{v} = (\sqrt{3}, -1)\) and \(\mathbf{w} = \mathbf{i} + \mathbf{j}\) means \(\mathbf{w} = (1, 1)\).
2Step 2: Calculate Dot Product
The dot product of \(\mathbf{v}\) and \(\mathbf{w}\) is given by the formula \(\mathbf{v} \cdot \mathbf{w} = v_{1} w_{1} + v_{2} w_{2}\). So, \(\sqrt{3} \cdot 1 + (-1) \cdot 1 = \sqrt{3} - 1\). Therefore, \(\mathbf{v} \cdot \mathbf{w} = \sqrt{3} - 1\).
3Step 3: Find Magnitudes of Vectors
Calculate the magnitudes \(\|\mathbf{v}\|\) and \(\|\mathbf{w}\|\). \(\|\mathbf{v}\| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2\). \(\|\mathbf{w}\| = \sqrt{1^2 + 1^2} = \sqrt{2}\).
4Step 4: Calculate Cosine of the Angle
Use the dot product formula to find the cosine of the angle \(\theta\). \(\cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|}\). Substituting we get \(\cos \theta = \frac{\sqrt{3} - 1}{2 \sqrt{2}}\).
5Step 5: Determine Type of Vectors
Check the conditions for parallelism or orthogonality. Vectors are orthogonal if \(\mathbf{v} \cdot \mathbf{w} = 0\). Here, \(\mathbf{v} \cdot \mathbf{w} = \sqrt{3} - 1 eq 0\), so they are not orthogonal. They are parallel if \(\theta = 0\) or \(\theta = \pi\). Since \(\cos \theta eq 1\) or \( -1 \), they are not parallel. So, they are neither.
Key Concepts
VectorsDot ProductAngle Between VectorsVector Magnitudes
Vectors
Vectors are mathematical objects that have both a magnitude (or length) and a direction.
They are commonly represented with an arrow in a coordinate system.
For this exercise, we have two vectors: \( \textbf{v} = \root 3 \textbf{i} - \textbf{j} \) and \( \textbf{w} = \textbf{i} + \textbf{j} \).
Each vector can be broken down into its components along the coordinate axes.
For instance, \( \textbf{v} \) has components \( ( \root 3, -1 ) \) and \( \textbf{w} \) has components \( (1, 1) \).
These components help us in various vector calculations such as addition, subtraction, and the dot product.
They are commonly represented with an arrow in a coordinate system.
For this exercise, we have two vectors: \( \textbf{v} = \root 3 \textbf{i} - \textbf{j} \) and \( \textbf{w} = \textbf{i} + \textbf{j} \).
Each vector can be broken down into its components along the coordinate axes.
For instance, \( \textbf{v} \) has components \( ( \root 3, -1 ) \) and \( \textbf{w} \) has components \( (1, 1) \).
These components help us in various vector calculations such as addition, subtraction, and the dot product.
Dot Product
The dot product (also known as the scalar product) is a way to multiply two vectors resulting in a scalar.
The formula for the dot product of two vectors \( \textbf{v} \) and \( \textbf{w} \) is: \[ \textbf{v} \bf{\bullet} \textbf{w} = v_{1}w_{1} + v_{2}w_{2} \]Here, you multiply the corresponding components of both vectors and then add the results.
For our vectors, we have: \[ \textbf{v} \bf{\bullet} \textbf{w} = (\root 3 \times 1) + (-1 \times 1) = \root 3 - 1 \]. This is a single number and tells us something about the relative orientation of \( \textbf{v} \) and \( \textbf{w} \).
The formula for the dot product of two vectors \( \textbf{v} \) and \( \textbf{w} \) is: \[ \textbf{v} \bf{\bullet} \textbf{w} = v_{1}w_{1} + v_{2}w_{2} \]Here, you multiply the corresponding components of both vectors and then add the results.
For our vectors, we have: \[ \textbf{v} \bf{\bullet} \textbf{w} = (\root 3 \times 1) + (-1 \times 1) = \root 3 - 1 \]. This is a single number and tells us something about the relative orientation of \( \textbf{v} \) and \( \textbf{w} \).
Angle Between Vectors
The angle between two vectors can be found using the dot product.
When you take the dot product of two vectors, it is related to the cosine of the angle between them.
The relationship is given by the formula: \[ \textbackslash cos(\theta) = \frac{\textbf{v} \bullet \textbf{w}}{orm{\textbf{v}}orm{\textbf{w}}} \], where \( \theta \) is the angle between vectors \( \textbf{v} \) and \( \textbf{w} \) and \( orm{\textbf{v}} \), \( orm{\textbf{w}} \) are the magnitudes of the vectors.
For our vectors: \[ \textbackslash cos(\theta) = \frac{\root 3 - 1}{2\textbackslash sqrt{2}} \] The value of \( \textbackslash cos(\theta) \) lets us find \( \theta \) using the inverse cosine function.
When you take the dot product of two vectors, it is related to the cosine of the angle between them.
The relationship is given by the formula: \[ \textbackslash cos(\theta) = \frac{\textbf{v} \bullet \textbf{w}}{orm{\textbf{v}}orm{\textbf{w}}} \], where \( \theta \) is the angle between vectors \( \textbf{v} \) and \( \textbf{w} \) and \( orm{\textbf{v}} \), \( orm{\textbf{w}} \) are the magnitudes of the vectors.
For our vectors: \[ \textbackslash cos(\theta) = \frac{\root 3 - 1}{2\textbackslash sqrt{2}} \] The value of \( \textbackslash cos(\theta) \) lets us find \( \theta \) using the inverse cosine function.
Vector Magnitudes
The magnitude (or length) of a vector is a measure of how long the vector is.
For a vector \( \textbf{v} = (v_{1}, v_{2}) \), the magnitude is given by \[ orm{\textbf{v}} = \textbackslash sqrt{v_{1}^{2} + v_{2}^{2}} \]It is calculated using the Pythagorean theorem on its components.
For our vector \( \textbf{v} \), the magnitude is \[ orm{\textbf{v}} = \textbackslash sqrt{( \root 3 )^2 + (-1)^2} = 2 \]. Similarly, for \( \textbf{w} \), the magnitude is \[ orm{\textbf{w}} = \textbackslash sqrt{1^2 + 1^2} = \textbackslash sqrt{2} \]. Understanding magnitudes is essential for calculating vector dot products and angles.
For a vector \( \textbf{v} = (v_{1}, v_{2}) \), the magnitude is given by \[ orm{\textbf{v}} = \textbackslash sqrt{v_{1}^{2} + v_{2}^{2}} \]It is calculated using the Pythagorean theorem on its components.
For our vector \( \textbf{v} \), the magnitude is \[ orm{\textbf{v}} = \textbackslash sqrt{( \root 3 )^2 + (-1)^2} = 2 \]. Similarly, for \( \textbf{w} \), the magnitude is \[ orm{\textbf{w}} = \textbackslash sqrt{1^2 + 1^2} = \textbackslash sqrt{2} \]. Understanding magnitudes is essential for calculating vector dot products and angles.
Other exercises in this chapter
Problem 12
Multiple Choice If \(z_{1}=r_{1} e^{i \theta_{1}}\) and \(z_{2}=r_{2} e^{i \theta_{2}}\) are complex numbers, then \(\frac{z_{1}}{z_{2}}, z_{2} \neq 0,\) equals
View solution Problem 12
True or False A cardioid passes through the pole.
View solution Problem 13
In Problems \(13-24,\) plot each complex number in the complex plane and write it in polar form and in exponential form. $$ 1+i $$
View solution Problem 14
(a) find the dot product v \(\cdot \mathbf{w} ;\) (b) find the angle between \(\mathbf{v}\) and \(\mathbf{w} ;\) (c) state whether the vectors are parallel, ort
View solution