Chapter 3: Problem 1
Determine the magnitude and the direction angle of each of the following vectors. (a) \(\mathbf{v}=3 \mathbf{i}+5 \mathbf{j}\) (c) \(\mathbf{a}=4 \mathbf{i}-7 \mathbf{j}\) (b) \(\mathbf{w}=\langle-3,6\rangle\) (d) \(\mathbf{u}=\langle-3,-5\rangle\)
Short Answer
Expert verified
The magnitudes and direction angles of the vectors are:
a) \(\| \mathbf{v} \| = \sqrt{34}\) and \(\theta \approx 59.04^\circ\)
b) \(\| \mathbf{w} \| = \sqrt{45}\) and \(\theta \approx 116.57^\circ\)
c) \(\| \mathbf{a} \| = \sqrt{65}\) and \(\theta \approx -60.26^\circ\)
d) \(\| \mathbf{u} \| = \sqrt{34}\) and \(\theta \approx 149.04^\circ\)
Step by step solution
01
To find the magnitude of \(\mathbf{v}=3\mathbf{i}+5\mathbf{j}\), we can use the formula \(\| \mathbf{v} \| = \sqrt{a^2 + b^2}\) with \(a=3\) and \(b=5\). Therefore, the magnitude is \[ \| \mathbf{v} \| = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \] #a) Direction angle of \(\mathbf{v}\)#
To find the direction angle \(\theta\) of \(\mathbf{v}\), we use the formula \(\theta = \tan^{-1}(\frac{b}{a})\) with \(a=3\) and \(b=5\). Thus, the direction angle is \[\theta = \tan^{-1} \left(\frac{5}{3} \right) \approx 59.04^\circ\]
#c) Magnitude of \(\mathbf{a}\)#
02
To find the magnitude of \(\mathbf{a}=4\mathbf{i}-7\mathbf{j}\), we can use the formula \(\| \mathbf{a} \| = \sqrt{a^2 + b^2}\) with \(a=4\) and \(b=-7\). Therefore, the magnitude is \[ \| \mathbf{a} \| = \sqrt{4^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65} \] #c) Direction angle of \(\mathbf{a}\)#
To find the direction angle \(\theta\) of \(\mathbf{a}\), we use the formula \(\theta = \tan^{-1}(\frac{b}{a})\) with \(a=4\) and \(b=-7\). Thus, the direction angle is \[\theta = \tan^{-1} \left(\frac{-7}{4} \right) \approx -60.26^\circ\]
#b) Magnitude of \(\mathbf{w}\)#
03
To find the magnitude of \(\mathbf{w}=\langle-3,6\rangle\), we can use the formula \(\| \mathbf{w} \| = \sqrt{a^2 + b^2}\) with \(a=-3\) and \(b=6\). Therefore, the magnitude is \[ \| \mathbf{w} \| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \] #b) Direction angle of \(\mathbf{w}\)#
To find the direction angle \(\theta\) of \(\mathbf{w}\), we use the formula \(\theta = \tan^{-1}(\frac{b}{a})\) with \(a=-3\) and \(b=6\). Thus, the direction angle is \[\theta = \tan^{-1} \left(\frac{6}{-3} \right) \approx 116.57^\circ\]
#d) Magnitude of \(\mathbf{u}\)#
04
To find the magnitude of \(\mathbf{u}=\langle-3,-5\rangle\), we can use the formula \(\| \mathbf{u} \| = \sqrt{a^2 + b^2}\) with \(a=-3\) and \(b=-5\). Therefore, the magnitude is \[ \| \mathbf{u} \| = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \] #d) Direction angle of \(\mathbf{u}\)#
To find the direction angle \(\theta\) of \(\mathbf{u}\), we use the formula \(\theta = \tan^{-1}(\frac{b}{a})\) with \(a=-3\) and \(b=-5\). Thus, the direction angle is \[\theta = \tan^{-1} \left(\frac{-5}{-3} \right) \approx 149.04^\circ\]
The magnitudes and direction angles of the vectors are as follows:
a) \(\| \mathbf{v} \| = \sqrt{34}\) and \(\theta \approx 59.04^\circ\)
b) \(\| \mathbf{w} \| = \sqrt{45}\) and \(\theta \approx 116.57^\circ\)
c) \(\| \mathbf{a} \| = \sqrt{65}\) and \(\theta \approx -60.26^\circ\)
d) \(\| \mathbf{u} \| = \sqrt{34}\) and \(\theta \approx 149.04^\circ\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude
The magnitude of a vector is essentially the vector's length. Imagine a straight line connecting the origin of a graph to the vector's endpoint. The magnitude tells you how long this line is. We calculate the magnitude using the Pythagorean theorem. This is because the vector forms a right triangle with the horizontal and vertical axes.
For any vector in the form \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\), the formula for magnitude is \(\|\mathbf{v}\| = \sqrt{a^2 + b^2}\). This formula results from the Pythagorean theorem, applied to the horizontal component \(a\) and the vertical component \(b\) of the vector.
For example:
For any vector in the form \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\), the formula for magnitude is \(\|\mathbf{v}\| = \sqrt{a^2 + b^2}\). This formula results from the Pythagorean theorem, applied to the horizontal component \(a\) and the vertical component \(b\) of the vector.
For example:
- For the vector \(\mathbf{v} = 3\mathbf{i} + 5\mathbf{j}\), the magnitude is calculated as \(\sqrt{3^2 + 5^2} = \sqrt{34}\).
- Similarly, for the vector \(\mathbf{w} = \langle -3, 6 \rangle\), the magnitude is \(\sqrt{(-3)^2 + 6^2} = \sqrt{45}\).
Direction Angle
The direction angle of a vector indicates the vector's angle away from a reference direction, usually the positive x-axis. It demonstrates the vector's orientation in space. To find the direction angle \(\theta\) of a vector, we use the tangent function, which relates an angle of a right triangle to its opposite and adjacent sides.
The formula to find the direction angle is \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\), where \(a\) and \(b\) are the horizontal and vertical components, respectively.
The formula to find the direction angle is \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\), where \(a\) and \(b\) are the horizontal and vertical components, respectively.
- For the vector \(\mathbf{v} = 3\mathbf{i} + 5\mathbf{j}\), the direction angle is \(\tan^{-1}\left(\frac{5}{3}\right) \approx 59.04^\circ\).
- For the vector \(\mathbf{a} = 4\mathbf{i} - 7\mathbf{j}\), the angle is \(\tan^{-1}\left(\frac{-7}{4}\right) \approx -60.26^\circ\).
Component Form
The component form of a vector gives us a clear view of its horizontal and vertical directions. Each vector can be written in component form using the notation \(\langle a, b \rangle\) or \(a\mathbf{i} + b\mathbf{j}\). These components tell exactly how far and in which direction they push or pull from the origin.
For example, consider the vector \(\mathbf{v} = 3\mathbf{i} + 5\mathbf{j}\). Here:
For example, consider the vector \(\mathbf{v} = 3\mathbf{i} + 5\mathbf{j}\). Here:
- The component \(3\mathbf{i}\) represents a move along the x-axis.
- The component \(5\mathbf{j}\) shows a move along the y-axis.
- The \(-3\) indicates a leftward move on the x-axis.
- The \(6\) suggests an upward move on the y-axis.
Tangent Function
The tangent function, \(\tan(\theta)\), is a trigonometric function that arises several times when discussing vectors, especially in finding the direction angle. It relates the opposite side to the adjacent side of a right triangle. In the context of vectors, these sides correspond to the vertical component \(b\) and horizontal component \(a\), respectively.
When we want to find a direction angle \(\theta\) for a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\), the tangent function gives us \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\). Solutions to such problems involve considering which quadrants the vectors lie in because tangent values change signs across the axes.
Characteristics of the tangent function include:
When we want to find a direction angle \(\theta\) for a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\), the tangent function gives us \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\). Solutions to such problems involve considering which quadrants the vectors lie in because tangent values change signs across the axes.
Characteristics of the tangent function include:
- The ratio is positive when both \(a\) and \(b\) are positive or both are negative.
- The ratio is negative when only one of \(a\) or \(b\) is negative.
- Tangent values can repeat in a cycle of 180° because of its period.