Problem 35
Question
Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\langle- 12,5\rangle $$
Step-by-Step Solution
Verified Answer
The vector has a magnitude of 13 and a direction of approximately 157.38°.
1Step 1: Finding the Magnitude
To find the magnitude of vector \( \mathbf{v} = \langle -12, 5 \rangle \), use the formula for the magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \):\[||\mathbf{v}|| = \sqrt{a^2 + b^2}\]Substitute the components of \( \mathbf{v} \):\[||\mathbf{v}|| = \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\]Thus, the magnitude of the vector is 13.
2Step 2: Calculating the Direction Angle
The direction angle \( \theta \) of the vector can be found using the tangent function:\[\tan \theta = \frac{b}{a}\]Substitute the components of \( \mathbf{v} \):\[\tan \theta = \frac{5}{-12}\]Calculate the angle using the arctangent function:\[\theta = \tan^{-1}\left(\frac{5}{-12}\right)\]This results in an angle of approximately \(-22.62^\circ\). However, since the vector is in the second quadrant (negative \( x \)-component, positive \( y \)-component), add 180°:\[\theta = -22.62^\circ + 180^\circ = 157.38^\circ\]Thus, the direction of the vector is approximately \( 157.38^\circ \).
Key Concepts
Vector MagnitudeVector Direction AngleTangent Function in VectorsArctangent FunctionQuadrant Analysis for Vectors
Vector Magnitude
A vector magnitude represents the length or size of the vector. It is always a non-negative value and provides a measure of how long the vector is. To find the magnitude of a vector, use the formula: \[||\mathbf{v}|| = \sqrt{a^2 + b^2}\] where \(a\) and \(b\) are the components of the vector.
- In our exercise, the vector is \(\mathbf{v} = \langle -12, 5 \rangle\).
- To find its magnitude: \[\sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\]
Vector Direction Angle
The direction angle of a vector points towards where the vector is headed on a coordinate plane. It is crucial for understanding the orientation of the vector. To find this angle \(\theta\), the tangent function is used:\[\tan \theta = \frac{b}{a}\]Given our vector \(\mathbf{v} = \langle -12, 5 \rangle\):
- Apply the formula: \[\tan \theta = \frac{5}{-12}\]
Tangent Function in Vectors
The tangent function relates an angle within a right triangle to the ratio of the opposite side to the adjacent side. In vector analysis, it helps us understand angles relative to the vector components.
- For our vector \(\mathbf{v} = \langle -12, 5 \rangle\), we calculate \(\tan \theta = \frac{5}{-12}\).
Arctangent Function
The arctangent function, or \( \tan^{-1} \), takes a tangent value and returns the angle corresponding to it. This is typically between \(-90^\circ\) and \(90^\circ\). For a broader range, adjustments are needed based on the vector’s orientation:
- From \(\tan \theta = \frac{5}{-12}\), \( \theta = \tan^{-1}\left( \frac{5}{-12} \right) \) corresponds to about \(-22.62^\circ\).
Quadrant Analysis for Vectors
Analyzing the quadrant where a vector resides helps determine its direction more accurately. The coordinate plane is divided into four quadrants:
- Quadrant I: Positive \(x\) and \(y\)
- Quadrant II: Negative \(x\), positive \(y\)
- Quadrant III: Negative \(x\) and \(y\)
- Quadrant IV: Positive \(x\), negative \(y\)
- \(-22.62^\circ + 180^\circ = 157.38^\circ\)
Other exercises in this chapter
Problem 34
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33-36 Let \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) be vectors and let \(a\) be a scalar. Prove the given property. $$(\mathbf{u}+\mathbf{v}) \cdot \mathbf
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Convert the rectangular coordinates to polar coordinates with \(r>0\) and \(0 \leq \theta
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