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Chapter 9: Problem 35

Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\langle- 12,5\rangle $$

Short Answer

Expert verified
The vector has a magnitude of 13 and a direction of approximately 157.38掳.

Step by step solution

01

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.
02

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掳:\[\theta = -22.62^\circ + 180^\circ = 157.38^\circ\]Thus, the direction of the vector is approximately \( 157.38^\circ \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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: \[\sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\]
This calculates to a magnitude of 13. The magnitude is crucial for understanding the scale of the vector's effect.
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: \[\tan \theta = \frac{5}{-12}\]
Generally, this gives only a reference angle. The true direction in degrees must take into account the vector's position around the coordinate axes.
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 \(\tan \theta = \frac{5}{-12}\).
This provides a clue about which quadrant the vector points towards. To find the angle itself, the inverse of the tangent function, or the arctangent, is used.
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鈥檚 orientation:
  • From \(\tan \theta = \frac{5}{-12}\), \( \theta = \tan^{-1}\left( \frac{5}{-12} \right) \) corresponds to about \(-22.62^\circ\).
Note, this is just a direction and requires quadrant analysis to finalize the true angle.
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 \(y\)
For \(\langle -12, 5 \rangle\):- The vector lies in Quadrant II (negative \(x\), positive \(y\)).- Adjust the direction by adding \(180^\circ\) since the calculated angle is relative to the negative \(x\)-axis:
  • \(-22.62^\circ + 180^\circ = 157.38^\circ\)
This makes the direction approximately \(157.38^\circ\), adjusting for the vector's quadrant.

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