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Magazine Chapter 3: Problem 8
$$ \vec{\mathbf{v}}_{1}=-6.0 \hat{\mathrm{i}}+8.0 \hat{\mathrm{j}} \text { and } \vec{\mathbf{v}}_{2}=4.5 \hat{\mathrm{i}}-5.0 \mathrm{j} . $$ mine the magnitude and direction of $$ (a) \vec{\mathbf{v}}_{1},(b) \vec{\mathbf{v}}_{2} $$ $$ (c) \vec{\mathbf{v}}_{1}+\vec{\mathbf{v}}_{2} \text { and }(d) \vec{\mathbf{v}}_{2}-\vec{\mathbf{v}}_{1-} $$
Short Answer
Expert verified
(a) Magnitude: 10.0, Direction: 126.87掳; (b) Magnitude: 6.73, Direction: 311.63掳; (c) Magnitude: 3.35, Direction: 116.57掳; (d) Magnitude: 16.71, Direction: 308.66掳.
Step by step solution
01
Find Magnitude of \( \vec{\mathbf{v}}_1 \)
To find the magnitude of a vector \( \vec{\mathbf{v}} = a \hat{\mathrm{i}} + b \hat{\mathrm{j}} \), use the formula \( |\vec{\mathbf{v}}| = \sqrt{a^2 + b^2} \). Here, \( a = -6.0 \) and \( b = 8.0 \), so: \[ |\vec{\mathbf{v}}_1| = \sqrt{(-6.0)^2 + (8.0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10.0 \]
02
Determine Direction of \( \vec{\mathbf{v}}_1 \)
The direction (angle \( \theta \) with the positive x-axis) of a vector \( \vec{\mathbf{v}} \) can be found using \( \tan^{-1}(b/a) \). Here, \( a = -6.0 \) and \( b = 8.0 \): \[ \theta = \tan^{-1}\left(\frac{8.0}{-6.0}\right) \approx \tan^{-1}(-1.33) \approx -53.13^\circ \] Since the vector is in the second quadrant, the final angle is \( 180^\circ - 53.13^\circ \approx 126.87^\circ \).
03
Find Magnitude of \( \vec{\mathbf{v}}_2 \)
For the vector \( \vec{\mathbf{v}}_2 = 4.5 \hat{\mathrm{i}} - 5.0 \hat{\mathrm{j}} \), use the same formula: \[ |\vec{\mathbf{v}}_2| = \sqrt{(4.5)^2 + (-5.0)^2} = \sqrt{20.25 + 25} = \sqrt{45.25} \approx 6.73 \]
04
Determine Direction of \( \vec{\mathbf{v}}_2 \)
The direction is given by: \[ \theta = \tan^{-1}\left(\frac{-5.0}{4.5}\right) \approx \tan^{-1}(-1.11) \approx -48.37^\circ \] Since the vector is in the fourth quadrant, the angle with the positive x-axis is \( 360^\circ - 48.37^\circ \approx 311.63^\circ \).
05
Find \( \vec{\mathbf{v}}_1 + \vec{\mathbf{v}}_2 \)
Use vector addition:\[ \vec{\mathbf{v}}_1 + \vec{\mathbf{v}}_2 = (-6.0 + 4.5) \hat{\mathrm{i}} + (8.0 - 5.0) \hat{\mathrm{j}} = -1.5 \hat{\mathrm{i}} + 3.0 \hat{\mathrm{j}} \]
06
Magnitude and Direction of \( \vec{\mathbf{v}}_1 + \vec{\mathbf{v}}_2 \)
Magnitude: \[ |\vec{\mathbf{v}}_1 + \vec{\mathbf{v}}_2| = \sqrt{(-1.5)^2 + (3.0)^2} = \sqrt{2.25 + 9.0} = \sqrt{11.25} \approx 3.35 \] Direction: \[ \theta = \tan^{-1}\left(\frac{3.0}{-1.5}\right) \approx \tan^{-1}(-2.0) \approx -63.43^\circ \] Since it's in the second quadrant: \( 180^\circ - 63.43^\circ \approx 116.57^\circ \).
07
Find \( \vec{\mathbf{v}}_2 - \vec{\mathbf{v}}_1 \)
Using vector subtraction: \[ \vec{\mathbf{v}}_2 - \vec{\mathbf{v}}_1 = (4.5 + 6.0) \hat{\mathrm{i}} + (-5.0 - 8.0) \hat{\mathrm{j}} = 10.5 \hat{\mathrm{i}} - 13.0 \hat{\mathrm{j}} \]
08
Magnitude and Direction of \( \vec{\mathbf{v}}_2 - \vec{\mathbf{v}}_1 \)
Magnitude: \[ |\vec{\mathbf{v}}_2 - \vec{\mathbf{v}}_1| = \sqrt{(10.5)^2 + (-13.0)^2} = \sqrt{110.25 + 169.0} = \sqrt{279.25} \approx 16.71 \] Direction: \[ \theta = \tan^{-1}\left(\frac{-13.0}{10.5}\right) \approx \tan^{-1}(-1.24) \approx -51.34^\circ \] Since it's in the fourth quadrant: \( 360^\circ - 51.34^\circ \approx 308.66^\circ \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Magnitude
The magnitude of a vector is essentially its length, which helps us understand how 'big' the vector is without considering its direction. To find it, we apply the Pythagorean theorem within the vector's components. If a vector \( \vec{\mathbf{v}} = a \hat{\mathrm{i}} + b \hat{\mathrm{j}} \), the magnitude is given by:
- Magnitude Formula: \[|\vec{\mathbf{v}}| = \sqrt{a^2 + b^2}\]
- Calculate \((-6.0)^2 = 36\) and \((8.0)^2 = 64\).
- Add these values: \(36 + 64 = 100\).
- Take the square root: \( \sqrt{100} = 10.0\).
Vector Direction
While magnitude gives us the size, direction tells us where the vector points. It's measured as an angle \( \theta \) from the positive x-axis. To find it for a vector \( \vec{\mathbf{v}} = a \hat{\mathrm{i}} + b \hat{\mathrm{j}} \), we use:
- Direction Formula: \[ \theta = \tan^{-1}\left(\frac{b}{a}\right)\]
- Calculate: \[ \theta = \tan^{-1}\left(\frac{8.0}{-6.0}\right) \approx -53.13^\circ\]
- The angle \(-53.13^\circ\) places it in the second quadrant, so adjust: \(180^\circ - 53.13^\circ \approx 126.87^\circ\).
Vector Addition
Adding vectors is akin to combining journeys, where both the path and resulting destination change. Vector addition involves adding their corresponding components. If vectors \( \vec{\mathbf{v}}_1 = a_1 \hat{\mathrm{i}} + b_1 \hat{\mathrm{j}} \) and \( \vec{\mathbf{v}}_2 = a_2 \hat{\mathrm{i}} + b_2 \hat{\mathrm{j}} \) are added, the result is:
- Resulting Vector: \[\vec{\mathbf{v}}_1 + \vec{\mathbf{v}}_2 = (a_1 + a_2) \hat{\mathrm{i}} + (b_1 + b_2) \hat{\mathrm{j}}\]
- Calculate \((-6.0 + 4.5) = -1.5\) and \((8.0 - 5.0) = 3.0\).
- Resulting vector: \( -1.5 \hat{\mathrm{i}} + 3.0 \hat{\mathrm{j}} \).
Vector Subtraction
Subtraction offers a different perspective 鈥� it's like finding the change in positioning or retracing steps. Subtraction is executed by reversing the direction and magnitude of the vector to subtract, and then adding it. Given vectors \( \vec{\mathbf{v}}_1 = a_1 \hat{\mathrm{i}} + b_1 \hat{\mathrm{j}} \) and \( \vec{\mathbf{v}}_2 = a_2 \hat{\mathrm{i}} + b_2 \hat{\mathrm{j}} \), the difference is:
- Subtraction Formula: \[\vec{\mathbf{v}}_2 - \vec{\mathbf{v}}_1 = (a_2 - a_1) \hat{\mathrm{i}} + (b_2 - b_1) \hat{\mathrm{j}}\]
- Compute \((4.5 + 6.0) = 10.5\) and \((-5.0 - 8.0) = -13.0\).
- Resulting vector difference: \( 10.5 \hat{\mathrm{i}} - 13.0 \hat{\mathrm{j}} \).
