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Answer :
To perform the operations and determine the magnitude and direction angle of the resultant vector, follow these steps:
1. Vector Operations: Start with the vectors given in component form:
- [tex]\( r = \langle 8, 4 \rangle \)[/tex]
- [tex]\( s = \langle -3, -4 \rangle \)[/tex]
- [tex]\( t = \langle -5, 1 \rangle \)[/tex]
You need to calculate [tex]\( 5r - 3s + 8t \)[/tex].
2. Calculate [tex]\( 5r \)[/tex]:
[tex]\[
5r = 5 \times \langle 8, 4 \rangle = \langle 40, 20 \rangle
\][/tex]
3. Calculate [tex]\(-3s\)[/tex]:
[tex]\[
-3s = -3 \times \langle -3, -4 \rangle = \langle 9, 12 \rangle
\][/tex]
4. Calculate [tex]\(8t\)[/tex]:
[tex]\[
8t = 8 \times \langle -5, 1 \rangle = \langle -40, 8 \rangle
\][/tex]
5. Add the results:
Combine all the resultant vectors calculated:
[tex]\[
\langle 40, 20 \rangle + \langle 9, 12 \rangle + \langle -40, 8 \rangle = \langle 9, 40 \rangle
\][/tex]
6. Magnitude of the Resultant Vector:
Use the formula for the magnitude of a vector [tex]\(\langle x, y \rangle\)[/tex], which is [tex]\(\sqrt{x^2 + y^2}\)[/tex]:
[tex]\[
\text{Magnitude} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41.0
\][/tex]
7. Direction Angle:
The direction angle [tex]\(\theta\)[/tex] is calculated using the arctangent function:
[tex]\[
\theta = \tan^{-1}\left(\frac{\text{y-component}}{\text{x-component}}\right) = \tan^{-1}\left(\frac{40}{9}\right)
\][/tex]
Converting the angle from radians to degrees:
[tex]\[
\theta \approx 77.3^\circ
\][/tex]
Based on the calculations, the magnitude of the resultant vector is [tex]\(41.0\)[/tex] and the direction angle is approximately [tex]\(77.3^\circ\)[/tex]. Therefore, the correct answer is:
- Magnitude: [tex]\(41.0\)[/tex]
- Direction Angle: [tex]\(77.3^\circ\)[/tex]
The option corresponding to this result is [tex]\(41.0, \theta = 77.3^\circ\)[/tex].
1. Vector Operations: Start with the vectors given in component form:
- [tex]\( r = \langle 8, 4 \rangle \)[/tex]
- [tex]\( s = \langle -3, -4 \rangle \)[/tex]
- [tex]\( t = \langle -5, 1 \rangle \)[/tex]
You need to calculate [tex]\( 5r - 3s + 8t \)[/tex].
2. Calculate [tex]\( 5r \)[/tex]:
[tex]\[
5r = 5 \times \langle 8, 4 \rangle = \langle 40, 20 \rangle
\][/tex]
3. Calculate [tex]\(-3s\)[/tex]:
[tex]\[
-3s = -3 \times \langle -3, -4 \rangle = \langle 9, 12 \rangle
\][/tex]
4. Calculate [tex]\(8t\)[/tex]:
[tex]\[
8t = 8 \times \langle -5, 1 \rangle = \langle -40, 8 \rangle
\][/tex]
5. Add the results:
Combine all the resultant vectors calculated:
[tex]\[
\langle 40, 20 \rangle + \langle 9, 12 \rangle + \langle -40, 8 \rangle = \langle 9, 40 \rangle
\][/tex]
6. Magnitude of the Resultant Vector:
Use the formula for the magnitude of a vector [tex]\(\langle x, y \rangle\)[/tex], which is [tex]\(\sqrt{x^2 + y^2}\)[/tex]:
[tex]\[
\text{Magnitude} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41.0
\][/tex]
7. Direction Angle:
The direction angle [tex]\(\theta\)[/tex] is calculated using the arctangent function:
[tex]\[
\theta = \tan^{-1}\left(\frac{\text{y-component}}{\text{x-component}}\right) = \tan^{-1}\left(\frac{40}{9}\right)
\][/tex]
Converting the angle from radians to degrees:
[tex]\[
\theta \approx 77.3^\circ
\][/tex]
Based on the calculations, the magnitude of the resultant vector is [tex]\(41.0\)[/tex] and the direction angle is approximately [tex]\(77.3^\circ\)[/tex]. Therefore, the correct answer is:
- Magnitude: [tex]\(41.0\)[/tex]
- Direction Angle: [tex]\(77.3^\circ\)[/tex]
The option corresponding to this result is [tex]\(41.0, \theta = 77.3^\circ\)[/tex].
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