Answer :
To solve this problem, we will first perform the vector operations and then find the magnitude and direction angle of the resultant vector. Let's do it step-by-step.
### Step 1: Perform the Vector Operation
Given the vectors:
- [tex]\( r = \langle 8, 4 \rangle \)[/tex]
- [tex]\( s = \langle -3, -4 \rangle \)[/tex]
- [tex]\( t = \langle -5, 1 \rangle \)[/tex]
The operation we need to perform is [tex]\( 5r - 3s + 8t \)[/tex].
#### Calculate each part individually:
1. 5 times vector [tex]\( r \)[/tex]:
[tex]\[
5r = 5 \times \langle 8, 4 \rangle = \langle 40, 20 \rangle
\][/tex]
2. -3 times vector [tex]\( s \)[/tex]:
[tex]\[
-3s = -3 \times \langle -3, -4 \rangle = \langle 9, 12 \rangle
\][/tex]
3. 8 times vector [tex]\( t \)[/tex]:
[tex]\[
8t = 8 \times \langle -5, 1 \rangle = \langle -40, 8 \rangle
\][/tex]
#### Combine these results:
Now add these vectors together:
[tex]\[
5r - 3s + 8t = \langle 40, 20 \rangle + \langle 9, 12 \rangle + \langle -40, 8 \rangle
\][/tex]
Perform the addition component-wise:
- For the first component: [tex]\( 40 + 9 - 40 = 9 \)[/tex]
- For the second component: [tex]\( 20 + 12 + 8 = 40 \)[/tex]
So, the resultant vector is [tex]\( \langle 9, 40 \rangle \)[/tex].
### Step 2: Calculate the Magnitude
The magnitude of a vector [tex]\( \langle x, y \rangle \)[/tex] is calculated using the formula:
[tex]\[
\text{Magnitude} = \sqrt{x^2 + y^2}
\][/tex]
For our resultant vector [tex]\( \langle 9, 40 \rangle \)[/tex]:
[tex]\[
\text{Magnitude} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41
\][/tex]
### Step 3: Calculate the Direction Angle
The direction angle [tex]\( \theta \)[/tex] is found using the tangent function:
[tex]\[
\tan \theta = \frac{y}{x}
\][/tex]
For our vector [tex]\( \langle 9, 40 \rangle \)[/tex]:
[tex]\[
\tan \theta = \frac{40}{9}
\][/tex]
Find [tex]\( \theta \)[/tex] by taking the arctan (inverse tangent):
[tex]\[
\theta = \arctan\left(\frac{40}{9}\right)
\][/tex]
Using a calculator, you find:
[tex]\[
\theta \approx 77.03^\circ
\][/tex]
### Conclusion
The magnitude of the resultant vector is 41, and the direction angle is approximately [tex]\( 77.03^\circ \)[/tex]. However, 773 degrees isn't a regular answer and seems off; the direction should be between 0 and 360 degrees. Double-check calculations or answer selection if needed.
Based on the choices you provided initially, the calculated magnitude correctly matches the resultant vector, but please ensure the answer selection or problem source is accurate.
### Step 1: Perform the Vector Operation
Given the vectors:
- [tex]\( r = \langle 8, 4 \rangle \)[/tex]
- [tex]\( s = \langle -3, -4 \rangle \)[/tex]
- [tex]\( t = \langle -5, 1 \rangle \)[/tex]
The operation we need to perform is [tex]\( 5r - 3s + 8t \)[/tex].
#### Calculate each part individually:
1. 5 times vector [tex]\( r \)[/tex]:
[tex]\[
5r = 5 \times \langle 8, 4 \rangle = \langle 40, 20 \rangle
\][/tex]
2. -3 times vector [tex]\( s \)[/tex]:
[tex]\[
-3s = -3 \times \langle -3, -4 \rangle = \langle 9, 12 \rangle
\][/tex]
3. 8 times vector [tex]\( t \)[/tex]:
[tex]\[
8t = 8 \times \langle -5, 1 \rangle = \langle -40, 8 \rangle
\][/tex]
#### Combine these results:
Now add these vectors together:
[tex]\[
5r - 3s + 8t = \langle 40, 20 \rangle + \langle 9, 12 \rangle + \langle -40, 8 \rangle
\][/tex]
Perform the addition component-wise:
- For the first component: [tex]\( 40 + 9 - 40 = 9 \)[/tex]
- For the second component: [tex]\( 20 + 12 + 8 = 40 \)[/tex]
So, the resultant vector is [tex]\( \langle 9, 40 \rangle \)[/tex].
### Step 2: Calculate the Magnitude
The magnitude of a vector [tex]\( \langle x, y \rangle \)[/tex] is calculated using the formula:
[tex]\[
\text{Magnitude} = \sqrt{x^2 + y^2}
\][/tex]
For our resultant vector [tex]\( \langle 9, 40 \rangle \)[/tex]:
[tex]\[
\text{Magnitude} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41
\][/tex]
### Step 3: Calculate the Direction Angle
The direction angle [tex]\( \theta \)[/tex] is found using the tangent function:
[tex]\[
\tan \theta = \frac{y}{x}
\][/tex]
For our vector [tex]\( \langle 9, 40 \rangle \)[/tex]:
[tex]\[
\tan \theta = \frac{40}{9}
\][/tex]
Find [tex]\( \theta \)[/tex] by taking the arctan (inverse tangent):
[tex]\[
\theta = \arctan\left(\frac{40}{9}\right)
\][/tex]
Using a calculator, you find:
[tex]\[
\theta \approx 77.03^\circ
\][/tex]
### Conclusion
The magnitude of the resultant vector is 41, and the direction angle is approximately [tex]\( 77.03^\circ \)[/tex]. However, 773 degrees isn't a regular answer and seems off; the direction should be between 0 and 360 degrees. Double-check calculations or answer selection if needed.
Based on the choices you provided initially, the calculated magnitude correctly matches the resultant vector, but please ensure the answer selection or problem source is accurate.
