Answer :
To solve the given problem, we need to perform vector operations and calculate the resultant vector, its magnitude, and the direction angle.
### Step-by-step Solution
Given:
- Vectors in component form:
- [tex]\( r = \langle 8, 4 \rangle \)[/tex]
- [tex]\( s = \langle -3, -4 \rangle \)[/tex]
- [tex]\( t = \langle -5, 1 \rangle \)[/tex]
Objective:
- Compute [tex]\( 5r - 3s + 8t \)[/tex]
- Find the magnitude and direction angle of the resultant vector.
### Step 1: Perform the vector operations
First, we calculate each scaled vector separately:
- [tex]\( 5r = 5 \times \langle 8, 4 \rangle = \langle 40, 20 \rangle \)[/tex]
- [tex]\( -3s = -3 \times \langle -3, -4 \rangle = \langle 9, 12 \rangle \)[/tex]
- [tex]\( 8t = 8 \times \langle -5, 1 \rangle = \langle -40, 8 \rangle \)[/tex]
Now, add these vectors together:
[tex]\[
5r - 3s + 8t = \langle 40, 20 \rangle + \langle 9, 12 \rangle + \langle -40, 8 \rangle
\][/tex]
Combine the components:
[tex]\[
= \langle (40 + 9 - 40), (20 + 12 + 8) \rangle
\][/tex]
[tex]\[
= \langle 9, 40 \rangle
\][/tex]
The resultant vector is [tex]\(\langle 9, 40 \rangle\)[/tex].
### Step 2: Calculate the magnitude of the resultant vector
The magnitude (or length) of a vector [tex]\(\langle a, b \rangle\)[/tex] is calculated using the formula:
[tex]\[
\text{Magnitude} = \sqrt{a^2 + b^2}
\][/tex]
For our resultant vector [tex]\(\langle 9, 40 \rangle\)[/tex]:
[tex]\[
= \sqrt{9^2 + 40^2}
\][/tex]
[tex]\[
= \sqrt{81 + 1600}
\][/tex]
[tex]\[
= \sqrt{1681}
\][/tex]
[tex]\[
= 41.0
\][/tex]
### Step 3: Determine the direction angle
The direction angle [tex]\(\theta\)[/tex] with the positive x-axis can be found using:
[tex]\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\][/tex]
Substitute [tex]\(a = 9\)[/tex] and [tex]\(b = 40\)[/tex]:
[tex]\[
\theta = \tan^{-1}\left(\frac{40}{9}\right)
\][/tex]
This gives approximately [tex]\(\theta = 77.3^\circ\)[/tex].
### Conclusion
The magnitude of the resultant vector is [tex]\(41.0\)[/tex], and the direction angle is [tex]\(77.3^\circ\)[/tex]. Therefore, the correct choice among the options provided is:
[tex]\(41.0, \theta=77.3^{\circ}\)[/tex]
### Step-by-step Solution
Given:
- Vectors in component form:
- [tex]\( r = \langle 8, 4 \rangle \)[/tex]
- [tex]\( s = \langle -3, -4 \rangle \)[/tex]
- [tex]\( t = \langle -5, 1 \rangle \)[/tex]
Objective:
- Compute [tex]\( 5r - 3s + 8t \)[/tex]
- Find the magnitude and direction angle of the resultant vector.
### Step 1: Perform the vector operations
First, we calculate each scaled vector separately:
- [tex]\( 5r = 5 \times \langle 8, 4 \rangle = \langle 40, 20 \rangle \)[/tex]
- [tex]\( -3s = -3 \times \langle -3, -4 \rangle = \langle 9, 12 \rangle \)[/tex]
- [tex]\( 8t = 8 \times \langle -5, 1 \rangle = \langle -40, 8 \rangle \)[/tex]
Now, add these vectors together:
[tex]\[
5r - 3s + 8t = \langle 40, 20 \rangle + \langle 9, 12 \rangle + \langle -40, 8 \rangle
\][/tex]
Combine the components:
[tex]\[
= \langle (40 + 9 - 40), (20 + 12 + 8) \rangle
\][/tex]
[tex]\[
= \langle 9, 40 \rangle
\][/tex]
The resultant vector is [tex]\(\langle 9, 40 \rangle\)[/tex].
### Step 2: Calculate the magnitude of the resultant vector
The magnitude (or length) of a vector [tex]\(\langle a, b \rangle\)[/tex] is calculated using the formula:
[tex]\[
\text{Magnitude} = \sqrt{a^2 + b^2}
\][/tex]
For our resultant vector [tex]\(\langle 9, 40 \rangle\)[/tex]:
[tex]\[
= \sqrt{9^2 + 40^2}
\][/tex]
[tex]\[
= \sqrt{81 + 1600}
\][/tex]
[tex]\[
= \sqrt{1681}
\][/tex]
[tex]\[
= 41.0
\][/tex]
### Step 3: Determine the direction angle
The direction angle [tex]\(\theta\)[/tex] with the positive x-axis can be found using:
[tex]\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\][/tex]
Substitute [tex]\(a = 9\)[/tex] and [tex]\(b = 40\)[/tex]:
[tex]\[
\theta = \tan^{-1}\left(\frac{40}{9}\right)
\][/tex]
This gives approximately [tex]\(\theta = 77.3^\circ\)[/tex].
### Conclusion
The magnitude of the resultant vector is [tex]\(41.0\)[/tex], and the direction angle is [tex]\(77.3^\circ\)[/tex]. Therefore, the correct choice among the options provided is:
[tex]\(41.0, \theta=77.3^{\circ}\)[/tex]