Answer :
Let's solve the problem step by step by performing vector operations and then finding the magnitude and direction of the resultant vector.
Given vectors:
- [tex]\( r = (8, 4) \)[/tex]
- [tex]\( s = (-3, -4) \)[/tex]
- [tex]\( t = (-5, 1) \)[/tex]
We need to perform the operation [tex]\( 5r - 3s + 8t \)[/tex].
Step 1: Multiply each vector by its scalar.
- [tex]\( 5r = 5 \times (8, 4) = (40, 20) \)[/tex]
- [tex]\( -3s = -3 \times (-3, -4) = (9, 12) \)[/tex]
- [tex]\( 8t = 8 \times (-5, 1) = (-40, 8) \)[/tex]
Step 2: Add the resulting vectors.
- Add the x-components: [tex]\( 40 + 9 - 40 = 9 \)[/tex]
- Add the y-components: [tex]\( 20 + 12 + 8 = 40 \)[/tex]
So, the resultant vector is [tex]\( (9, 40) \)[/tex].
Step 3: Calculate the magnitude of the resultant vector.
The magnitude [tex]\( |v| \)[/tex] of a vector [tex]\( (a, b) \)[/tex] is given by the formula:
[tex]\[ |v| = \sqrt{a^2 + b^2} \][/tex]
Substitute the components of the resultant vector:
[tex]\[ |v| = \sqrt{9^2 + 40^2} \][/tex]
[tex]\[ |v| = \sqrt{81 + 1600} \][/tex]
[tex]\[ |v| = \sqrt{1681} \][/tex]
[tex]\[ |v| = 41.0 \][/tex]
Step 4: Determine the direction angle [tex]\(\theta\)[/tex] of the resultant vector.
The direction angle [tex]\(\theta\)[/tex] with respect to the positive x-axis can be found using the tangent:
[tex]\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \][/tex]
For the resultant vector [tex]\( (9, 40) \)[/tex]:
[tex]\[ \theta = \tan^{-1}\left(\frac{40}{9}\right) \][/tex]
Calculating this gives:
[tex]\[ \theta \approx 77.3^\circ \][/tex]
Therefore, the magnitude and direction angle of the resultant vector are [tex]\( 41.0 \)[/tex] and [tex]\( 77.3^\circ \)[/tex], respectively.
Given vectors:
- [tex]\( r = (8, 4) \)[/tex]
- [tex]\( s = (-3, -4) \)[/tex]
- [tex]\( t = (-5, 1) \)[/tex]
We need to perform the operation [tex]\( 5r - 3s + 8t \)[/tex].
Step 1: Multiply each vector by its scalar.
- [tex]\( 5r = 5 \times (8, 4) = (40, 20) \)[/tex]
- [tex]\( -3s = -3 \times (-3, -4) = (9, 12) \)[/tex]
- [tex]\( 8t = 8 \times (-5, 1) = (-40, 8) \)[/tex]
Step 2: Add the resulting vectors.
- Add the x-components: [tex]\( 40 + 9 - 40 = 9 \)[/tex]
- Add the y-components: [tex]\( 20 + 12 + 8 = 40 \)[/tex]
So, the resultant vector is [tex]\( (9, 40) \)[/tex].
Step 3: Calculate the magnitude of the resultant vector.
The magnitude [tex]\( |v| \)[/tex] of a vector [tex]\( (a, b) \)[/tex] is given by the formula:
[tex]\[ |v| = \sqrt{a^2 + b^2} \][/tex]
Substitute the components of the resultant vector:
[tex]\[ |v| = \sqrt{9^2 + 40^2} \][/tex]
[tex]\[ |v| = \sqrt{81 + 1600} \][/tex]
[tex]\[ |v| = \sqrt{1681} \][/tex]
[tex]\[ |v| = 41.0 \][/tex]
Step 4: Determine the direction angle [tex]\(\theta\)[/tex] of the resultant vector.
The direction angle [tex]\(\theta\)[/tex] with respect to the positive x-axis can be found using the tangent:
[tex]\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \][/tex]
For the resultant vector [tex]\( (9, 40) \)[/tex]:
[tex]\[ \theta = \tan^{-1}\left(\frac{40}{9}\right) \][/tex]
Calculating this gives:
[tex]\[ \theta \approx 77.3^\circ \][/tex]
Therefore, the magnitude and direction angle of the resultant vector are [tex]\( 41.0 \)[/tex] and [tex]\( 77.3^\circ \)[/tex], respectively.