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Answer :
To solve the given vector operation and find the magnitude and direction angle of the resultant vector, follow these steps:
1. Understand the vectors:
- Vector [tex]\( r = \langle 8, 4 \rangle \)[/tex]
- Vector [tex]\( s = \langle -3, -4 \rangle \)[/tex]
- Vector [tex]\( t = \langle -5, 1 \rangle \)[/tex]
2. Perform the operations:
We need to calculate [tex]\( 5r - 3s + 8t \)[/tex].
- First, multiply each vector by its scalar:
- [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 results together:
- [tex]\( \langle 40, 20 \rangle + \langle 9, 12 \rangle + \langle -40, 8 \rangle \)[/tex]
- Combine the x-components: [tex]\( 40 + 9 - 40 = 9 \)[/tex]
- Combine the y-components: [tex]\( 20 + 12 + 8 = 40 \)[/tex]
So, the resultant vector is [tex]\( \langle 9, 40 \rangle \)[/tex].
3. Calculate the magnitude:
The magnitude of 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]\[
\text{Magnitude} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41.0
\][/tex]
4. Calculate the direction angle:
The direction angle [tex]\( \theta \)[/tex] of a vector [tex]\( \langle a, b \rangle \)[/tex] is given by:
[tex]\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\][/tex]
For [tex]\( \langle 9, 40 \rangle \)[/tex]:
[tex]\[
\theta = \tan^{-1}\left(\frac{40}{9}\right)
\][/tex]
Converting this angle from radians to degrees gives approximately [tex]\( 77.3^{\circ} \)[/tex].
Thus, the magnitude and direction angle of the resultant vector are [tex]\( 41.0 \)[/tex] and [tex]\( 77.3^{\circ} \)[/tex], respectively. The correct option is [tex]\( 41.0, \theta=77.3^{\circ} \)[/tex].
1. Understand the vectors:
- Vector [tex]\( r = \langle 8, 4 \rangle \)[/tex]
- Vector [tex]\( s = \langle -3, -4 \rangle \)[/tex]
- Vector [tex]\( t = \langle -5, 1 \rangle \)[/tex]
2. Perform the operations:
We need to calculate [tex]\( 5r - 3s + 8t \)[/tex].
- First, multiply each vector by its scalar:
- [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 results together:
- [tex]\( \langle 40, 20 \rangle + \langle 9, 12 \rangle + \langle -40, 8 \rangle \)[/tex]
- Combine the x-components: [tex]\( 40 + 9 - 40 = 9 \)[/tex]
- Combine the y-components: [tex]\( 20 + 12 + 8 = 40 \)[/tex]
So, the resultant vector is [tex]\( \langle 9, 40 \rangle \)[/tex].
3. Calculate the magnitude:
The magnitude of 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]\[
\text{Magnitude} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41.0
\][/tex]
4. Calculate the direction angle:
The direction angle [tex]\( \theta \)[/tex] of a vector [tex]\( \langle a, b \rangle \)[/tex] is given by:
[tex]\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\][/tex]
For [tex]\( \langle 9, 40 \rangle \)[/tex]:
[tex]\[
\theta = \tan^{-1}\left(\frac{40}{9}\right)
\][/tex]
Converting this angle from radians to degrees gives approximately [tex]\( 77.3^{\circ} \)[/tex].
Thus, the magnitude and direction angle of the resultant vector are [tex]\( 41.0 \)[/tex] and [tex]\( 77.3^{\circ} \)[/tex], respectively. The correct option is [tex]\( 41.0, \theta=77.3^{\circ} \)[/tex].
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