Let's convert the vector [tex]\(\mathbf{t} = -5\mathbf{i} + 12\mathbf{j}\)[/tex] to its trigonometric form.
### Step-by-Step Solution:
1. Identify the components [tex]\( t_i \)[/tex] and [tex]\( t_j \)[/tex]:
- The vector given is [tex]\(\mathbf{t} = -5\mathbf{i} + 12\mathbf{j}\)[/tex].
- Here, [tex]\( t_i = -5 \)[/tex] is the [tex]\(i\)[/tex]-component (or the horizontal component).
- [tex]\( t_j = 12 \)[/tex] is the [tex]\(j\)[/tex]-component (or the vertical component).
2. Calculate the magnitude [tex]\( r \)[/tex] of the vector:
- The magnitude [tex]\( r \)[/tex] of a vector [tex]\(\mathbf{t} = t_i\mathbf{i} + t_j\mathbf{j}\)[/tex] is given by:
[tex]\[
r = \sqrt{t_i^2 + t_j^2}
\][/tex]
- Substituting the values [tex]\( t_i = -5 \)[/tex] and [tex]\( t_j = 12 \)[/tex]:
[tex]\[
r = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\][/tex]
- Therefore, the magnitude [tex]\( r \)[/tex] of the vector is [tex]\( 13.0 \)[/tex].
3. Calculate the angle [tex]\( \theta \)[/tex] in radians:
- The angle [tex]\( \theta \)[/tex] that the vector makes with the positive [tex]\( x \)[/tex]-axis (the horizontal axis) can be found using the arctangent function:
[tex]\[
\theta = \arctan\left(\frac{t_j}{t_i}\right)
\][/tex]
- Substituting the values [tex]\( t_i = -5 \)[/tex] and [tex]\( t_j = 12 \)[/tex]:
[tex]\[
\theta = \arctan\left(\frac{12}{-5}\right)
\][/tex]
- The arctangent function will return a value in the correct quadrant where the vector lies. Since [tex]\( t_i \)[/tex] is negative and [tex]\( t_j \)[/tex] is positive, the vector lies in the second quadrant.
- Therefore, the angle [tex]\( \theta \)[/tex] is approximately [tex]\( 1.965587446494658 \)[/tex] radians.
4. Express the vector [tex]\(\mathbf{t}\)[/tex] in trigonometric form:
- The trigonometric form of a vector is given by [tex]\( r (\cos(\theta) \mathbf{i} + \sin(\theta) \mathbf{j}) \)[/tex].
- Here, [tex]\( r = 13.0 \)[/tex] and [tex]\( \theta = 1.965587446494658 \)[/tex].
Therefore, the trigonometric form of the vector [tex]\(\mathbf{t} = -5\mathbf{i} + 12\mathbf{j}\)[/tex] is:
[tex]\[
\mathbf{t} = 13.0 (\cos(1.965587446494658) \mathbf{i} + \sin(1.965587446494658) \mathbf{j})
\][/tex]
