Certainly! Let's go through the process step-by-step to find the magnitude and direction of the given vector [tex]\( \mathbf{v} = \langle 30, 16 \rangle \)[/tex].
### Step 1: Calculate the Magnitude of the Vector
The magnitude of a vector [tex]\( \mathbf{v} \)[/tex] with components [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is given by the formula:
[tex]\[
\|\mathbf{v}\| = \sqrt{x^2 + y^2}
\][/tex]
For the vector [tex]\( \mathbf{v} = \langle 30, 16 \rangle \)[/tex]:
[tex]\[
\|\mathbf{v}\| = \sqrt{30^2 + 16^2}
\][/tex]
Next, we compute the squares of the components:
[tex]\[
30^2 = 900 \quad \text{and} \quad 16^2 = 256
\][/tex]
Then, we add these squares:
[tex]\[
900 + 256 = 1156
\][/tex]
Finally, we take the square root of this sum to find the magnitude:
[tex]\[
\|\mathbf{v}\| = \sqrt{1156} = 34
\][/tex]
So, the magnitude of the vector is:
[tex]\[
\|\mathbf{v}\| = 34
\][/tex]
### Step 2: Calculate the Direction (Angle θ) of the Vector
The direction or angle [tex]\( \theta \)[/tex] of a vector can be found using the arctangent function:
[tex]\[
\theta = \tan^{-1} \left( \frac{y}{x} \right)
\][/tex]
For the vector [tex]\( \mathbf{v} = \langle 30, 16 \rangle \)[/tex]:
[tex]\[
\theta = \tan^{-1} \left( \frac{16}{30} \right)
\][/tex]
Next, we evaluate the arctangent:
[tex]\[
\theta = \tan^{-1} \left( 0.5333\ldots \right)
\][/tex]
This gives us an angle in radians, which we then convert to degrees:
[tex]\[
\theta \approx 28.07^{\circ}
\][/tex]
(Note: Make sure to round the direction to two decimal places as stated in the problem.)
### Final Answer
Summarizing our results, we have:
- The magnitude of the vector is:
[tex]\[
\|\mathbf{v}\| = 34
\][/tex]
- The direction of the vector is:
[tex]\[
\theta = 28.07^{\circ}
\][/tex]
So, the complete vector description in polar form is:
[tex]\[
\mathbf{v} = 34 \quad \text{at} \quad 28.07^{\circ}
\][/tex]
