Find the magnitude and direction angle (in degrees) of the following vector.v=7(cos(82°)i+sin(82°)j)

1. Break Down the Problem

We need to find two things for the vector $\mathbf{v} = 7(\cos(82^\circ) \mathbf{i} + \sin(82^\circ) \mathbf{j})$:

1. The magnitude of the vector $\mathbf{v}$.
2. The direction angle of the vector $\mathbf{v}$.

2. Relevant Concepts

• Magnitude of a Vector: The magnitude of a vector $\mathbf{v} = a\mathbf{i} + b\mathbf{j}$ is given by: $|\mathbf{v}| = \sqrt{a^2 + b^2}$

• Direction Angle: The direction angle $\theta$ can be found using: $\theta = \tan^{-1}\left(\frac{b}{a}\right)$

3. Analysis and Detail

1. Identify Components:

   • The vector can be expressed as: $\mathbf{v} = 7\cos(82^\circ) \mathbf{i} + 7\sin(82^\circ) \mathbf{j}$
   • Let $a = 7\cos(82^\circ)$ and $b = 7\sin(82^\circ)$.

2. Calculate Magnitude:

   • First, compute $a$ and $b$: $a = 7 \cos(82^\circ) \\ b = 7 \sin(82^\circ)$
   • Now compute the magnitude: $|\mathbf{v}| = \sqrt{(7\cos(82^\circ))^2 + (7\sin(82^\circ))^2} \\ = 7\sqrt{(\cos(82^\circ))^2 + (\sin(82^\circ))^2}$
   • Using the Pythagorean identity $(\sin^2 + \cos^2 = 1)$: $|\mathbf{v}| = 7\sqrt{1} = 7$

3. Calculate Direction Angle:

   • Now use the components $a$ and $b$: $\theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{7\sin(82^\circ)}{7\cos(82^\circ)}\right) = \tan^{-1}(\tan(82^\circ))$
   • Therefore: $\theta = 82^\circ$

4. Verify and Summarize

• We have calculated the magnitude of the vector $\mathbf{v}$ to be $7$.
• We also found the direction angle to be $82^\circ$, which corresponds to the angle given in the vector definition.

Final Answer

• Magnitude: $7$
• Direction Angle: $82^\circ$