Given two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ in standard position, what is the angle $\Theta$ between them?

• A. ${tan}^{-1}\left(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)$
• B. $\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|}$
• C. ${sin}^{-1}\left(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)$
• D. ${cos}^{-1}\left(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)$

Answer: (D)

Verified step by step guidance

1. Recall that the angle \( \Theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) can be found using the dot product formula: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\Theta) \).
2. Rearrange the formula to solve for \( \cos(\Theta) \): \(\n\)\(\n\)\( \cos(\Theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \).
3. To find the angle \( \Theta \), take the inverse cosine (arccos) of both sides: \(\n\)\(\n\)\( \Theta = \cos^{-1} \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \right) \).
4. Note that the dot product \( \vec{a} \cdot \vec{b} \) is calculated as the sum of the products of the corresponding components of \( \vec{a} \) and \( \vec{b} \).
5. Make sure to compute the magnitudes \( |\vec{a}| \) and \( |\vec{b}| \) correctly as \( |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \) (for 3D vectors) or similarly for 2D vectors, then substitute all values into the formula to find \( \Theta \).