Question

Find the measure of \( \angle VDA \).

Hint:

For angles between vectors, always use the dot product formula and ensure the magnitude is correctly computed.

Answer 1

Step 1: Formula for the angle between vectors
The angle \( \theta \) between vectors \( \overrightarrow{VD} \) and \( \overrightarrow{DA} \) is calculated using: \[ \cos \theta = \frac{\overrightarrow{VD} \cdot \overrightarrow{DA}}{|\overrightarrow{VD}| \cdot |\overrightarrow{DA}|}. \]
Step 2: Calculate vectors \( \overrightarrow{VD} \) and \( \overrightarrow{DA} \)
Based on prior calculations: \[ \overrightarrow{VD} = \overrightarrow{V} - \overrightarrow{D} = (-3\hat{i} + 7\hat{j} + 11\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5\hat{i} + 4\hat{j} + 7\hat{k}. \] \[ \overrightarrow{DA} = 5\hat{i} + 2\hat{j} + 4\hat{k}. \]
Step 3: Calculate the dot product \( \overrightarrow{VD} \cdot \overrightarrow{DA} \)
\[ \overrightarrow{VD} \cdot \overrightarrow{DA} = (-5)(5) + (4)(2) + (7)(4) = -25 + 8 + 28 = 11. \]
Step 4: Calculate the magnitudes of \( \overrightarrow{VD} \) and \( \overrightarrow{DA} \)
\[ |\overrightarrow{VD}| = \sqrt{(-5)^2 + 4^2 + 7^2} = \sqrt{25 + 16 + 49} = \sqrt{90}. \] \[ |\overrightarrow{DA}| = \sqrt{(5)^2 + (2)^2 + (4)^2} = \sqrt{25 + 4 + 16} = \sqrt{45}. \]
Step 5: Calculate \( \cos \theta \)
\[ \cos \theta = \frac{11}{\sqrt{90} \cdot \sqrt{45}} = \frac{11}{\sqrt{4050}} = \frac{11}{\sqrt{90} \sqrt{45}}. \]
Step 6: Final Result
The measure of \( \angle VDA \) is: \[ \theta = \cos^{-1} \left( \frac{11}{\sqrt{90} \sqrt{45}} \right). \]