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: Recall the formula for the angle between vectors
The angle \( \theta \) between two vectors \( \overrightarrow{VD} \) and \( \overrightarrow{DA} \) is given by: \[ \cos \theta = \frac{\overrightarrow{VD} \cdot \overrightarrow{DA}}{|\overrightarrow{VD}| \cdot |\overrightarrow{DA}|}. \] Step 2: Compute \( \overrightarrow{VD} \) and \( \overrightarrow{DA} \)
From previous 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: Compute \( \overrightarrow{VD} \cdot \overrightarrow{DA} \)
\[ \overrightarrow{VD} \cdot \overrightarrow{DA} = (-5)(5) + (4)(2) + (7)(4) = -25 + 8 + 28 = 11. \] Step 4: Compute 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: Compute \( \cos \theta \)
\[ \cos \theta = \frac{11\sqrt{2}}{\sqrt{90} \cdot \sqrt{45}} = \frac{11\sqrt{2}}{\sqrt{4050}} = \frac{11\sqrt{2}}{90}. \] Step 6: Final result
The measure of \( \angle VDA \) is: \[ \theta = \cos^{-1} \left( \frac{11\sqrt{2}}{90} \right). \]