Questions: Find the angle, in radians, between v=4i-4sqrt(3)j and w=-3sqrt(3)i+3j. The angle between v=4i-4sqrt(3)j and w=-3sqrt(3)i+3j is . (Type an exact answer using pi as required.)

Solution Steps

To find the angle between two vectors \(\mathbf{v}\) and \(\mathbf{w}\), we can use the dot product formula: \[ \mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\| \|\mathbf{w}\| \cos(\theta) \] where \(\theta\) is the angle between the vectors. Solving for \(\theta\), we get: \[ \theta = \arccos\left(\frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|}\right) \] We need to calculate the dot product \(\mathbf{v} \cdot \mathbf{w}\), the magnitudes \(\|\mathbf{v}\|\) and \(\|\mathbf{w}\|\), and then use the arccos function to find the angle.

Given vectors: \[ \mathbf{v} = 4 \mathbf{i} - 4 \sqrt{3} \mathbf{j} = [4, -6.9282] \] \[ \mathbf{w} = -3 \sqrt{3} \mathbf{i} + 3 \mathbf{j} = [-5.1962, 3] \]

The dot product of \(\mathbf{v}\) and \(\mathbf{w}\) is: \[ \mathbf{v} \cdot \mathbf{w} = 4 \cdot (-5.1962) + (-6.9282) \cdot 3 = -20.7848 - 20.7846 = -41.5692 \]

The magnitude of \(\mathbf{v}\) is: \[ \|\mathbf{v}\| = \sqrt{4^2 + (-6.9282)^2} = \sqrt{16 + 48} = 8.0000 \] The magnitude of \(\mathbf{w}\) is: \[ \|\mathbf{w}\| = \sqrt{(-5.1962)^2 + 3^2} = \sqrt{27 + 9} = 6.0000 \]

Using the dot product and magnitudes, the angle \(\theta\) between the vectors is: \[ \theta = \arccos\left(\frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|}\right) = \arccos\left(\frac{-41.5692}{8.0000 \cdot 6.0000}\right) = \arccos\left(\frac{-41.5692}{48.0000}\right) = \arccos(-0.8660) \] \[ \theta \approx 2.6180 \text{ radians} \]

Final Answer