Question

Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors \( \mathbf{a} = 3 \hat{i} + \hat{j} + 2 \hat{k} \) and \( \mathbf{b} = 2 \hat{i} - 2 \hat{j} + 4 \hat{k} \). Determine the angle formed between the kite strings. Assume there is no slack in the strings.

Hint:

The angle between two vectors can be found using the formula \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \), where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product and \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are the magnitudes of the vectors.

Answer 1

To find the angle \( \theta \) between the vectors \( \mathbf{a} \) and \( \mathbf{b} \), we use the formula: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \] First, calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \): \[ \mathbf{a} \cdot \mathbf{b} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + = 12 \] Now, calculate the magnitudes of the vectors \( \mathbf{a} \) and \( \mathbf{b} \): \[ |\mathbf{a}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \] \[ |\mathbf{b}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} \] Now, substitute into the cosine formula: \[ \cos \theta = \frac{12}{\sqrt{14} \times \sqrt{24}} = \frac{12}{\sqrt{336}} = \frac{12}{\sqrt{336}} = \frac{12}{1.33} \approx 0.654 \] Thus, \( \theta \approx \cos^{-1}(0.654) \approx 60^\circ \).