Vector Algebra
Cross product and angles
Grade 12
Question:
<p>The sine of the angle between the vectors <span style='font-family:monospace'>a = 3<strong>i</strong> + <strong>j</strong> + <strong>k</strong></span> and <span style='font-family:monospace'>b = 2<strong>i</strong> - 2<strong>j</strong> + <strong>k</strong></span> is</p>
<p>(a) \(\frac{74}{99}\)</p>
<p>(b) \(\frac{25}{99}\)</p>
<p>(c) \(\frac{37}{99}\)</p>
<p>(d) \(\frac{5}{41}\)</p>
Step-by-Step Solution
Key Concept: The sine of angle between two vectors is given by |a × b| / (|a||b|). Compute the cross product, find its magnitude and the individual vector magnitudes.
Step 1: Compute the cross product \(\mathbf{a} \times \mathbf{b}\): \[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 1 \\ 2 & -2 & 1 \end{vmatrix} = 3\mathbf{i} - \mathbf{j} - 8\mathbf{k}\] Step 2: Find magnitudes: \(|\mathbf{a} \times \mathbf{b}| = \sqrt{9 + 1 + 64} = \sqrt{74}\) \(|\mathbf{a}| = \sqrt{9 + 1 + 1} = \sqrt{11}\) \(|\mathbf{b}| = \sqrt{4 + 4 + 1} = 3\) Step 3: Apply formula \(\sin \theta = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}\): \(\sin \theta = \frac{\sqrt{74}}{\sqrt{11} \cdot 3} = \frac{\sqrt{74}}{3\sqrt{11}} = \frac{74}{99}\) ∴ Answer is (a).
Correct Answer: a