Question

If \(\theta\) is the angle between the vectors \(4\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and \( \mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\), then \(\sin 2\theta\) is:

• A. \(\frac{\sqrt{3}}{\sqrt{95}}\)
• B. \(-\frac{\sqrt{3}}{\sqrt{95}}\)
• C. \(-\frac{\sqrt{285}}{49}\)
• D. \(\frac{\sqrt{258}}{49}\)

Hint:

For vectors, always check calculations for dot products and cross products, as sign errors can lead to incorrect results.

Answer 1

The Correct Option is C

Let \(\mathbf{a} = 4\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{b} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\). We know that \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\). \[\mathbf{a} \cdot \mathbf{b} = (4)(1) + (-1)(3) + (2)(-2) = 4 - 3 - 4 = -3\] \[|\mathbf{a}| = \sqrt{4^2 + (-1)^2 + 2^2} = \sqrt{16 + 1 + 4} = \sqrt{21}\] \[|\mathbf{b}| = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}\] \[\cos \theta = \frac{-3}{\sqrt{21}\sqrt{14}} = \frac{-3}{\sqrt{294}} = \frac{-3}{\sqrt{49 \times 6}} = \frac{-3}{7\sqrt{6}}\] We need to find \(\sin 2\theta = 2\sin \theta \cos \theta\). We know \(\cos \theta = \frac{-3}{7\sqrt{6}}\). We have \(\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{-3}{7\sqrt{6}}\right)^2 = 1 - \frac{9}{49 \times 6} = 1 - \frac{3}{98} = \frac{98-3}{98} = \frac{95}{98}\). Therefore, \(\sin \theta = \pm \sqrt{\frac{95}{98}} = \pm \frac{\sqrt{95}}{7\sqrt{2}}\). \[\sin 2\theta = 2\sin \theta \cos \theta = 2\left(\pm \frac{\sqrt{95}}{7\sqrt{2}}\right)\left(\frac{-3}{7\sqrt{6}}\right) = \mp \frac{6\sqrt{95}}{49\sqrt{12}} = \mp \frac{6\sqrt{95}}{49 \times 2\sqrt{3}} = \mp \frac{3\sqrt{95}}{49\sqrt{3}}\] \[\sin 2\theta = \mp \frac{3\sqrt{95}\sqrt{3}}{49 \times 3} = \mp \frac{\sqrt{285}}{49}\] Since \(\cos \theta = \frac{-3}{7\sqrt{6}}<0\), \(\theta\) is in the second quadrant. If \(\theta\) is in the second quadrant, \(2\theta\) can be in the first or second quadrant. Since \(\sin 2\theta = \mp \frac{\sqrt{285}}{49}\), we must determine the sign. We have \(\cos \theta = \frac{-3}{7\sqrt{6}}\). Since \(\cos \theta<0\), \(\theta\) is in the second quadrant. Thus, \(\frac{\pi}{2}<\theta<\pi\), which implies \(\pi<2\theta<2\pi\). Therefore, \(2\theta\) is in the third or fourth quadrant. Since \(\sin 2\theta\) is negative in the third and fourth quadrants, \(\sin 2\theta = -\frac{\sqrt{285}}{49}\). Final Answer: The final answer is $\boxed{(3)}$