Vector Algebra
Cross Product
Grade 12
Question:
<p>In a parallelogram ABCD, \(\vec{AB} = \vec{i} + \vec{j} + \vec{k}\) and diagonal \(\vec{AC} = \vec{i} - \vec{j} + \vec{k}\) and area of parallelogram is 8 sq units, then \(\angle BAC\) is equal to</p>
<p>(a) \(\frac{\pi}{6}\)</p>
<p>(b) \(\frac{\pi}{3}\)</p>
<p>(c) \(\sin^{-1}\left(\frac{8}{3}\right)\)</p>
<p>(d) \(\cos^{-1}\left(\frac{8}{3}\right)\)</p>
Step-by-Step Solution
Key Concept: Use the cross product formula to find the sine of the angle between two vectors.
Step 1: We have \(\vec{AB} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{AC} = \hat{i} - \hat{j} + \hat{k}\) Step 2: Compute the cross product: \(\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} = 2\hat{i} - 2\hat{k}\) Step 3: Therefore \(|\vec{AB} \times \vec{AC}| = \sqrt{4 + 4} = 2\sqrt{2}\) Step 4: Using \(\sin \theta = \frac{|\vec{AB} \times \vec{AC}|}{|\vec{AB}||\vec{AC}|}\), we get \(\sin \theta = \frac{2\sqrt{2}}{\sqrt{3} \cdot \sqrt{3}} = \frac{2\sqrt{2}}{3}\) ∴ Answer is (c).
Correct Answer: C