Vector Algebra
Rotations in Planes
Grade 12
Question:
<p>Two adjacent sides of a parallelogram ABCD are given by \(\overrightarrow{AB} = 2\vec{i} + 10\vec{j} + 11\vec{k}\) and \(\overrightarrow{AD} = -\vec{i} + 2\vec{j} + 2\vec{k}\). The side AD is rotated by an acute angle \(\alpha\) in the plane of the parallelogram so that AD becomes \(\overrightarrow{AD'}\). If \(\overrightarrow{AD'}\) makes a right angle with the side AB, then the cosine of the angle \(\alpha\) is given by</p>
<p>(a) \(\frac{8}{9}\)</p>
<p>(b) \(\frac{17}{9}\)</p>
<p>(c) \(\frac{1}{3}\)</p>
<p>(d) \(\frac{4}{5}\)</p>
Step-by-Step Solution
Key Concept: Rotation in a plane preserves vector magnitude; use orthogonality condition and linear combination in the plane to find the rotation angle.
Step 1: Calculate the magnitudes and dot product of the given vectors. \(|\overrightarrow{AB}| = \sqrt{4 + 100 + 121} = \sqrt{225} = 15\) \(|\overrightarrow{AD}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\) \(\overrightarrow{AB} \cdot \overrightarrow{AD} = -2 + 20 + 22 = 40\) Step 2: \(\overrightarrow{AD'}\) is in the plane of the parallelogram and \(\overrightarrow{AD'} \perp \overrightarrow{AB}\). The plane of the parallelogram is spanned by \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\). Let \(\overrightarrow{AD'} = x\overrightarrow{AB} + y\overrightarrow{AD}\). From \(\overrightarrow{AD'} \cdot \overrightarrow{AB} = 0\): \(x|\overrightarrow{AB}|^2 + y(\overrightarrow{AD} \cdot \overrightarrow{AB}) = 0\) \(225x + 40y = 0 \Rightarrow y = -\frac{225x}{40} = -\frac{45x}{8}\) Step 3: The angle \(\alpha\) between \(\overrightarrow{AD}\) and \(\overrightarrow{AD'}\) satisfies: \(\cos\alpha = \frac{\overrightarrow{AD} \cdot \overrightarrow{AD'}}{|\overrightarrow{AD}||\overrightarrow{AD'}|}\) With constraint \(|\overrightarrow{AD'}| = |\overrightarrow{AD}| = 3\) (rotation preserves length), we find \(\cos\alpha = \frac{8}{9}\). ∴ Answer is (a).
Correct Answer: A