Vector Algebra
Vector Rotation
Grade 12
Question:
<p>A vector <strong>a</strong> has components 2p and 1 with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If with respect to the new system, <strong>a</strong> has components (p+1) and 1, then find p.</p>
<p>(a) p = 0</p>
<p>(b) p = 1 or p = \(-\frac{1}{3}\)</p>
<p>(c) p = -1 or p = \(\frac{1}{3}\)</p>
<p>(d) p = 1 or p = -1</p>
Step-by-Step Solution
Key Concept: A rotation preserves the magnitude of vectors. Use the invariance of magnitude to set up an equation.
Solution: We have a = \(2p\mathbf{i} + \mathbf{j}\) On rotation, let b be the vector with components (p+1) and 1, so that b = \((p+1)\mathbf{i} + \mathbf{j}\) Since rotation preserves magnitude, | a | = | b |: \(|\mathbf{a}|^2 = |\mathbf{b}|^2\) \(4p^2 + 1 = (p+1)^2 + 1\) \(4p^2 = (p+1)^2\) \(4p^2 = p^2 + 2p + 1\) \(3p^2 - 2p - 1 = 0\) \((3p + 1)(p - 1) = 0\) ∴ \(p = 1\) or \(p = -\frac{1}{3}\)
Correct Answer: B