Vector Algebra
Magnitude and Direction of Vectors
Grade 12
Question:
<p>Let two non-collinear unit vectors \(\vec{a}\) and \(\vec{b}\) form an acute angle. A point P moves, so that at any time \(t\) the position vector \(\overrightarrow{OP}\) (where O is the origin) is given by \(\overrightarrow{OP} = \vec{a}\cos t + \vec{b}\sin t\). When P is farthest from origin O, let M be the length of \(\overrightarrow{OP}\) and \(\vec{u}\) be the unit vector along \(\overrightarrow{OP}\). Then find \(\vec{u}\) and \(M\).</p>
<p>(a) \(\vec{u} = \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}\) and \(M = (1 + \vec{a} \cdot \vec{b})^{1/2}\)</p>
<p>(b) \(\vec{u} = \frac{\vec{a} - \vec{b}}{|\vec{a} - \vec{b}|}\) and \(M = (1 - \vec{a} \cdot \vec{b})^{1/2}\)</p>
<p>(c) \(\vec{u} = \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}\) and \(M = (1 - \vec{a} \cdot \vec{b})^{1/2}\)</p>
<p>(d) \(\vec{u} = \frac{\vec{a} - \vec{b}}{|\vec{a} - \vec{b}|}\) and \(M = (1 + \vec{a} \cdot \vec{b})^{1/2}\)</p>
Step-by-Step Solution
Key Concept: To find maximum distance, maximize |OP|² by finding critical points using calculus, then the unit vector is obtained by normalizing the position vector at that critical point.
Step 1: Find \(|\overrightarrow{OP}|^2\): \[|\overrightarrow{OP}|^2 = (\vec{a}\cos t + \vec{b}\sin t) \cdot (\vec{a}\cos t + \vec{b}\sin t)\] \[= |\vec{a}|^2\cos^2 t + |\vec{b}|^2\sin^2 t + 2(\vec{a} \cdot \vec{b})\cos t \sin t\] Since \(|\vec{a}| = |\vec{b}| = 1\): \[= \cos^2 t + \sin^2 t + 2(\vec{a} \cdot \vec{b})\cos t \sin t = 1 + (\vec{a} \cdot \vec{b})\sin 2t\] Step 2: Maximize \(|\overrightarrow{OP}|^2\): Maximum occurs when \(\sin 2t = 1\), giving \(|\overrightarrow{OP}|^2_{max} = 1 + \vec{a} \cdot \vec{b}\) Therefore \(M = \sqrt{1 + \vec{a} \cdot \vec{b}}\) Step 3: Find unit vector: When \(\sin 2t = 1\), we have \(\cos t = \sin t = \frac{1}{\sqrt{2}}\) \(\overrightarrow{OP} = \frac{1}{\sqrt{2}}(\vec{a} + \vec{b})\) Unit vector: \(\vec{u} = \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}\) ∴ Answer is (a).
Correct Answer: A