Vector Algebra
Magnitude and Projections
Grade 12
Question:
<p>Let <strong>u</strong>, <strong>v</strong> and <strong>w</strong> be such that \(|\mathbf{u}| = 1\), \(|\mathbf{v}| = 2\), \(|\mathbf{w}| = 3\). If the projection of <strong>v</strong> along <strong>u</strong> is equal to that of <strong>w</strong> along <strong>u</strong> and <strong>v</strong>, <strong>w</strong> are perpendicular to each other, then \(|\mathbf{u} - \mathbf{v} + \mathbf{w}|\) equals</p>
<p>(a) \(\sqrt{14}\)</p>
<p>(b) \(\sqrt{7}\)</p>
<p>(c) \(2\)</p>
<p>(d) \(14\)</p>
Step-by-Step Solution
Key Concept: Use the conditions that equal projections give <strong>v</strong> ⋅ <strong>u</strong> = <strong>w</strong> ⋅ <strong>u</strong> and perpendicularity gives <strong>v</strong> ⋅ <strong>w</strong> = 0, then expand the magnitude squared formula.
Given: \(|\mathbf{u}| = 1\), \(|\mathbf{v}| = 2\), \(|\mathbf{w}| = 3\) Condition 1: Projection of v along u = Projection of w along u \(\frac{\mathbf{v} ⋅ \mathbf{u}}{|\mathbf{u}|} = \frac{\mathbf{w} ⋅ \mathbf{u}}{|\mathbf{u}|}\) \(\therefore \mathbf{v} ⋅ \mathbf{u} = \mathbf{w} ⋅ \mathbf{u}\) ... (i) Condition 2: v and w are perpendicular: \(\mathbf{v} ⋅ \mathbf{w} = 0\) ... (ii) Step 1: Calculate \(|\mathbf{u} - \mathbf{v} + \mathbf{w}|^2\) \(= |\mathbf{u}|^2 + |\mathbf{v}|^2 + |\mathbf{w}|^2 - 2(\mathbf{u} ⋅ \mathbf{v}) + 2(\mathbf{u} ⋅ \mathbf{w}) - 2(\mathbf{v} ⋅ \mathbf{w})\) \(= 1 + 4 + 9 - 2(\mathbf{u} ⋅ \mathbf{v}) + 2(\mathbf{u} ⋅ \mathbf{w}) - 0\) Step 2: From condition (i): \(\mathbf{u} ⋅ \mathbf{v} = \mathbf{u} ⋅ \mathbf{w}\) \(= 14 - 2(\mathbf{u} ⋅ \mathbf{v}) + 2(\mathbf{u} ⋅ \mathbf{v}) = 14\) \(\therefore |\mathbf{u} - \mathbf{v} + \mathbf{w}| = \sqrt{14}\) ∴ Answer is (a): \(\sqrt{14}\)
Correct Answer: a