Vector Algebra
Projection of vectors
Grade None
Question:
<p>Let \(\vec{u}\), \(\vec{v}\), \(\vec{w}\) be such that \(|\vec{u}|=1\), \(|\vec{v}|=2\), \(|\vec{w}|=3\). If the projection \(\vec{v}\) along \(\vec{u}\) is equal to that of \(\vec{w}\) along \(\vec{u}\) and \(\vec{v}\), \(\vec{w}\) are perpendicular to each other then \(|\vec{u}-\vec{v}+\vec{w}|\) equals</p>
<p>2</p>
<p>\(\sqrt{7}\)</p>
<p>\(\sqrt{14}\)</p>
<p>14</p>
Step-by-Step Solution
Key Concept: Use the projection condition to establish a relationship between dot products, then leverage perpendicularity (v⊥w) to find |u−v+w|² by expanding and substituting the constraints systematically.
Step 1: Interpret the projection condition Projection of v along u = Projection of w along u This means: (v·u)/|u| = (w·u)/|u| Since |u| = 1: v·u = w·u Step 2: Set up the magnitude calculation | u − v + w |^2 = ( u − v + w )·( u − v + w ) Expanding: = | u |^2 + | v |^2 + | w |^2 − 2 u · v + 2 u · w − 2 v · w Step 3: Substitute known values | u |^2 = 1, | v |^2 = 4, | w |^2 = 9 v · w = 0 (perpendicular) v · u = w · u (from projection condition) = 1 + 4 + 9 − 2 u · v + 2 u · w − 0 = 14 + 2( u · w − u · v ) = 14 + 2(0) = 14 Step 4: Final answer | u − v + w | = √14 ∴ Answer: C
Correct Answer: C