Vector Algebra
Vector Addition
Grade 12
Question:
<p>Five points given by A, B, C, D and E are in a plane. Three forces $\vec{AC}$, $\vec{AD}$ and $\vec{AE}$ act at A and three forces $\vec{CB}$, $\vec{DB}$ and $\vec{EB}$ act at B. Then, their resultant is</p>
<p>(a) $2\vec{AC}$</p>
<p>(b) $3\vec{AB}$</p>
<p>(c) $3\vec{DB}$</p>
<p>(d) $2\vec{BC}$</p>
Step-by-Step Solution
Key Concept: Group the vectors to form chains that simplify using the triangle rule of vector addition. Each pair combines to give $\vec{AB}$.
Solution: Points A, B, C, D and E are in a plane. Resultant = $(\vec{AC} + \vec{AD} + \vec{AE}) + (\vec{CB} + \vec{DB} + \vec{EB})$ $= (\vec{AC} + \vec{CB}) + (\vec{AD} + \vec{DB}) + (\vec{AE} + \vec{EB})$ $= \vec{AB} + \vec{AB} + \vec{AB} = 3\vec{AB}$
Correct Answer: b