Vector Algebra
Dot Product and Projection
Grade 12
Question:
<p>Given a parallelogram \(OACB\) with \(\overrightarrow{OA}=\vec{a}\) and \(\overrightarrow{OB}=\vec{b}\). The length of the diagonal \(\overrightarrow{OC}\) is \(a\), \(b\), \(\sqrt{a^2+b^2-ab}\) or \(\sqrt{a^2+b^2+ab}\). If angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{2\pi}{3}\), then \(|\vec{a}+\vec{b}|\) is</p>
<li>\(\sqrt{a^2+b^2+ab}\)</li>
<li>\(\sqrt{a^2+b^2}\)</li>
<li>\(\sqrt{a^2+b^2-ab}\)</li>
<li>\(\sqrt{a^2-ab+b^2}\)</li>
Step-by-Step Solution
Key Concept: |a+b|^2=|a|^2+2a \cdot b+|b|^2 with a \cdot b=|a||b|cos(2\pi/3)=-|a||b|/2=-ab/2. So |a+b|^2=a^2-ab+b^2.
$|\vec{a}+\vec{b}|^2=|\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2=a^2+2ab\cos\!\tfrac{2\pi}{3}+b^2=a^2-ab+b^2$.
So $|\vec{a}+\vec{b}|=\sqrt{a^2-ab+b^2}$. Answer: (C)
Correct Answer: C