Vector Algebra
Cross Product
Grade 12
Question:
<p>[JEE Main 2021] If \(|\vec{a}|=|\vec{b}|=|\vec{a}-\vec{b}|=1\), then \(|\vec{a}+\vec{b}|\) is</p>
<li>1</li>
<li>\(\sqrt2\)</li>
<li>\(\sqrt3\)</li>
<li>2</li>
Step-by-Step Solution
Key Concept: |a-b|^2=|a|^2-2a \cdot b+|b|^2=1-2a \cdot b+1=2-2a \cdot b=1 \to a \cdot b=½. Then |a+b|^2=2+2a \cdot b=2+1=3.
\(|\vec{a}-\vec{b}|^2=|\vec{a}|^2-2\vec{a}\cdot\vec{b}+|\vec{b}|^2=1-2\vec{a}\cdot\vec{b}+1=2-2\vec{a}\cdot\vec{b}=1\Rightarrow\vec{a}\cdot\vec{b}=\frac12\).
\(|\vec{a}+\vec{b}|^2=1+2\cdot\frac12+1=3\Rightarrow|\vec{a}+\vec{b}|=\sqrt3\). Answer: (C)
Correct Answer: C