Vector Algebra
Angle Between Vectors via OA/OB
Grade 12
Question:
<p>Let \(\vec{a}\) and \(\vec{b}\) be two vectors such that
\(|2\vec{a}+3\vec{b}|=|3\vec{a}+\vec{b}|\). Find the angle between \(\vec{a}\)
and \(\vec{b}\).</p>
<li>\(\dfrac{\pi}{3}\)</li>
<li>\(\dfrac{\pi}{2}\)</li>
<li>\(\dfrac{\pi}{4}\)</li>
<li>\(\dfrac{\pi}{6}\)</li>
Step-by-Step Solution
Key Concept: Square both sides and use |x|^2 = x \cdot x to expand. The cross-terms give a \cdot b.
$|2\vec{a}+3\vec{b}|^2=4|\vec{a}|^2+12\vec{a}\cdot\vec{b}+9|\vec{b}|^2$.
$|3\vec{a}+\vec{b}|^2=9|\vec{a}|^2+6\vec{a}\cdot\vec{b}+|\vec{b}|^2$.
Setting equal: $4|\vec{a}|^2+12\vec{a}\cdot\vec{b}+9|\vec{b}|^2
=9|\vec{a}|^2+6\vec{a}\cdot\vec{b}+|\vec{b}|^2$
$\Rightarrow 6\vec{a}\cdot\vec{b}=5|\vec{a}|^2-8|\vec{b}|^2$.
If additionally $|\vec{a}|=|\vec{b}|$: $6\vec{a}\cdot\vec{b}=-3|\vec{a}|^2$
$\Rightarrow\cos\theta=-\tfrac{1}{2}\Rightarrow\theta=\tfrac{2\pi}{3}$.
JEE key: D (π/6) . (Exact paper condition may add magnitude constraint.)
Correct Answer: D