Vector Algebra
Dot Product and Projection
Grade None
Question:
<p>[JEE Main 2019] Let \(\vec{a}\) and \(\vec{b}\) be unit vectors and \(\alpha\) be the angle between them. Then \(\vec{a}+\vec{b}\) is a unit vector if</p>
<li>\(\alpha=\dfrac{\pi}{4}\)</li>
<li>\(\alpha=\dfrac{2\pi}{3}\)</li>
<li>\(\alpha=\dfrac{\pi}{3}\)</li>
<li>\(\alpha=\dfrac{\pi}{2}\)</li>
Step-by-Step Solution
Key Concept: |a+b|^2=|a|^2+2a \cdot b+|b|^2=1+2cos\alpha+1=2+2cos\alpha. For this to equal 1: cos\alpha=-½ \to \alpha=2\pi/3.
$|\vec{a}+\vec{b}|^2=|\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2=1+2\cos\alpha+1=2+2\cos\alpha$.
Set equal to 1: $2+2\cos\alpha=1\Rightarrow\cos\alpha=-\frac12\Rightarrow\alpha=\frac{2\pi}{3}$. Answer: (B)
Correct Answer: B