Q.Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a}+\vec{b}$ is a unit vector if (A) $\theta=\frac{\pi}{4}$ (B) $\theta=\frac{\pi}{3}$ (C) $\theta=\frac{\pi}{2}$ (D) $\theta=\frac{2\pi}{3}$

Concept: Unit Vector Verification — For $\vec{a}+\vec{b}$ to be a unit vector, its magnitude must equal 1.

Step 1: Write the magnitude squared of the sum:

Step 2: Since $\vec{a}$ and $\vec{b}$ are unit vectors, $|\vec{a}|=|\vec{b}|=1$ and $\vec{a}\cdot\vec{b} = \cos\theta$. So:

Step 3: For $\vec{a}+\vec{b}$ to be a unit vector, set $|\vec{a}+\vec{b}|^2 = 1$: …

The key idea is simple: we know the lengths of $\vec{a}$ and $\vec{b}$ individually — each is 1 — and we know the angle between them. The magnitude of their sum depends only on these three numbers. So we can write an expression for $|\vec{a}+\vec{b}|$ in terms of $\theta$, set it equal to 1, and solve.

Why does this work? Because the magnitude of a vector sum is not just the sum of magnitudes — it depends on how the vectors are aligned. When they point in the same direction ($\theta=0$), the sum has length 2; when opposite ($\theta=\pi$), length 0. Somewhere in between, the sum can be exactly 1. We just need to find that angle.

Let’s go step by step.

1. Write the magnitude squared. For any two vectors,

Since $\vec{a}$ and $\vec{b}$ are unit vectors, $\vec{a}\cdot\vec{a}=1$ and $\vec{b}\cdot\vec{b}=1$. The dot product is $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = 1\cdot1\cdot\cos\theta = \cos\theta$.

So

2. Set the condition. We want $\vec{a}+\vec{b}$ to be a unit vector, meaning $|\vec{a}+\vec{b}| = 1$. Squaring both sides gives $|\vec{a}+\vec{b}|^2 = 1$. Therefore,

3. Solve for $\cos\theta$.

4. Find $\theta$ in the given options. The angle between two vectors is conventionally taken in $[0,\pi]$. $\cos\theta = -\frac12$ gives $\theta = \frac{2\pi}{3}$ (or $120^\circ$). Checking the options:
   • $\theta = \pi/4$ gives $\cos\theta = \sqrt2/2$, too large. …

Method: Angle Condition for the Sum of Two Unit Vectors

Use this when two unit vectors are given and their sum must have a specified magnitude.

Steps

Step 1: Expand the magnitude squared

For unit vectors $|\vec a|=|\vec b|=1$ and $\vec a\cdot\vec b=\cos\theta$, so $|\vec a+\vec b|^2=2+2\cos\theta$.

Step 2: Impose the magnitude condition (squared)

'$\vec a+\vec b$ is a unit vector' means $|\vec a+\vec b|=1$, so $|\vec a+\vec b|^2=1$:

$2+2\cos\theta=1.$ …

Common Mistakes

Mistake 1: Setting $|\vec a+\vec b|=1$ without squaring

Why it's wrong: keeping the root $\sqrt{2+2\cos\theta}=1$ is easy to mishandle. Correct approach: square first to get the clean equation $2+2\cos\theta=1$.

Mistake 2: Using $\sin\theta$ in the expansion

Why it's wrong: the expansion of $|\vec a+\vec b|^2$ uses the dot product $2\,\vec a\cdot\vec b=2\cos\theta$, never $\sin\theta$. Correct approach: expand with $\cos\theta$.

Mistake 3: Rejecting the negative cosine as impossible …