Q.State True or False: The formula $|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2\vec{a}\times\vec{b}$ is valid for non-zero vectors $\vec{a}$ and $\vec{b}$.

The key idea is that the magnitude squared of a sum uses the dot product, not the cross product.

Step 1: The correct identity is

$|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}$.

Step 2: The cross product $\vec{a}\times\vec{b}$ is a vector, not a scalar, so adding it to a scalar sum is dimensionally invalid. …

Why This Question Tests a Fundamental Distinction

The problem isn't about computation — it's about recognising the type of multiplication involved. When you square the magnitude of a sum of vectors, you're really doing:

That dot product is the only way to get a scalar from two vectors. The cross product $\vec{a} \times \vec{b}$ gives a vector perpendicular to both — you cannot add a vector to a scalar. That alone makes the statement false.

But let's walk through it carefully so the reasoning sticks.

1. Write the definition of magnitude squared For any vector $\vec{v}$, $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$. So:

2. Expand using the distributive property of the dot product The dot product is bilinear, so:

3. Simplify using symmetry Since $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}$, and $\vec{a}\cdot\vec{a} = |\vec{a}|^2$, $\vec{b}\cdot\vec{b} = |\vec{b}|^2$, we get:

4. Compare with the given formula The problem claims the term is $2\,\vec{a}\times\vec{b}$. But $\vec{a}\times\vec{b}$ is a vector, while $|\vec{a}+\vec{b}|^2$, $|\vec{a}|^2$, and $|\vec{b}|^2$ are all scalars. You cannot add a vector to a scalar — the expression is not even dimensionally consistent. …

Method: Type-Checking a Vector Identity (Scalar vs Vector)

Use this to judge whether a proposed vector formula can even be valid, before computing anything.

Steps

Step 1: Classify each term as scalar or vector

$|\vec{a}+\vec{b}|^2$, $|\vec{a}|^2$, $|\vec{b}|^2$ are scalars. A dot product $\vec{a}\cdot\vec{b}$ is a scalar, but a cross product $\vec{a}\times\vec{b}$ is a vector.

Step 2: Check type consistency …

Common Mistakes

Mistake 1: Not noticing a vector is being added to scalars

Why it's wrong: $\vec{a}\times\vec{b}$ is a vector while $|\vec{a}|^2$ and $|\vec{b}|^2$ are scalars, so the sum is meaningless. Correct approach: type-check first — the formula fails on that alone.

Mistake 2: Confusing dot and cross in the expansion

Why it's wrong: $|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})$ produces $2\,\vec{a}\cdot\vec{b}$, never a cross product. Correct approach: the correct middle term is $2\,\vec{a}\cdot\vec{b}$. …