Q.If |𝑎⃗| = 3, |𝑏⃗⃗| = 4 and |𝑎⃗ + 𝑏⃗⃗| =5, then |𝑎⃗ − 𝑏⃗⃗| =
(A) 3
(B) 4
(C) 5
(D) 8

The key idea is that the given magnitudes satisfy the Pythagorean relation, so the vectors are perpendicular.

Step 1 — Square the magnitude of the sum:

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

Step 2 — Substitute the given values:

$5^2 = 3^2 + 4^2 + 2\vec{a}\cdot\vec{b}$

$25 = 9 + 16 + 2\vec{a}\cdot\vec{b}$

$25 = 25 + 2\vec{a}\cdot\vec{b} \implies \vec{a}\cdot\vec{b} = 0$. …

The problem gives you three magnitudes: $|\vec{a}| = 3$, $|\vec{b}| = 4$, and $|\vec{a}+\vec{b}| = 5$. You need $|\vec{a}-\vec{b}|$. The numbers 3, 4, 5 are a Pythagorean triple, which hints that $\vec{a}$ and $\vec{b}$ are perpendicular — but you don’t need to assume that. There’s a clean algebraic relation that handles any angle.

The key is to square the magnitudes. For any two vectors, $|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}$ and $|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b}$. Adding these eliminates the dot product, giving a direct link between the sum and difference magnitudes.

This is the parallelogram law — it holds for any two vectors in any dimension.

Now apply it step by step.

1. Write the known squares.

   $|\vec{a}|^2 = 9$, $|\vec{b}|^2 = 16$, $|\vec{a}+\vec{b}|^2 = 25$.

2. Plug into the parallelogram law.

3. Solve for the unknown. $|\vec{a}-\vec{b}|^2 = 50 - 25 = 25$ …

Method: Relating $|\vec{a}+\vec{b}|$ and $|\vec{a}-\vec{b}|$ through squares

Use this whenever you are given some of the magnitudes $|\vec{a}|$, $|\vec{b}|$, $|\vec{a}+\vec{b}|$, $|\vec{a}-\vec{b}|$ and asked for a missing one — no components are given, so you work with lengths alone.

Steps

Step 1: Turn every magnitude into a square using $|\vec{v}|^2=\vec{v}\cdot\vec{v}$.

Expanding the dot product,

Step 2: Combine the two to remove the unknown angle.

Adding them cancels the $\vec{a}\cdot\vec{b}$ term — this is the parallelogram law: …

Common Mistakes

Mistake 1: Assuming $|\vec{a}-\vec{b}|=|\vec{a}+\vec{b}|$ because $3,4,5$ looks symmetric.

Why it's wrong: that equality is only true when $\vec{a}\perp\vec{b}$; guessing it skips the reasoning even though it happens to give the right number here. Correct approach: derive it from $|\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2=2(|\vec{a}|^2+|\vec{b}|^2)$.

Mistake 2: Adding the magnitudes directly, e.g. $|\vec{a}-\vec{b}|=|\vec{a}|-|\vec{b}|=3-4$. …