Q.Find $|\vec{a}|$ and $|\vec{b}|,$ if $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|.$

We are given three magnitudes: $|\vec{a}| = 4$, $|\vec{b}| = 5$, and $|\vec{a} - \vec{b}| = 3$. The angle between $\vec{a}$ and $\vec{b}$ is $\theta$, and we need $\cot^2\theta$.

Concept & Intuition

The key is the relationship between the magnitude of a difference of vectors and the dot product. For any two vectors,

This is essentially the law of cosines in vector form. Once we have $\cos\theta$, we can find $\sin^2\theta = 1 - \cos^2\theta$, and then $\cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta}$.

Step-by-step

1. Write the magnitude-squared relation

Substitute the given numbers:

2. Simplify

3. Solve for $\cos\theta$

4. Find $\sin^2\theta$

5. Compute $\cot^2\theta$ $\cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta} = \frac{\frac{16}{25}}{\frac{9}{25}} = \frac{16}{9}.$ …

Concept. For two vectors with angle $\theta$ between them,

Step 1 — find $\cos\theta$ from the sum.

Step 2 — get $\sin^2\theta$.

Step 3 — find $k^2$ from the difference. …

The key idea is geometric: the modulus of a sum of two complex numbers equals the sum of their moduli only when the vectors representing them are parallel and point the same way. This is the complex-number version of the triangle inequality becoming an equality.

Why this works:

For any two vectors, the triangle inequality says $|z_1+z_2| \le |z_1|+|z_2|$, with equality if and only if the vectors are collinear and have the same direction. In complex terms, “same direction” means their arguments (amplitudes) are equal modulo $2\pi$, so the difference in amplitudes is $0$ (or an integer multiple of $2\pi$, but the smallest non‑negative difference is $0$).

Step‑by‑step reasoning:

1. Recall the triangle inequality for complex numbers: For any $z_1, z_2 \in \mathbb{C}$,

with equality if and only if $z_1$ and $z_2$ are non‑negative real multiples of each other — i.e., one is a non‑negative real scalar times the other.

2. Translate “non‑negative real multiple” into polar form: Write $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$ with $r_1, r_2 \ge 0$. The condition $z_2 = k z_1$ for some $k \ge 0$ means

Since $e^{i\phi}$ is a positive real number only when $\phi = 2\pi n$ (for integer $n$), we get $\theta_2 - \theta_1 = 2\pi n$.

3. Find the difference in amplitudes: …

All observations are negative and arranged in ascending order:

so $x_1$ is the most negative (largest magnitude) and $x_n$ is the least negative (smallest magnitude).

Changing the signs of the first and last terms replaces $x_1$ by $-x_1=|x_1|$ and $x_n$ by $-x_n=|x_n|$. The new data set is

$\{\,|x_1|,\; x_2,\; x_3,\;\ldots,\; x_{n-1},\; |x_n|\,\}.$ …