[Basic Vector Operations] — PROVE-IT

P — Problem types: [what the exam asks; 3–5 names] Scalar Multiplication, R — Rules & relations: [definitions, identities, theorems you actually use] O — Overview idea (why it works): [2–4 lines intuition or proof sketch] V — Variants: [easy→twist; what changes in the plan] E — Exemplars: [2–3 worked examples with one-line reasons for steps] IT — Invariants & Traps: [must-hold properties; classic errors; quick checks]

[Angles between vectors] — PROVE-IT

P — Problem types: Angles between vectors - Angle $0$ between $u$ and $v$ R — Rules & relations: Review Pythagorean rules O — Overview idea (why it works): [2–4 lines intuition or proof sketch] V — Variants: [easy→twist; what changes in the plan] E — Exemplars:

Proof: For an angle where ∠

The formula is:

$$u\ast v=|u||v|\cos (\theta )for0<\theta <\pi$$

Derived from: Co-sine rule

$$c^2=a^2+b^2-2ab\cos (\theta )$$ $$a=|u|,b=|v|,c=|v-u|$$ $$|v-u{|}^2=|u{|}^2+|v{|}^2-2|u||v|\cos (\theta )$$

Expand and you’re left with the original formula

Examples: Find the angle $\theta$ between the vectors $u=(2,1,3)$ and $v=(-1,1,1)$

$$u\ast v=|u||v|\cos (\theta )$$ $$u\ast v=(2\ast -1,1\ast 1,3\ast 1)$$ $$u\ast v=2$$ $$|u|=\sqrt{2\ast 2+1\ast 1+3\ast 3}$$ $$|u|=\sqrt{14}$$ $$|v|=\sqrt{-1\ast -1+1\ast 1+1\ast 1}$$ $$|v|=\sqrt{3}$$

Substitute

$$2=\sqrt{14}\ast \sqrt{3}\cos (\theta )$$ $$\cos (\theta )=2/\sqrt{42}$$ $$\theta =1.2571radians$$

If Orthogonal/ Perpendicular |_

IT — Invariants & Traps: [must-hold properties; classic errors; quick checks]