Revise: Vectors in Three Dimensions

|v| = √(x²+y²+z²).

v=(2,3,6) → |v|=7.

a·b = a₁b₁+a₂b₂+a₃b₃ — a scalar.

(1,2,3)·(4,−5,6)=12.

cosθ = (a·b)/(|a||b|).

(1,0,0),(1,1,0) → θ=45°.

a×b produces a NEW vector, perpendicular to both.

i×j=k.

Area = |a×b|.

(3,0,0),(0,4,0) → area 12.

r = r₀ + t·v.

r₀=(1,2,3), v=(2,−1,4) → r=(1+2t,2−t,3+4t).

n·(r−r₀) = 0.

r₀=(1,0,2), n=(3,1,−2) → 3x+y−2z=−1.

Volume = |a·(b×c)|.

Unit vectors → volume 1.

The 3D extension of the Pythagorean theorem, applied twice.

v=(1,4,8) → |v|=9.

a·b = 0 exactly when a and b are perpendicular.

(2,0,3)·(−1,4,2)=4.

Comes from rearranging a·b = |a||b|cosθ.

(3,4,0),(0,0,5) → θ=90°.

Follows a specific cyclic index pattern — order matters.

(2,1,0)×(1,3,0) → z-component 5.

|a×b| = |a||b|sinθ, the area formula in disguise.

(5,0,0),(0,2,0) → area 10.

r₀ is a fixed point; v is the direction — do not confuse them.

r=(1+2t,2−t,3+4t) at t=5 → y=−3.

Expands into the familiar a(x−x₀)+b(y−y₀)+c(z−z₀)=0.

r₀=(2,1,0), n=(1,1,1) → d=3.

Combines cross product (base area) with dot product (height).

(2,0,0),(0,3,0),(0,0,4) → volume 24.