Vector Operations, Dot and Cross Products, and Lines

Written at December 23, 2024 on English with a size of 10.69 KB.

Vector Determination by Length and Angle

V = <||V|| Cosθ, ||V|| Sinθ> ---> ||V||Cosθi + ||V||Sinθj

Example:

a) Find the vector of length 2 that makes an angle of π/4 with the positive x-axis.

b) Find the angle that the vector V = - $\sqrt{\ }$ ₃ i + j makes with the positive x-axis.

a) <||V||Cosθ , ||V||Sinθ> = <2cos45, 2sin45> ---> < $\sqrt{\ }$ ₂, $\sqrt{\ }$ ₂>

b) Normalize... ||V|| = $\sqrt{\ }$ _{(-3)² + 1²} = $\sqrt{\ }$ ₄ = 2 -----> V/||V|| = <- $\sqrt{\ }$ _3/2 , 1/2> = <cosθ, sinθ> ----> cosθ = - $\sqrt{\ }$ _3/2, sinθ = 1/2 ---> θ = 5π/6

Dot Product

If U = <U₁, U₂> and V = <V₁, V₂>, then the dot product is U•V = U₁V₁ + U₂V₂.

Example:

a) U = <3, 5>, V = <-1, 2> -----> U•V = (3)(-1) + (5)(2) ---> U•V = -3 + 10 --> U•V = 7

b) U = <1, -3, 4>, V = <1, 5, 2> ----> U•V = (1)(1) + (5)(-3) + (2)(4) ---> U•V = -6

Angles Between Vectors

Cosθ = U•V / ||U|| ||V||

Example:

Find the angle between the vector U = i - 2j + k and V = -3i + 6j + 2k

U = <1, -2, 1>; V = <-3, 6, 2> ----> ||U|| = $\sqrt{\ }$ _{1² + (-2)² + 1²} = $\sqrt{\ }$ ₆

||V|| = $\sqrt{\ }$ _{(-3)² + 6² + 2²} = $\sqrt{\ }$ ₄₉ = 7 ---> U•V = 1(-3) + (-2)(6) + (1)(2) = -13

Cosθ = U•V / ||U|| ||V|| = -13 / (7 $\sqrt{\ }$ ₆); θ = Cos^-1(-13 / (7 $\sqrt{\ }$ ₆))

Decomposing Vectors into Orthogonal Components

V = <V₁, V₂> = <||V||Cosθ, ||V||Sinθ>

e₁ = <1, 0> e₂ = <0, 1>

V•e₁ = V₁(1) + V₂(0) = V₁ V•e₂ = V₁(0) + V₂(1) = V₂

V = V₁e₁ + V₂e₂ V = (V•e₁)e₁ + (V•e₂)e₂

Example:

V = <2, 3> ; e₁ = <1/ $\sqrt{\ }$ ₂ , 1/ $\sqrt{\ }$ ₂>, e₂ = <-1/ $\sqrt{\ }$ ₂ , 1/ $\sqrt{\ }$ ₂>

V•e₁ = 2(1/ $\sqrt{\ }$ ₂) + 3(1/ $\sqrt{\ }$ ₂) = 5/ $\sqrt{\ }$ ₂

V•e₂ = 2(-1/ $\sqrt{\ }$ ₂) + 3(1/ $\sqrt{\ }$ ₂) = 1/ $\sqrt{\ }$ ₂

V = 5/ $\sqrt{\ }$ ₂e₁ + 1/ $\sqrt{\ }$ ₂e₂ -----> 5/ $\sqrt{\ }$ ₂(e₁) = 5/ $\sqrt{\ }$ ₂<1/ $\sqrt{\ }$ ₂, 1/ $\sqrt{\ }$ ₂> = <5/2, 5/2>

1/ $\sqrt{\ }$ ₂(e₂) = 1/ $\sqrt{\ }$ ₂<-1/ $\sqrt{\ }$ ₂, 1/ $\sqrt{\ }$ ₂> = <-1/2, 1/2>

Work

Work = Force × distance = ||F||d W = ||FCosθ|| ||PQ||

Cross Product

Matrix 3x3 and 2x2 (the numbers in the first row go outside) (if there are two rows with two equal columns, it equals 0, and you change and multiply by -1) (I AND K matrix +, J -)

Geometric Properties of the Cross Product

U•(U × V) = 0 V•(U × V) = 0

Scalar Triple Product

Volume of a parallelepiped: V = |U•(V × W)| (the same as the cross product)

Algebraic Properties of the Scalar Triple Product

The same as the scalar triple product, but you have to see if it equals 0 (lie in the same plane).

Parametric Equations of Lines

Example:

Find the parametric equation of the line passing through the point (1, 2, -3) and parallel to V = 4i + 5j - 7k.... x = 1 + 4t, y = 2 + 5t, z = -3 - 7t.

Example:

Let L₁ and L₂ be the lines:

L₁: x = 1 + 4t, y = 5 - 4t, z = -1 + 5t L₂: x = 2 + 8t, y = 4 - 3t, z = 5 + t

a) Are the lines parallel?

b) Do the lines intersect?

a) L₁ is parallel to the vector V = <4, -4, 5>

L₂ is parallel to the vector V = <8, -3, 1> (not parallel because neither is a scalar multiple of the other).

b) 1 + 4t₁ = 2 + 8t₂

5 - 4t₁ = 4 - 3t₂

-1 + 5t₁ = 5 + t₂

(Eliminate 4t₁) 6 = 6 + 5t₂ ---> 5t₂ = 0, t₂ = 0

(Put it in the first equation) 1 + 4t₁ = 2 ---> 4t₁ = 1 ---> t₁ = 1/4

(Put it in the third equation) See if it is equal.

Line Segments

Example:

Find the parametric equation describing the line segment joining the points P₁(2, 4, -1) and P₂(5, 0, 7).

V = P₁P₂ = <3, -4, 8> ----> (Find x, y, and z by multiplying P₁ with P₁P₂ and adding t) x = 2 + 3t, y = 4 - 4t, z = -1 + 8t

Entradas relacionadas:

Tags: