Unit 2 3D Cartesian Vectors

Question 1

The Projections of parallel and equal line segments on a plane are

Answer 1

Parallel and equal

Question 2

Explain how to find the distance from (3,7,-5) to each of the planes:

xy-plane
yz-plane
xz-plane

Answer 2

The distance from each plane is the coordinate dimension which is excluded from the plane:

xy = z (5)
yz = x (3)
xz = y (7)

Question 3

Explain how to find the distance of a coordinate from an axis, example:

(3,7,-5) from x-axis

Answer 3

Take the two excluded values then use this distance formula:

x-axis = √(y² + z²)

@ (3,7,-5) = √(7² + (-5)²)

Question 4

Explain the projections of (x,y,z) on each plane:

xy-plane
xz-plane
yz-plane

Answer 4

xy = (x , y , 0)
xz = (x , 0 , z)
yz = (0 , y , z)

Question 5

What is the Force Vector in Cartesian Vector Form ?

Answer 5

F = { xi+ yj+ zk }

Question 6

How do you find Position Vectors in Cartesian Vector Form when given two positions? Ex: rᵤᵥ when u = {3i + 4j + 6k}m and v = {2i + 1j + 2k}m

Answer 6

rᵤᵥ = v - b

rᵤᵥ = {(2 - 3)i + (1 - 4)j + (2 - 6)k}m

Question 7

What is the standard form of Unit Vectors (û)

Answer 7

û = { cosαi + cosβj + cosγk }

Question 8

How is the coordinate angle found?

Answer 8

α = cos⁻¹(ûi)
β = cos⁻¹(ûj)
γ = cos⁻¹(ûk)

Question 9

How is the Unit Vector Found using force or position vectors?

Answer 9

û = { (x/|F|)i + ( y/|F|)j + ( z/|F|)k }

or

û = { { (x/|rᵤᵥ|)i + ( y/|rᵤᵥ|)j + ( z/|rᵤᵥ|)k }

Question 10

How to find Cartesian Vector Form of a Force Vector when given Force Magnitude and Coordinates?

Answer 10

Start by finding the Position Coordinates in CVF using: rᵤᵥ = v - b

Then find the magnitude of rᵤᵥ using:
|rᵤᵥ| = √( xi² + yj² + zk²)

Now find the unit vector using:
û = { ( x /|rᵤᵥ|)i + ( y /|rᵤᵥ|)j + ( z /|rᵤᵥ|)k }

Lastly use F = |F|û:

F = |F|{ ûi+ ûj+ ûk }

Question 11

If given values of F₁ how can you find values of F₂ such that Fᵣ is at a minimum ?

Note: F₁ + F₂ = Fᵣ

Answer 11

We know that Fᵣ is min/max when its first derivative is = 0

Find the CVF form of F₁ then set up the equation:

|Fᵣ| = √((Fx₂ + Fx₁)²i + (Fy₂ + Fy₁)²j)

Then take the sq of both sides so that function is cleaner and easier to work with.

|Fᵣ|² = (Fx₂ + Fx₁)²i + (Fy₂ + Fy₁)²j

Solve the squares take the derivative then find the critical point setting [Fᵣ²]’ = 0 and finding the values of F₁.

Plug these new values into the original function and solve for|Fᵣ|.