7. Vectors in 3-D Space

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Magnitude of a 3-D Vector
Adding 3-D Vectors
Dot Product of 3-D Vectors
Direction Cosines
Angle Between Vectors
Application

We saw earlier how to represent 2-dimensional vectors on the x-y plane.

Now we extend the idea to represent 3-dimensional vectors using the x-y-z axes. (See The 3-dimensional Co-ordinate System for background on this).

Example

The vector OP has initial point at the origin O (0, 0, 0) and terminal point at P (2, 3, 5). We can draw the vector OP as follows:

3D vector

Magnitude of a 3-Dimensional Vector

We saw above that the distance between 2 points in 3-dimensional space is

distance AB = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]

For the vector OP above, the magnitude of the vector is given by:

| OP | = √(2² + 3² + 5²) = 6.16 units

Adding 3-dimensional Vectors

Earlier we saw how to add 2-dimensional vectors. We now extend the idea for 3-dimensional vectors.

We simply add the i components together, then the j components and finally, the k components.

Example

Two anchors are holding a ship in place and their forces acting on the ship are represented by vectors A and B as follows:

A = 2i + 5j − 4k and B = −2i − 3j − 5k

If we were to replace the 2 anchors with 1 single anchor, what vector represents that single vector?

Answer

Dot Product of 3-dimensional Vectors

To find the dot product (or scalar product) of 3-dimensional vectors, we just extend the ideas from the dot product in 2 dimensions that we met earlier.

Example - Dot Product Using Magnitude and Angle

Find the dot product of the vectors P and Q given that the angle between the two vectors is 35° and

| P | = 25 units and | Q | = 4 units

Answer

Example - Dot Product if Vectors are Multiples of Unit Vectors

Find the dot product of the vectors A and B (these come from our anchor example above):

A = 2i + 5j − 4k and B = −2i − 3j − 5k

Answer

Direction Cosines

Suppose we have a vector OA with initial point at the origin and terminal point at A.

Suppose also that we have a unit vector in the same direction as OA. (Go here for a reminder on unit vectors).

Let our unit vector be:

u = u₁ i + u₂ j + u₃ k

On the graph, u is the unit vector (in black) pointing in the same direction as vector OA, and i, j, and k (the unit vectors in the x-, y- and z-directions respectively) are marked in green.

unit vector

We now zoom in on the vector u, and change orientation slightly, as follows:

cosines

Now, if in the diagram above,

α is the angle between u and the x-axis (in dark red),
β is the angle between u and the y-axis (in green) and
γ is the angle between u and the z-axis (in pink),

then we can use the scalar product and write:

u₁

= u • i

= 1 × 1 × cos α

= cos α

u₂

= u • j

= 1 × 1 × cos β

= cos β

u₃

= u • k

= 1 × 1 × cos γ

= cos γ

So we can write our unit vector u as:

u = cos α i + cos β j + cos γ k

These 3 cosines are called the direction cosines.

Angle Between 3-Dimensional Vectors

Earlier, we saw how to find the angle between 2-dimensional vectors. We use the same formula for 3-dimensional vectors:

Example

Find the angle between the vectors P = 4i + 0j + 7k and Q = -2i + j + 3k.

Answer

Exercise

Find the angle between the vectors P = 3i + 4j − 7k and Q = -2i + j + 3k.

Answer

Application

We have a cube ABCO PQRS which has a string along the cube's diagonal B to S and another along the other diagonal C to P

What is the angle between the 2 strings?

Answer

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