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:
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?
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
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
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.
We now zoom in on the vector u, and change orientation slightly, as follows:
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.
Exercise
Find the angle between the vectors P = 3i + 4j − 7k and Q = -2i + j + 3k.
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?
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