7. Vectors in 3-D Space
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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).
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 = sqrt ((x_2-x_1)^2+ (y_2-y_1)^2+ (z_2-z_1)^2)`
For the vector OP above, the magnitude of the vector is given by:
`| OP | = sqrt(2^2+ 3^2+ 5^2) = 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.
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 2 - 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 3 - 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
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:
= u `*` i
= 1 × 1 × cos α
= cos α
= u`*` j
= 1 × 1 × cos β
= cos β
= 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:
`theta=arccos((P * Q)/(|P||Q|))`
Find the angle between the vectors P = 4i + 0j + 7k and Q = -2i + j + 3k.
Find the angle between the vectors P = 3i + 4j − 7k and Q = -2i + j + 3k.
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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