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(A dot B) = 4.2494

We will make use of this value a little later in this module.

Compute the projection

Suppose your problem calls for the projection of vector A onto vector B. Let's compute the value of that projection. For this,we need to know the magnitude of vector B, which we can compute using the Pythagorean theorem:

Bmag = sqrt(2.9*2.9 + 0.78*0.78), or

Bmag = 3.0

The projection of A onto B is equal to

Projection = (A dot B)/Bmag, or

Projection = 4.294/3 = 1.43

We will also have more to say about this value a little later in this module.

What do we mean by the projection?

Using the vectors on your graph board, draw a line segment beginning at the tip of vector A. Make that line perpendicular to vector B and mark the point on vector B wherethat line intersects vector B.

The distance from the origin to that point is the value of the projection of vector A onto vector B.

According to the above arithmetic, that distance should be equal to 1.43 units. Hopefully when youmeasure it on your graph board, you will get approximately the same value.

Compute the angle between the vectors

Suppose instead that your problem calls for the angle between the two vectors. Given that the dot product is equal to the product of the magnitudes and thecosine of the angle between the vectors, the cosine of the angle between the vectors is equal to

cosine(angle) = (A dot B)/(Amag *Bmag)

You may or may not know this from your earlier experience with trigonometry, but the angle in the above expression is the arccosine of the cosine value.

To compute the angle, we first need to compute the magnitude of vector A. Once again using the Pythagorean theorem,

Amag = sqrt(1.0*1.0+1.73*1.73), or

Amag = 2

Now compute the angle between the vectors

We now have all of the information that we need to computer the angle between the vectors. Using Google calculator nomenclature,

Angle = arccos((A dot B)/(Amag *Bmag)) in degrees, or

angle = arccos(4.294/(2*3)) in degrees, or

angle = 44.30 degrees

(Actually, I chose the original values in hopes of causing this final answer to come out to 45 degrees, but the round off errors along the way threw things offa bit.)

What have we learned?

For these two vectors, we have learned that

  • The angle between them is 44.3 degrees
  • The projection of vector A onto vector B is equal to 1.43 units

Perpendicular or parallel vectors

Now consider what happens as the angle varies between 90 degrees (perpendicular vectors) and 0 degrees (parallel vectors) for a given pair ofvectors.

As the angle approaches 90 degrees, the cosine of the angle approaches 0, and the dot product of the vectors approaches 0.

As the angle approaches 0 degrees, the cosine of the angle approaches 1.0 and the dot product approaches a value that is the product of the magnitudes of thetwo vectors.

For a given pair of vectors, the dot product can be thought of as a measure of the extent to which they are parallel. The closer they are toparallel, the greater will be the value of the dot product.

A check on the projection value

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Source:  OpenStax, Accessible physics concepts for blind students. OpenStax CNX. Oct 02, 2015 Download for free at https://legacy.cnx.org/content/col11294/1.36
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