Finding angles between vectors

In summary: The third side is the vector from P to E. Find the dot product of the two adjacent sides. Find the magnitudes of the two adjacent sides. Find the angle between them using the dot product formula.
  • #1
adrimare
33
0

Homework Statement



I have two related questions related because they both ask to find angles between vectors.

The first one says to determine the angles between the sides of a parallelogram determined by the vectors OA = (3,2,-6) and OB = (-6,6,-2).

The second one asks to determine the angle between OP and AE, where OP = (3,4,5) and AE = (-3,4,5).

Homework Equations



?

The Attempt at a Solution



For the first one, I tried to use the cosine law, but got 1.5 degrees instead of 84.4 degrees.
These are my steps.

magnitude of OA = [tex]\sqrt{}(32 + 22 + -62) [/tex]= [tex]\sqrt{}49[/tex] = 7

magnitude of OB = [tex]\sqrt{}](-62 + 62 + -22) [/tex]= [tex]\sqrt{}76[/tex]

vector AB = (-9,4,4)

magnitude of AB = [tex]\sqrt{}(-92 + 42 + 42)[/tex] = [tex]\sqrt{}113[/tex]

[tex]\Theta[/tex] = cos-1 ((72 + [tex]\sqrt{}76[/tex] - [tex]\sqrt{}113[/tex]) / (2*7*[tex]\sqrt{}76[/tex])) = approximately 1.5 degrees

I'm confused as to what to do here.

The next question has me confused as well, as I seem to be missing something key.

Please help!
 
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  • #2
adrimare said:

Homework Statement



I have two related questions related because they both ask to find angles between vectors.

The first one says to determine the angles between the sides of a parallelogram determined by the vectors OA = (3,2,-6) and OB = (-6,6,-2).

The second one asks to determine the angle between OP and AE, where OP = (3,4,5) and AE = (-3,4,5).

Homework Equations



?

The Attempt at a Solution



For the first one, I tried to use the cosine law, but got 1.5 degrees instead of 84.4 degrees.
These are my steps.

magnitude of OA = [tex]\sqrt{}(32 + 22 + -62) [/tex]= [tex]\sqrt{}49[/tex] = 7
Don't put [ sup] or [ sub] tags inside [ tex] tags. The reason should be obvious.
adrimare said:
magnitude of OB = [tex]\sqrt{}](-62 + 62 + -22) [/tex]= [tex]\sqrt{}76[/tex]

vector AB = (-9,4,4)

magnitude of AB = [tex]\sqrt{}(-92 + 42 + 42)[/tex] = [tex]\sqrt{}113[/tex]

[tex]\Theta[/tex] = cos-1 ((72 + [tex]\sqrt{}76[/tex] - [tex]\sqrt{}113[/tex]) / (2*7*[tex]\sqrt{}76[/tex])) = approximately 1.5 degrees

I'm confused as to what to do here.

The next question has me confused as well, as I seem to be missing something key.

Please help!
Your mixed tex and sup stuff is too hard to read, so I won't try to pick out the exact problem. Using the Law of Cosines (there's an easier way if you know about the dot product) and simplifying things a bit, you should get cos(theta) = 6/(7sqrt(76)). From this I get theta ~ 84.4 degrees. Make sure your calculator is in degree mode.
 
  • #3
thanks. It was in radian mode. whoops. But what about my second related question? Can I do the same thing? And I do know about dot product, but I'm supposed to do the question without it.
 
  • #4
Sure, the 2nd problem is almost exactly the same. The two sides adjacent to your angle are OP and AE.
 

FAQ: Finding angles between vectors

What are vectors and how are they used to find angles?

Vectors are quantities that have both magnitude and direction. They are used in mathematics and physics to represent physical quantities such as force, velocity, and acceleration. To find angles between vectors, we can use techniques such as the dot product and the cross product.

What is the dot product and how does it help find angles?

The dot product is a mathematical operation that takes two vectors and produces a scalar quantity. It is used to determine the angle between two vectors by taking the cosine of the angle between them. The dot product is calculated by multiplying the magnitudes of the two vectors and the cosine of the angle between them.

Can the dot product be used for any type of vector?

The dot product can only be used for vectors in the same dimension. This means that if we have two vectors in three-dimensional space, we can use the dot product to find the angle between them. However, if we have two vectors in different dimensions, such as one in two-dimensional space and one in three-dimensional space, we cannot use the dot product to find the angle between them.

Are there any other methods for finding angles between vectors?

Yes, the cross product is another method for finding angles between vectors. Unlike the dot product, the cross product results in a vector quantity rather than a scalar quantity. The magnitude of the resulting vector is equal to the product of the magnitudes of the two vectors and the sine of the angle between them. The direction of the resulting vector is perpendicular to the original two vectors and follows the right-hand rule.

What is the difference between the dot product and the cross product?

The dot product results in a scalar quantity, while the cross product results in a vector quantity. Additionally, the dot product gives the cosine of the angle between the vectors, while the cross product gives the sine of the angle and the direction of the resulting vector. The dot product can only be used for vectors in the same dimension, while the cross product can be used for vectors in any dimension.

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