Triangle and dot product to find angle

In summary, Vector A is composed of negative 2.32 units, 5.16 units, and 7.88 units. Vector B is composed of negative 1.12 units, 3.45 units, and 12.8 units. The vectors are perpendicular to each other and have a magnitude of 5.00 units.
  • #1
bigbk92
10
0
Vector A and vector B are expressed in component form.
A = [2.32,-5.16,7.88]
B = [-1.12,3.45,-12.8]

The standard arrow representation of these vectors and that of can be arranged to form a triangle in a plane that represents the geometric equivalent of the subtraction operation. The following items refer to this triangle.

(a) Please state, to three significant figures, the three angles of the triangle.

(b) Please express, in component form to three significant figures, a vector of magnitude 5.00 units that is perpendicular to the plane of the triangle.

(c) Please express, in component form to three significant figures, a second vector of magnitude 5.00 units that is perpendicular to the plane of the triangle.




I think the way I am supposed to appraoch this problem is to make a third vector B-A which will connect to B and A to make a triangle. So I found B-A to be [-3.44,8.61,-20.68].

Then I found the angle between A and B by using the dot product(equal to -121.3) and found the angle to be 160 degrees. I did the same thing to find the anggle between A and B-A but the angle was 169 degrees. So this is not a triangle as the angles do not add to 180 degrees.

Can someone help me figure out where i went wrong and how to approach this problem?
 
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  • #2
correction


of these vectors and that of B - A can be arranged to form
 
  • #3
bigbk92 said:
I think the way I am supposed to appraoch this problem is to make a third vector B-A which will connect to B and A to make a triangle. So I found B-A to be [-3.44,8.61,-20.68].

Then I found the angle between A and B by using the dot product(equal to -121.3) and found the angle to be 160 degrees. I did the same thing to find the anggle between A and B-A but the angle was 169 degrees. So this is not a triangle as the angles do not add to 180 degrees.

Can someone help me figure out where i went wrong and how to approach this problem?

Everything is correct, but you calculated the angle φ shown in the picture instead of the angle of the triangle (the yellow one). That is the angle enclosed by -a and b-a. Subtract 169° from 180°.

ehild
 

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1) How do I find the angle between two vectors using the dot product?

The angle between two vectors can be found using the dot product formula: θ = cos^-1((a · b) / (|a| * |b|)), where a and b are the two vectors and |a| and |b| represent their magnitudes.

2) What is the relationship between the dot product and the angle between two vectors?

The dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them. This relationship is expressed in the dot product formula: a · b = |a| * |b| * cos(θ).

3) Can the dot product be used to find the angle between non-perpendicular vectors?

Yes, the dot product can be used to find the angle between any two vectors, regardless of whether they are perpendicular or not. However, the angle obtained will be the smallest angle between the two vectors.

4) How is the dot product related to the cross product?

The dot product and the cross product are two different operations with different results. The dot product results in a scalar (a single number) while the cross product results in a vector. However, the angle between two vectors can be found using the dot product and the cross product, as shown in the formula: sin(θ) = |a x b| / (|a| * |b|).

5) Can the dot product be used to find the angle between more than two vectors?

No, the dot product can only be used to find the angle between two vectors. If you want to find the angle between more than two vectors, you can use the dot product to find the angle between each pair of vectors and then use the law of cosines to calculate the overall angle.

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