Find the angle between two vectors

In summary: Therefore, using Pythagoras' Theorem: $$|A|^2 + |B|^2 = |C|^2$$$$5.25^2 + 5.25^2 = |C|^2$$$$27.6 = |C|^2$$$$|C| = 5.25$$Therefore, the triangle is a right angled triangle and using basic trigonometry:$$\sin \theta = \frac{|A|}{|C|} = \frac{5.25}{5.25} = 1$$$$\theta = \sin^{-1} 1 = \boxed{90^{\circ}}$$In summary,
  • #1
jamesbrewer
13
0

Homework Statement



Two vectors A and B have the same magnitude of 5.25. If the sum of these two vectors gives a third vector equal to 6.73j, determine the angle between A and B.

Homework Equations



For some vector [itex]\vec{R}[/itex]: [itex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/itex]

The Attempt at a Solution



I feel like I need to find the components of A and B, but I can't figure out how to do that with only the information given -- is it even possible?

I think the magnitude of the third vector is 6.73. Since there was no [itex]\hat{i}[/itex] term given for [itex]\vec{R}[/itex], I would say that [itex]|\vec{R}| = \sqrt{R_x^2 + R_Y^2} = \sqrt{6.73^2} = 6.73[/itex].

Given the components of A and B, this would be easy to solve. Without them I am utterly lost. What am I missing?
 
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  • #2
jamesbrewer said:

Homework Statement



Two vectors A and B have the same magnitude of 5.25. If the sum of these two vectors gives a third vector equal to 6.73j, determine the angle between A and B.

Homework Equations



For some vector [itex]\vec{R}[/itex]: [itex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/itex]

The Attempt at a Solution



I feel like I need to find the components of A and B, but I can't figure out how to do that with only the information given -- is it even possible?

I think the magnitude of the third vector is 6.73. Since there was no [itex]\hat{i}[/itex] term given for [itex]\vec{R}[/itex], I would say that [itex]|\vec{R}| = \sqrt{R_x^2 + R_Y^2} = \sqrt{6.73^2} = 6.73[/itex].

Given the components of A and B, this would be easy to solve. Without them I am utterly lost. What am I missing?

The vectors A and B, along with the resultant will form a triangle - probably not right angled - so you can use the cosine Rule to solve.

a2 = b2 + c2 - 2.b.c.cos(A)
 
  • #3
PeterO said:
The vectors A and B, along with the resultant will form a triangle - probably not right angled - so you can use the cosine Rule to solve.

a2 = b2 + c2 - 2.b.c.cos(A)

What are a, b, and c? The lengths of the sides of the triangle? If so, would that mean that a = |A|, b = |B|, and c = |C| (the resultant)?
 
  • #4
jamesbrewer said:
What are a, b, and c? The lengths of the sides of the triangle? If so, would that mean that a = |A|, b = |B|, and c = |C| (the resultant)?

a, b & c are the sides,
A is the angle opposite side a

[so a, b & c could represent vectors A, B and C but not necessarily, depends which angle you are looking for]
 
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  • #5
How can I form a triangle if I don't know anything other than the vector's magnitude? I have nothing to tell me what direction it points in.
 
  • #6
jamesbrewer said:
How can I form a triangle if I don't know anything other than the vector's magnitude? I have nothing to tell me what direction it points in.

You can stand a triangle up any way you like - its sides are still the same length and its angles are still the same size. A name like "the base" might apply to a different side, depending which way you arrange it, but that should not be a problem .. what's in a name?
 
  • #7
Here's what I've got:

[itex]|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|cos\theta[/itex]

[itex]6.73^2 = 5.25^2 + 5.25^2 - 2(5.25)^2 cos \theta[/itex]

[itex]45.29 = 27.56 + 27.56 - 2(27.56) cos \theta[/itex]

[itex]45.29 = 55.12 - 55.12 cos \theta[/itex]

[itex]45.29 - 55.12 = - 55.12 cos \theta[/itex]

[itex]-9.83 = -55.12 cos \theta[/itex]

[itex]\frac{-9.83}{-55.12} = cos \theta[/itex]

[itex]\theta = cos^-1 0.178[/itex]

[itex]\theta = 79.75^o[/itex]

My answer wasn't correct though, where did I go wrong?
 
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  • #8
One approach which you might want to consider is to look at the relationships between the vector components in terms of simultaneous equations. Suppose that the two initial vectors are A and B and the resultant is C.

Since C contains no x-component and has only a y-component, the sum Ax + Bx must be zero, or in other words, Bx = -Ax. Similarly, Ay + By must be Cy. You also have |A| = |B| = 5.25, so that's another pair of relationships. You have four unknowns (really three when you consider that the x-components of A and B are equal and opposite) and plenty of interrelationships to use to solve for them.
 
  • #9
jamesbrewer said:
Here's what I've got:

[itex]|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|cos\theta[/itex]

[itex]6.73^2 = 5.25^2 + 5.25^2 - 2(5.25)^2 cos \theta[/itex]

[itex]45.29 = 27.56 + 27.56 - 2(27.56) cos \theta[/itex]

[itex]45.29 = 55.12 - 55.12 cos \theta[/itex]

[itex]45.29 - 55.12 = - 55.12 cos \theta[/itex]

[itex]-9.83 = -55.12 cos \theta[/itex]

[itex]\frac{-9.83}{-55.12} = cos \theta[/itex]

[itex]\theta = cos^-1 0.178[/itex]

[itex]\theta = 79.75^o[/itex]

My answer wasn't correct though, where did I go wrong?

How incorrect was your answer? You have rounded off all the way through which could make a small difference
 
  • #10
PeterO said:
How incorrect was your answer? You have rounded off all the way through which could make a small difference

Unfortunately I have no idea. All I was told was "Incorrect answer."
 
  • #11
Let's step back from the problem and see if we can make additional simplifying deductions before invoking formulas.

Since the resultant of vectors A and B is a vector with only a y-component (6.73j), then A and B must have equal and opposite x-components. Further, since A and B have equal magnitudes (5.25), this then forces their y-components to be equal also. Why not equal and opposite you say? Because then their sum would be zero rather that +6.73.

attachment.php?attachmentid=39204&stc=1&d=1316810186.gif


So A and B "straddle" the positive y-axis, and make equal angles with that axis. The sum of their y-components is 6.73. Now it's time to write formulas. If you let [itex]\theta[/itex] be the angle between either A or B and the Y-axis, what is an expression for the y-component of A or B? If double that is 6.73, can you solve for [itex]\theta[/itex]? What then is the angle between A and B?
 

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  • #12
jamesbrewer said:
Unfortunately I have no idea. All I was told was "Incorrect answer."

Draw a diagram of the vectors, like gneil's and you will see that they actually want (180 - 79) degrees - with the decimal bits.
Rounding off needs to be addressed too.
gneil's method is excellent also - even preferable!
 
  • #13
Three vectors sum to zero. The magnitude of 2 vectors is equal and the third one is root 2 times the magnitude of the equal vectors. Find the angle between the three vectors. Plzz help me on dis one..:)
 
  • #14
Karthik k said:
Three vectors sum to zero. The magnitude of 2 vectors is equal and the third one is root 2 times the magnitude of the equal vectors. Find the angle between the three vectors. Plzz help me on dis one..:)

Since the three vectors will make up a triangle - so you end up where you started with a vector sum of zero - you should recognise the magnitudes 1,1,√2 as the sides of a very common triangle in trigonometry, from which we derive the exact vale of sin, cos and tan of a particular angle.
The other triangle used in trigonometry is known as the 2,1,√3 triangle [not that it relates directly to this question].
 

1. What is the formula for finding the angle between two vectors?

The formula for finding the angle between two vectors is:
θ = cos-1 ((a · b) / (|a| * |b|)), where a and b are the two vectors and |a| and |b| are the magnitudes of those vectors.

2. How do I calculate the dot product of two vectors?

The dot product (a · b) of two vectors can be calculated by multiplying the corresponding components of the vectors and adding the products together. For example, if a = [a1, a2, a3] and b = [b1, b2, b3], then a · b = a1 * b1 + a2 * b2 + a3 * b3.

3. Can the angle between two vectors be negative?

No, the angle between two vectors is always positive. If the dot product of two vectors is negative, the angle between them is greater than 90 degrees and the cosine value will be negative, but the actual angle is still positive and can be found using the formula mentioned in the first question.

4. Do the two vectors have to be in the same dimension to find the angle between them?

Yes, the two vectors must be in the same dimension (both 2D or both 3D) in order to find the angle between them. If the vectors are in different dimensions, the concept of an angle between them does not apply.

5. Can the angle between two vectors be greater than 180 degrees?

No, the angle between two vectors is always less than or equal to 180 degrees. This is because the dot product of two vectors is equal to the product of their magnitudes multiplied by the cosine of the angle between them. Since the cosine of any angle is always between -1 and 1, the dot product can never be greater than the product of the magnitudes, which means the angle can never be greater than 180 degrees.

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