Find the angle between two vectors

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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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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)
 
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)?
 
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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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.
 
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?
 
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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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.
 
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
 
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."
 
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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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!
 
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..:)
 
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].