# Vector product and angles

• jegues
In summary, the homework statement states that when the angle between two vectors is increased by 20 degrees their vector product doubles.
jegues

## Homework Statement

Two vectors lie their tails at the same point. When the angle between them is increased by 20 degrees the magnitude of their vector product doubles. The original angle between them was about _____ ?

## Homework Equations

$$\overline{a}$$$$\bullet$$$$\overline{b}$$ = |$$\overline{a}$$| |$$\overline{b}$$| cos$$\theta$$

## The Attempt at a Solution

Let the original angle be c and let the final angle be d, defined as follows:

d = 20 + c;

a * b = |a||b| cos(c)

a * b = 2[ |a| |b| ] cos(d)

cos(d) = cos(20 + c) = cos(20)cos(c) - sin(20)sin(c)

I get stuck about here, I can't figure it out...

I think the author of the problem wants you to do it numerically. That's why it says 'about.'

jegues said:
Let the original angle be c and let the final angle be d, defined as follows:

d = 20 + c;

a * b = |a||b| cos(c)

a * b = 2[ |a| |b| ] cos(d)

cos(d) = cos(20 + c) = cos(20)cos(c) - sin(20)sin(c)

I get stuck about here, I can't figure it out...

Okay, so now you basically have

$$\cos x=2\left(\cos20^\circ\cos x-\sin20^\circ\sin x\right)[/itex] (using $x$ instead of 'c') Can you think of a relation between $\sin x$ and $\cos x$ that would allow you to express this equation as a polynomial in the variable $u\equiv\cos x$? Also, doesn't 'vector product' refer to the cross product? Can you think of a relation between and that would allow you to express this equation as a polynomial in the variable ? The only thing I can think of is if u = cosx, then -du = sinx. How can I end solving my u with that du sticking around... You might just have to reword that last line for me ;) EDIT: This is ruining me! The only only relevant relationship would be sin^2 + cos^2 = 1 identity. If I had any hair on my head I'd be tearing it out by now. Last edited: Don't get du involved because this is not an integration problem, just an algebra problem. sin^2 x + cos^2 x= 1 is almost certainly what MaxL was thinking of! If u=cosx, u^2=cos^2x. What would sin^2 x be? BTW, I agree with MaxL: the question is intended to be solved numerically, i.e. using a graphing calculator or by guessing different values of theta. sin^2 x + cos^2 x= 1 is almost certainly what MaxL was thinking of! If u=cosx, u^2=cos^2x. What would sin^2 x be? sin^2x = 1-u^2; I don't have a sin^2 x though, can I get it somehow from sin(20)sinx, I don't know how to turn this into sin^2(BLAH) ? jegues said: sin^2x = 1-u^2; I don't have a sin^2 x though, can I get it somehow from sin(20)sinx, I don't know how to turn this into sin^2(BLAH) ? Isolate $\sin x$ on one side of the equation, and then square both sides... Also, as MaxL pointed out, "vector product" is usually interpreted as the cross product, not the dot product...so if that's what the question asks for... ideasrule said: BTW, I agree with MaxL: the question is intended to be solved numerically, i.e. using a graphing calculator or by guessing different values of theta. I don't necessarily agree with this...an exact answer is possible by means of the quadratic formula..so unless this is part of a unit on numerical analysis, I think that by using the word "about" the questioner is just asking for a numerical value (the exact answer is arccos(some garbage) ), not that the problem be solved by numerical methods. I stand corrected! There is absolutely a way to get the answer in the form of Arctan(stuff you know). Just remember Sin/Cos=Tan. I did it in about five lines of algebra, and didn't have to use the quadratic formula. Hang in there, Jegues! Also...vector products are cross products. So the dot product relation you put in your original post won't help you. (slaps self in face) Of course! Can't believe I missed that. So your solving this line for sinx, [tex]\cos x=2\left(\cos20^\circ\cos x-\sin20^\circ\sin x\right)$$

I got Tan^-1(BLAH) = -68.7, again clearly not the answer... D:

jegues said:
So your solving this line for sinx,

I got Tan^-1(BLAH) = -68.7, again clearly not the answer... D:

If you show us your work, we should be able to see where you are going wrong...

Jegues-you're still using dot product relations. The problem is talking about cross products.
|(a x b)| = |a||b|Sin(theta)

You're close!

## 1. What is a vector product?

A vector product, also known as the cross product, is a mathematical operation between two vectors that results in a third vector that is perpendicular to both of the original vectors. It is denoted by the symbol "x" or "⨯".

## 2. How do you calculate the vector product?

To calculate the vector product, you can use the following formula:
V1 ⨯ V2 = (V1yV2z - V1zV2y, V1zV2x - V1xV2z, V1xV2y - V1yV2x)
where V1 and V2 are the two vectors being multiplied. Alternatively, you can use the determinant method or the geometric method to calculate the vector product.

## 3. What is the physical meaning of the vector product?

The vector product has several physical meanings, including calculating the torque or moment of a force, determining the area of a parallelogram or triangle, and finding the direction of a magnetic field when a current flows through a wire.

## 4. Can you take the vector product of more than two vectors?

No, the vector product is only defined for two vectors. However, you can take the vector product of multiple vectors by performing the operation between two vectors at a time. For example, if you have three vectors A, B, and C, you can find the vector product of A and B first, and then take the vector product of the result with C.

## 5. How are angles related to the vector product?

The magnitude of the vector product is equal to the product of the magnitudes of the two vectors multiplied by the sine of the angle between them. Additionally, the direction of the vector product is perpendicular to both the original vectors, forming an angle of 90 degrees with them.

Replies
4
Views
882
Replies
4
Views
1K
Replies
9
Views
1K
Replies
1
Views
919
Replies
5
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
16
Views
2K
Replies
2
Views
1K
Replies
2
Views
1K