# Dot product, cross product

1. Jun 7, 2005

### jeanf

i need help with the following:
note that the big dots represents the dot product

1. suppose that $$a \bullet b = c \bullet b$$ for all vectors $$\overrightarrow{b}$$. show that $$\overrightarrow{a} = \overrightarrow{c}$$.

i suppose i can't simply divide out the b, right? anyway, i tried writing out the components of each vector - for example a=(a1, a2, a3), b=(b1, b2, b3), c=(c1, c2, c3). i got as far as:

(b1, b2, b3) $$\bullet$$(a1-c1, a2-c2, a3-c3) = 0

but i don't know how to simplify this further to eliminate the b. any help is appreciated.

2. Jun 7, 2005

### jeanf

cross product

my second question is:

2. suppose that $$\overrightarrow{a} X \overrightarrow{b} = \overrightarrow{c} X \overrightarrow{b}$$ for all vectors $$\overrightarrow{b}.$$ how are $$\overrightarrow{a}$$ and $$\overrightarrow{c}$$ related?.

i wrote the vectors in terms of its components, took the cross product, and i got as far as:

(b3(a2-c1) + b2(a3-c3), b1(a3-c3) + b3(a1-c1), b2(a1-c1) + b1(a2-c2)) = 0

again, i'm not sure how to simplify this further or what conclusion to draw about the relationship between a and c. any help is greatly appreciated.

3. Jun 7, 2005

### whozum

For the first, did you know that $\vec{a} \bullet \vec{b} = |a||b|cos(\theta)$? That should be all you need.

There is a similar identity for the cross product.

4. Jun 7, 2005

### jeanf

i have considered that, but the angles between a and b, and c and b, are different (presumably). so i end up with

|a| cos(t) = |c| cos (s)

...which doesn't seem to help very much.

5. Jun 7, 2005

### Curious3141

For the dot product, here's a hint. Consider the cases where $$\vec{b}$$ is a unit vector first in the direction of $$\vec{a}$$ then in the direction of $$\vec{c}$$. Draw up some equations and see what you can conclude about the angle between vectors $$\vec{a}$$ and $$\vec{c}$$, then what that implies about the relationship between their magnitudes.

6. Jun 7, 2005

### Curious3141

Here's the solution for the dot product :

Let the angle between vectors $$\vec{a}$$ and $$\vec{c}$$ be $$\theta$$.

Let $$\vec{b}$$ be a unit vector in the same direction as $$\vec{a}$$

Then $$\vec{a} \bullet \vec{b} = |\vec{a}|$$

$$\vec{c} \bullet \vec{b} = |\vec{c}|\cos\theta$$

These two are equal, so we have :

$$|\vec{a}| = |\vec{c}|\cos\theta$$ --eqn(1)

Now let $$\vec{b}$$ be a unit vector in the direction of $$\vec{c}$$

Working through it in the same way gives :

$$|\vec{c}| = |\vec{a}|\cos\theta$$ --eqn(2)

Solving those we get :

$$|\vec{c}|\cos^2{\theta} = |\vec{c}|$$

Either $$|\vec{c}| = |\vec{a}| = 0$$, meaning both of them are null vectors (in which case they're equal in any case),

or $$\cos^2{\theta} = 1$$

giving $$\cos{\theta} = 1$$ (the negative root is inadmissible because if you look at equations 1 and 2, the magnitudes are positive, so the cosine term can't be negative).

So we have $$\theta = 0$$, and putting that back into either equation 1 or 2, we get $$|\vec{a}| = |\vec{c}|$$.

So the two vectors $$\vec{a}$$ and $$\vec{c}$$ are of the same magnitude and oriented in the same direction. By definition, therefore, $$\vec{a} = \vec{c}$$ (QED)

7. Jun 7, 2005

### Galileo

Write the equation in a different form:

$$(\vec a -\vec c)\times \vec b=\vec 0$$
for any vector $\vec b$.

(You can do the same thing with the dot product also).
What can you infer about $\vec a-\vec c$ from this?

8. Jun 7, 2005

### dextercioby

The first exercise is incorrect.

$$\vec{a}\cdot\vec{b}=\vec{c}\cdot\vec{b}$$

Using the distributivity,one gets

$$\left(\vec{a}-\vec{c}\right)\cdot\vec{b}=0$$

Which has the infinite set of solutions:

$$\vec{a}-\vec{c}=\left\{\vec{d}\times\vec{b}|,\vec{d}\in V,\vec{b}\in V \right\}$$.

The "V" is a linear space and the $\vec{d}$ and $\vec{b}$ are arbitrary vectors.

Daniel.

9. Jun 7, 2005

### Galileo

The exercise is to show that $\vec a \cdot \vec b=\vec c\cdot \vec b\forall \vec b \Rightarrow \vec a = \vec c$.
This is clearly true, you just have to pick a clever value for b, say b=a-c

Last edited: Jun 7, 2005
10. Jun 7, 2005

### dextercioby

That's not true.That difference is not zero,but an arbitrary vector.Zero vector is not arbitrary.

Daniel.

11. Jun 7, 2005

### Galileo

Are you sure you have the same understanding of the problem as I do?
$\vec b$ can be any vector.

12. Jun 7, 2005

### dextercioby

Yes.

HINT:$$\vec{c}\cdot\vec{b}=\vec{c}\cdot\left(\vec{b}+\vec{c}\times \vec{d}\right)$$

,where \vec{d} is an arbitrary vector...

Daniel.

13. Jun 7, 2005

### Crosson

here is how I would handle the two dimentional case:

a * b = (ax + ay)(bx + by) = ax*bx + ay*by

c * b = (cx + cy)(bx + by) = cx*bx + cy*by

Two vectors are equal only if their corresponding components are equal. So if a * b = c *b then we have:

ax*bx = cx*bx

ay*by=cy*by

Which shows that a and c are equal.

14. Jun 7, 2005

### arildno

You are wrong, Daniel.
$$\vec{a},\vec{c}$$ are fixed vectors, and the $$\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}$$ is supposed to hold for ALL $$\vec{b}$$
This is equivalent to:
$$(\vec{a}-\vec{c})\cdot\vec{b}=0$$
In particular, this must hold for the choice $$\vec{b}=\vec{a}-\vec{c}$$ from which follows $$\vec{a}=\vec{c}$$

Last edited: Jun 7, 2005
15. Jun 7, 2005

### dextercioby

That's wrong

$$\vec{a}\cdot\vec{b}$$ is a scalar

and hence

$$a_{x}b_{x}+a_{y}b_{y}=c_{x}b_{x}+c_{y}b_{y} \nRightarrow \left\{\begin{array}{c}a_{x}b_{x}=c_{x}b_{x}\\a_{y}b_{y}=c_{y}b_{y} \end{array}\right$$

Okay??

Daniel.

16. Jun 7, 2005

### dextercioby

I'm not wrong,my solution does not exclude the 0 vector.That "d" is arbitrary,just as "b",so it can be the 0 vector.

Daniel.

17. Jun 7, 2005

### arildno

$$\vec{a},\vec{c}$$ are FIXED vectors; their difference do not change if $$\vec{b}$$ changes.

18. Jun 7, 2005

### dextercioby

Who said "a" and "c" are fixed?

Daniel.

Last edited: Jun 7, 2005
19. Jun 7, 2005

### arildno

It says for all b here.

20. Jun 7, 2005

### dextercioby

Yes,and my solution exploits the equality of the 2 scalar products.

Daniel.