# Really fundamental cross/dot product questions

1. Sep 6, 2014

So I've been working on physics homework and we have some vector/dot product questions.
This is really long, but the questions I have really are rudimentary at best.

I have seven total questions.

You're given two vectors that only have an x and y component, A, and B, and the positive Z axis is out of the page.

A points roughly (no numbers given) 45 degrees (north-east). B point roughly 315 degrees (north-west).

I've tried to read the class notes on vectors and things, however I really can't seem to understand it.

There are basically three things I'm trying to understand.
Vectors, which I understand, and the difference between cross products and dot products.

Cross products seem to be |a||b|sine($\theta$).
Dot products seem to be abCosine($\theta$).

So please explain if what I'm doing is right-

24
What direction is vector axb? (A times B)
So since it's a cross product, I guess it's the absolute magnitude of a and b times sine.
However I really don't understand the angle thing. How is the angle measured? Is the $\theta$ measured depending on the x or y axis? Or just THE angle between A and B?
So I wrote down the direction is upwards with no x-component? I have no better guess. Approximately the angle between a and b looks to be 90-120 degrees, 90 degrees, which would appear to sort of "add" a and b, just with larger magnitude.

25
The next question is how is bxa related to axb in magnitude and direction?
I'm sure that I'm wrong, however i can only think that cross products are commutative and that they are both identical. I guess there could be a thing with the angle, but still I'm not sure.

26
How could you change the direction of a, leaving it in the x, y plane to make axb zero?
Sine(180)=0, so you would just make a paralell to B.

27
If we rotate the coordinate axis so that x points towards the bottom of the paper, which I assume means to rotate the grid by 90 degrees clockwise, "how will axb change?"
This really doesn't seem to be relevant? I'm supposing a and b don't move with the graph, but either ways since first of all cross products are absolute values and that the angle between a and b is the same no matter how you rotate the xy axis, 27 seems to be a pointless question.

I think I sort of understand dot products after reading the class notes.
So one question I have is the result of a dot product, such as a$\bullet$b, is C, which is a scalar. So a scalar is just a number, but I'm confused what the difference between a scalar and a vector is. A scalar has an angle and a magnitude, I think, so isn't that just a vector?

And the only way to get a negative dot product is with the angle>90, right? Since a and b are both magnitudes, which cannot be negative?

28. How could you change the direction of a, leaving it in the xy plane to make a dot b as large as possible?
Since cosine 0 or 180= maximum, making A to be parallel would give the largest dot product.

29
How could you change a's direction to make a dot b as small as possible? Can c be zero when a and b are not?
When a is perpendicular to b, axb is zero. C can be zero any time a and b are nonzero but perpendicular.

30 Can you assign a direction to C? If so, what direction would c be for the vectors a and b as shown?
Since A is roughly 45 degrees and B is 315 degrees roughly, the dot product is close to zero. I am not sure if C is just a number or has an angle. I think it doesn't?

2. Sep 6, 2014

You need to read up on the differences between scalars and vectors. You also need to read about cross products.

A scalar is just a (real) number.

A vector has both magnitude and direction. You should think of it as an arrow in the coordinate system. It has a certain length (its magnitude) and it points somewhere (its direction).

The dot product takes two vectors and produces a scalar.

The cross product takes two vectors and produces a new vector.

The formula you have for the cross product is only for its magnitude. $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin \theta$, where $\theta$ indeed is the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$.

Your book should also tell you something about how you determine the direction of the vector produced by the cross product.

3. Sep 6, 2014

### Fredrik

Staff Emeritus
Since this is the homework forum, I can only give you hints, not complete answers.

The only part of that that's relevant here is that they're not parallel, and not 0.

The dot product $x\cdot y$ is $|x||y|\cos\theta$ where $\theta$ is the angle between the vectors, but the cross product $x\times y$ is a vector with magnitude $|x||y|\sin\theta$.

Note that the definitions make it clear that $x\cdot y$ is a number and $x\times y$ is a vector:
\begin{align}
&x\cdot y=x_1y_1+x_2y_2+x_3y_3\\
&x\times y=(x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_3)
\end{align}

That angle only contributes to the magnitude, not the direction. The direction is always perpendicular to the plane that contains x and y. A lot of people use a "right-hand rule" to remember which of the two perpendicular directions it is. http://en.wikipedia.org/wiki/Right-hand_rule

The angle has nothing to do with coordinate axes. There's a unique plane that contains x,y and 0. The angle is measured in that plane.

You have a formula for the magnitude, so I suggest that you use it. Cross products aren't commutative. They're not even associative. Use the definition that I included above, or the right-hand rule, to figure out the direction.

That's a good start, but there are actually two directions that work (two angles).

The point is that the axes are irrelevant, or equivalently, that your choice of basis for $\mathbb R^3$ is irrelevant. This isn't obvious, since the definition I included above can give the impression that the basis matters. The numbers $x_1,x_2,x_3,y_1,y_2,y_3$ all depend on it.

Scalars don't have angles (except when we're talking about complex scalars, but there's no need to talk about that now). A scalar is a number. A vector in $\mathbb R^3$ is an ordered triple of numbers $(x_1,x_2,x_3)$.