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[jaked122]

Yeah. The dot product is super mega useful. The cross product really isn't. It's R3 only, serviceable in R2 by using 0 for Z.

Cross products are really weird, because they don't make much intuitive sense. You have an L shape, and the cross product of them is in the 3rd dimension. It always points "out" compared to the vectors it's a cross product of. You're finding the normal of the plane formed by those two vectors.

That makes more sense, though if I remember right, it's not commutative, correct?

Which is why the wedge product makes so much more sense.

Can I use the wedge product in place of a cross product?

[frostshotgg]

That makes more sense, though if I remember right, it's not commutative, correct?

Correct. Specifically, it's anticommutative so U x V = - (V x U).

[Spehss _]

Well, for part a it's just a vector projection. To get the vector rejection (b), take the original vector a and subtract its projection.

Haven't heard of vector projection and rejection terminology. I suspect if that is going to be covered by my class it's in the next chapter and we just haven't gotten to it yet. Skimming the wikipedia article on it and it looks like it's just what I'm doing in this problem. So nothing new, I suppose.

Magnitude of the first vector is 8.8, answer to part a is -3.2, which you're saying is the projection, and the answer for part b is 8.2, which you're saying is the rejection.

8.8-3.2=/=8.2

I guess by subtracting the projection from the original vector I need to separate the projection into x,y, and z components, then subtract those components from the original vector's components, then calculate the new magnitude using those resulting components.

[frostshotgg]

It doesn't.

Okay, that's not true. The cross product works in 3 dimensions and 7 dimensions. That's it. If you want something more general, look at wedge products and, hence, exterior algebra.

Missed this earlier, but R7's cross product is whack as fuck. There's not just one cross product per pair of vectors in R7.

Other interesting, but ultimately useless knowledge about cross products is that technically there is a cross product operation for R0 and R1, but it's always 0. The fact that there are only cross products for R0, R1, R3, and R7 is because there's only composition algebra in 1-, 2-, 4-, and 8-space.

[Spehss _]

Well, I'm talking vector subtraction, not magnitude subtraction. Big difference. <0,1>-<1,0> is not zero.

Am I right that to find the solution I need to "separate the projection into x,y, and z components, then subtract those components from the original vector's components, then calculate the new magnitude using those resulting components"? Because I am trying that and I'm not getting the answer in the book.

Using the magnitude in the previous part, -3.2, and the angle, 111, the x and y components would be -3.2cos(111)=1.146 or ~1.1 and -3.2sin(111)=-2.987 or ~3.0. The z component can be reverse calculated using 3.2=squareroot(1.1²+3²+z²), so z²=3.2²-(1.1²+3²) or 10.24-10.21=0.03 for z² which seems incredibly small.  But the three values together result in 3.2 for the magnitude, so it must be right. Square root of 0.03 is 0.19 or ~0.2. Then subtracting the x,y,z components form the original (4-1.1)i+(5+3)j+(-6+0.2) results in (2.9)i+(8)j+(-5.8)k.

The three components squared, added together, then the sum square root-ed results in 10.29. The answer in the book is 8.2.

[frostshotgg]

As an aside before I start: Don't approximate stuff until the end. It makes things really hard to follow and causes a lot of error in the long run.

You can't just use cos(theta) and sin(theta) to find the X and Y there. Those should always come out to your whole squared. The only reason your Z didn't come out to 0 there is because of calculator funniness.

Step one is to find the magnitude of A along the direction of B, so you rearrange your dotproduct formula for |A|*cos(theta), which leaves you with A·B/|B|. At this point you now have the answer to A, although usually questions would ask for the full components of Aproj so you won't be able to stop at this point.

The next step to what you need to do is called the unit vector. You're trying to find out what 1 unit in the direction of B is. This means you find the magnitude of B, then divide each component of B by it. You already know the magnitude is root(13), so now you find the unit vector of B. (-1/root(13),2/root(13),3/root(13)).

Now that you know the direction, you multiply the magnitude of the projection of A onto it and you're able to find out the components of Aproj.

To find B, since you now have done A properly, you subtract A-Aproj (not |A|-|Aproj|), and you're left with the rejection. Find the magnitude of that and you're done.


It's worth noting that if all you need is the magnitude of the rejection you could in theory find theta with the dotproduct, then |A|*sin(theta) would be the magnitude of the rejection, then you could find the normal to B in the plane defined by AB and multiply that by the magnitude of Arej to find the components, then A-Arej would be Aproj, but at that point you're just going in an ugly circle and could've solved it the sane way already and are just having fun.

[TheBiggerFish]

I have no clue what is happening right now and it is strangely comforting.  I don't know everything!  Yay!

[H4zardZ1]

I don't know which thread this one should i post, so:
Is (x^y)-((x^y-1)*(x-1))=(x^y-1)?

[crazysheep]

Only if (x^y-1) means x^(y-1)

[H4zardZ1]

Only if (x^y-1) means x^(y-1)

Sorry, i forgot that bracket right there. Thank you.

[hops]

Me and my tutor don't know how to solve b.

[TheBiggerFish]

Which 'b'?

13's literally taking the first function from a, and substituting it in, then simplifying.  Then you plug in some points and connect them in a vaguely appropriate line/curve.

As for 12.b), you have a trapezoid...you need to find the' height' between c and d to use that equation though.  Though I think you can solve it with all the sides...Wait, is 'd' the entire side?  It is.  Similar triangles!!!!  And areas of!!!!

[Helgoland]

Those exercises are shoddily made, and are a disgrace to mathematics as a field. Approximately? Seriously? And it isn't clear at all what line segments the letters correspond to. You should simply refuse to solve this and instead use it as toilet paper before handing it in.

Assuming the person who came up with his shit had even a tiny bit of common sense, here's an approach:

The bottom line segment of the small triangle to the top right is parallel to the bottom line because of the parallel postulate. That means you can just calculate b from \alpha and e (as you did in part a) ) and use A = b*c (because that's how parallelograms work).

Whatever they pay your tutor, they're paying too much.

[TheBiggerFish]

I agree with Helgoland on the whole tutor thing.

[hops]

Those exercises are shoddily made, and are a disgrace to mathematics as a field. Approximately? Seriously? And it isn't clear at all what line segments the letters correspond to. You should simply refuse to solve this and instead use it as toilet paper before handing it in.

Assuming the person who came up with his shit had even a tiny bit of common sense, here's an approach:

The bottom line segment of the small triangle to the top right is parallel to the bottom line because of the parallel postulate. That means you can just calculate b from \alpha and e (as you did in part a) ) and use A = b*c (because that's how parallelograms work).

Whatever they pay your tutor, they're paying too much.

Unfortunately, said exercise is a sample exam for my Germany scholarship.