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# Vectors - Proof watch

1. Hey,
Anyone know how to crack these questions? I can't seem to get to them

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2. Got this for C2 - not sure if its at all right
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3. (Original post by Fatts13)
Got this for C2 - not sure if its at all right
First line of your proof doesn't make sense with the |AC| |AC| cos theta, that would only equal |AC| |AC| if if theta were 90 degrees.

So, AC= BC - BA (flipping the AB around)

Now dot product.

AC.AC = (BC-BA).(BC-BA)

and carry on.

For C1. Just substitute into the formula for the function.
For part c, you want to aim to show |s(v)|^2 = |v|^2, and the result will follow.

Use what you know about the properties of the dot product.
4. (Original post by ghostwalker)
First line of your proof doesn't make sense with the |AC| |AC| cos theta, that would only equal |AC| |AC| if if theta were 90 degrees.

So, AC= BC - BA (flipping the AB around)

Now dot product.

AC.AC = (BC-BA).(BC-BA)

and carry on.

For C1. Just substitute into the formula for the function.
For part c, you want to aim to show |s(v)|^2 = |v|^2, and the result will follow.

Use what you know about the properties of the dot product.
So for C2:
AC.AC = (BC.BA).(BC-BA)
= (BC)^2 - (BC)(BA) - (BC)(BA) + (BA)^2
= (BC)^2 + (BA)^2 - 2(BC)(BA)
Do I just put in cos (angle ABC) at the end? And is BA the same as AB?

Sorry I'm asking the most stupidest questions.
5. Got this for C2
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6. (Original post by Fatts13)
So for C2:
AC.AC = (BC.BA).(BC-BA)
= (BC)^2 - (BC)(BA) - (BC)(BA) + (BA)^2
= (BC)^2 + (BA)^2 - 2(BC)(BA)
Do I just put in cos (angle ABC) at the end? And is BA the same as AB?

Sorry I'm asking the most stupidest questions.
You need to be careful which operation you're performing at any time.
I was a bit lax in my post, however this appears to be an issue, so I'll try and be accurate.

is the dot product, and it's equal to

(BC.BA).(BC-BA) should be

which gives you

Then from the defintion of the dot product, as you were intending:

Although BA is not the same as AB, as vectors they point in opposite directions and , but their lengths are the same and

You can use LaTex on TSR for mathematical symbols - see entry in Useful Resources widget to the right on the forum.

Edit: I've assumed all the overlines are meant to be arrows, but just didn't print very well in the question.
7. (Original post by ghostwalker)
You need to be careful which operation you're performing at any time.
I was a bit lax in my post, however this appears to be an issue, so I'll try and be accurate.

is the dot product, and it's equal to

(BC.BA).(BC-BA) should be

which gives you

Then from the defintion of the dot product, as you were intending:

Although BA is not the same as AB, as vectors they point in opposite directions and , but their lengths are the same and

You can use LaTex on TSR for mathematical symbols - see entry in Useful Resources widget to the right on the forum.

Edit: I've assumed all the overlines are meant to be arrows, but just didn't print very well in the question.
I'm so sorry for having made you do all this long working out. Thank you so much, I understand it now.
No my lecturer prefers we don't use arrows - bit weird but hey.
Thanks for letting me know about LaTex

Heres what i did for C1: Sorry for being a pest
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8. Not sure about C3 :s
9. (Original post by Fatts13)
Heres what i did for C1: Sorry for being a pest
No problem.

Yep, C1 looks fine, though one of your "." s should be a "-" in the first line of B - very minor point, and it's obvious you meant "-".

Geometric interpretation:
Draw a diagram, mark a v and an e (I suggest an acute angle between them), and apply the function.

One thing you may find helpful if you're not already aware of it, is (v.e)e is the component of v in the direction of e, when e is a unit vector.

Answer in spoiler, BUT have a go first:
Spoiler:
Show

The function s reflects v in a line perpendicular to e (I think).

Not got time for C3 at the moment.
10. (Original post by ghostwalker)
No problem.

Yep, C1 looks fine, though one of your "." s should be a "-" in the first line of B - very minor point, and it's obvious you meant "-".

Geometric interpretation:
Draw a diagram, mark a v and an e (I suggest an acute angle between them), and apply the function.

One thing you may find helpful if you're not already aware of it, is (v.e)e is the component of v in the direction of e, when e is a unit vector.

Answer in spoiler, BUT have a go first:
Spoiler:
Show

The function s reflects v in a line perpendicular to e (I think).

Not got time for C3 at the moment.
Thank you so much. I get it now. Thank you for your time and sorry for wasting so much of your time.
11. (Original post by Fatts13)
...
Nope - can't see how to do C3.
12. (Original post by ghostwalker)
Nope - can't see how to do C3.
I think thie following works:

Write .

The rest is essentially mechanical - I'll put an outline behind the spoiler:

Spoiler:
Show

Then becomes

and the LHS of the equality becomes

Expand everything out and nearly everything cancels to leave as required.

(Always possible I've made a stupid mistake somewhere, but as I got the right answer I'm guessing I didn't...)

13. (Original post by DFranklin)
I think thie following works:

Write .

The rest is essentially mechanical - I'll put an outline behind the spoiler:

Spoiler:
Show

Then becomes

and the LHS of the equality becomes

Expand everything out and nearly everything cancels to leave as required.

(Always possible I've made a stupid mistake somewhere, but as I got the right answer I'm guessing I didn't...)

Very nice - PRSOM.

I don't think that method even made it to my list of things to try. Got diverted by the hint and previous working - that's my excuse, and I'm sticking to it

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