# Thread: Vectors help!

1. ## Vectors help!

Hello. I'm currently having some problems on this question on vectors. If you could help me, it'll be great.

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Vector AC is [2,4,-3] and it forms the hypotenuse of a right angled triangle of trangle ABC. Vector AB is parallel to vector [1,1,0]. Find AB.

Well, I've done a little diagram to show you what it looks like. For my working out, I deduced that vector AB would be [k,k,0] for some scalar k because it's parallel to [1,1,0]. Is that the right way to go about it?

First, I tried finding the angle between A and C. I managed to get Cosθ = 6k/squareroot(58k²). From there, I deduced that AB = 6k and AC = sqaureroot(58k²). However, that doesn't really seem to help me much at all.

I also tried using vector subtraction so vector BC = vector AC - vecor AB but I'm not too sure how to go about that. Also, right-angled triangles seem to indicate Pythagoras so maybe that has something to do with it?

Anyway, can somebody please show me how to solve this problem or perhaps provide a starting point?

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Thankyou. Any help is of course, appreciated.

2. $\overrightarrow {BC} = \overrightarrow {AC} - \overrightarrow {AB}$

$\left\| {\overrightarrow {AC} } \right\|^2 = \left\| {\overrightarrow {AB} } \right\|^2 + \left\| {\overrightarrow {BC} } \right\|^2$

3. Alright, I managed to get BC as [2-k,4-k,-3].

From there, I used Pythagoras like you mentioned:

(√29)² = (√2k²)² + (√2k² - 12k + 29)²
29 = 2k² + 2k² - 12k + 29
4k² - 12k = 0
k² - 3k = 0

k = 0 & 3

So therefore, my answer is 3 meaning vector AB = [3,3,0]?
I cannot use 0 because a zero vector won't be possible.

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It looks right but there's still something that's bothering me...

BC = AC - AB right?
So if I simply pop the vectors back in and replace k with 3, it'll work out:

[2-3,4-3,-3] = [2,4,-3] - [3,3,0]
[-1,1,-3] = [-1,1,-3]

But however, this seems to work with ANY integer. For example, if I take 8 and replace all k's with 8:

[2-8,4-8,-3] = [2,4,-3] - [8,8,0]
[-6,-4,-3] = [-6,-4,-3]

So does that mean [8,8,0] is also an answer as for [k,k,0] where k is ANY integer? Or am I just complicating things more and doing it wrong?

I thought there's only suppose to be one answer...

Thanks!

4. Originally Posted by sqleung
...

But however, this seems to work with ANY integer. For example, if I take 8 and replace all k's with 8:

[2-8,4-8,-3] = [2,4,-3] - [8,8,0]
[-6,-4,-3] = [-6,-4,-3]

...
Keep in mind that $\overline{AC}$ is the hypotenuse in a right triangle which is the longest of the sides of the triangle.

Since $|\overrightarrow{AC}| = \sqrt{29}$ the sum of the squares of the 2 other legs can't exceed 29.

5. Originally Posted by earboth
Keep in mind that $\overline{AC}$ is the hypotenuse in a right triangle which is the longest of the sides of the triangle.

Since $|\overrightarrow{AC}| = \sqrt{29}$ the sum of the squares of the 2 other legs can't exceed 29.
Ah, okay. Thankyou very much for that