Vectors and Right-angled triangles

[sqleung]

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 Cos? = 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.

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.

 

[soroban]

Hello, sqleung"!

You were almost there!

\(\displaystyle \text{Vector }\overrightarrow{AC} \:= \:\langle2,4,-3\rangle\text{, and is the hypotenuse of right triangle }ABC.\)
\(\displaystyle \text{Vector }\overrightarrow{AB}\text{ is parallel to vector }\langle1,1,0\rangle.\;\;\text{Find }AB.\)

\(\displaystyle \text{I deduced that vector }AB\text{ would be }\langle k,k,0\rangle\text{ for some scalar }k\text{ because it's parallel to }\langle1,1,0\rangle.\)
Is that the right way to go about it? . . . . Yes!

\(\displaystyle \text{First, I tried finding }\angle A.\)
\(\displaystyle \text{I managed to get: }\;\cos A \:=\:\frac{6k}{\sqrt{59k^2}}\) . . . . Good!

\(\displaystyle \text{Reduce that fraction: }\;\cos A \;=\;\frac{6k}{\sqrt{59}\cdot k} \;=\;\frac{6}{\sqrt{59}}\)


\(\displaystyle \text{Then: }\;|AB| \;=\;|AC|\cdot\cos A\)

 

[sqleung]

|AB| = |AC| * cos A
6 = ?58 * cos A
Cos A = 6 / ?58
A = 38.016 degrees (3 DP)

So I managed to get the angle but I'm not too certain how that would help me find k. :? Any suggestions?

Thanks for your help though

 

[Deleted member 4993]

|AB| = |AC| * cos A
6 = ?58 * cos A
Cos A = 6 / ?58
A = 38.016 degrees (3 DP)

So I managed to get the angle but I'm not too certain how that would help me find k. :? Any suggestions?

Thanks for your help though

Use soroban's last equation to find the numerical value of |AB|

Equate that to |AB| calculated from <k,k,0>