# Do you know vectors and calculating them buggers?



## MatsNorway (Jul 4, 2011)

(3s-5) * (8,s-2)=0

Music while you ponder.


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## bigtex1989 (Feb 7, 2011)

I'm going to assume that is a dot product. I'm also going to assume that the first vector is (3,s-5). 

Now the solution!

(3,s-5) * (8,s-2) = 0

The definition of the dot product (with similar vector size) is

(a1,b1) * (a2,b2) = a1a2 + b1b2

so we know that 3*8 + (s-5)(s-2) = 0
24 + s^2 -7s + 10 = 0
s^2 - 7s + 34 = 0

we have a quadratic equation, so use the quadratic formula

s = 7+/-((49-136)^(1/2)) / 2

both answers are clearly imaginary. That is the process though in case you typed it wrong or something. 

It may also be worth noting that the dot product is used to tell if vectors are perpendicular or not (among many other functions). By saying the dot product of 2 vectors is zero, you are saying the vectors are perpendicular.

EDIT

If instead you are looking for a scalar (3s-5) times the vector (8,s-2) to equal zero, it is actually a bit easier.

The only way to get the zero is if the scalar is zero. So 3s-5 = 0 or s = 5/3


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## JohnGalt (Nov 5, 2011)

He could have also meant (3s, -5) * (8, s-2).

Or cross product!

So many possibilities.


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## MatsNorway (Jul 4, 2011)

quote from the book.

find s with calculations so that (3s,-5) stands directly (90deg i guess) on to(8,s-2)

sorry if its critical. i did not se the text until now..

You know what they say.

"if you want something right you have to do it yourself"


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## bigtex1989 (Feb 7, 2011)

ALRIGHT 

Use the dot product function.

from there, you are looking for the dot product to be zero.


(a1,b1) * (a2,b2) = a1a2 + b1b2 = 0

(3s)*8 + (-5)(s-2) = 0
24s - 5s + 10 = 0
19s = -10 
s = -10/19

Turns out your correction was pretty critical


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## MatsNorway (Jul 4, 2011)

thank you.


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