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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: spongebob-7 on August 15, 2012, 09:28:58 pm

Title: vector question
Post by: spongebob-7 on August 15, 2012, 09:28:58 pm: Hey,

Having problem with this question

b= 2i+3j-K
c= 4i-j+5k

n=xi+yj+zk

solve for x,y and z so that both b and c are perpendicular to n

Thanks for any help
Title: Re: vector question
Post by: b^3 on August 15, 2012, 09:59:23 pm: Remembering that if two vectors are perpendicular then their dot products is zero.
$\begin{alignedat}{1}\underset{\sim}{b}.\underset{\sim}{n} & =0 \\ (2\underset{\sim}{i}+3\underset{\sim}{j}-\underset{\sim}{k}).(x\underset{\sim}{i}+y\underset{\sim}{j}+z\underset{\sim}{k}) & =0 \\ 2x+3y-z & =0....[1] \\ \underset{\sim}{c}.\underset{\sim}{n} & =0 \\ (4\underset{\sim}{i}-\underset{\sim}{j}+5\underset{\sim}{k}).(x\underset{\sim}{i}+y\underset{\sim}{j}+z\underset{\sim}{k}) & =0 \\ 4x-y+5z & =0....[2] \\ z & =2x+3y....\left(from\:[1]\right) \\ 8-3-54x-y+5(2x+3y) & =0 \\ 4x-y+10x+15y & =0 \\ 14x+14y & =0 \\ \therefore x & =-y \\ z & =2(-y)+3y \\ \therefore z & =y \end{alignedat}$
So that leaves us with $x\underset{\sim}{i}-x\underset{\sim}{j}-x\underset{\sim}{k}=x\left(\underset{\sim}{i}-\underset{\sim}{j}-\underset{\sim}{k}\right)$ where $x\epsilon R$
This happens because the vector can be any length, so lets take the most basic one which will be $\underset{\sim}{i}-\underset{\sim}{j}-\underset{\sim}{k}$

EDIT: a few lines were missed, fixed it up, they're added now.

(also could be done using the cross product, but thats out of the course)
Title: Re: vector question
Post by: Jenny_2108 on August 15, 2012, 11:19:31 pm: Quote from: b^3 on August 15, 2012, 09:59:23 pm
also could be done using the cross product

Whats the cross product? Is it faster than this method?
Title: Re: vector question
Post by: paulsterio on August 15, 2012, 11:25:08 pm: Quote from: Jenny_2108 on August 15, 2012, 11:19:31 pm
Whats the cross product? Is it faster than this method?

It's the other form of vector multiplication. In order to evaluate a cross product of two vectors, multiply the two magnitudes and have the answer vector perpendicular to the two other vectors.

Look you don't need to know it, it really does you no good to know it, but if you're keen look at any Calculus book like Stewart's and you'll be able to understand it - it's not hard - it's just off the VCE course.

Evaluating a cross product does take a bit of time (but that might be because I'm not a Maths student at uni so I'm just slow) - but it should be faster.
Title: Re: vector question
Post by: spongebob-7 on August 15, 2012, 11:29:48 pm: thanks tons b^3
Title: Re: vector question
Post by: b^3 on August 15, 2012, 11:37:29 pm: Quote from: paulsterio on August 15, 2012, 11:25:08 pm
Quote from: Jenny_2108 on August 15, 2012, 11:19:31 pm
Whats the cross product? Is it faster than this method?
Evaluating a cross product does take a bit of time (but that might be because I'm not a Maths student at uni so I'm just slow) - but it should be faster.
I wouldn't really say it takes that long, its just a matter of not making a silly mistake or missing the negative on the second determinate for the j component. I'm kinda hesitant on explaining it as its not on the course (as said above). It was the quicker method for me, (I did it the cross product way first and got the answer before I'd realised the VCE way of doing it). But yeh as Paul's said, don't worry about it unless you want to learn something a little bit extra.