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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: VeryCrazyEdu. on April 21, 2010, 09:12:59 pm

Title: Perpendicular Vector
Post by: VeryCrazyEdu. on April 21, 2010, 09:12:59 pm: Hi,

I was just wondering if there was any way to find a vector that is perpendicular to 2 OTHER vectors..

I know that you can use the cross product however this isn't on the spec course...any ideas?

Thanks :)
Title: Re: Perpendicular Vector
Post by: kamil9876 on April 21, 2010, 09:44:25 pm: Suppose your two vectors are $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$ .

then by the dot product you want to solve:

$a_1x + a_2y + a_3z=0$

$b_1x + b_2y + b_3z=0$

Which isn't hard. (there are infinitely many solutions, as is obvious from geometry, but you can just select one, ussually you can do that easily by letting x=0 and solving two equations with two unknowns)
Title: Re: Perpendicular Vector
Post by: VeryCrazyEdu. on April 21, 2010, 09:56:15 pm: oh haha thats so easy compared to what i was thinking! Thanks heaps :)
Title: Re: Perpendicular Vector
Post by: rainbows. on April 24, 2010, 12:15:07 am: This is a related question for vectors,

Im confuse with it tho...

Find the unit vector which bisects the angel between a = 2i - j + 2k and b = 4i +3k. Hint: First find the unit vectors in the directions of the given vectors) >:S

Thanks :)
Title: Re: Perpendicular Vector
Post by: theuncle on April 24, 2010, 05:01:33 pm: Quote from: rainbows. on April 24, 2010, 12:15:07 am
Find the unit vector which bisects the angel between a = 2i - j + 2k and b = 4i +3k. Hint: First find the unit vectors in the directions of the given vectors) >:S
Here's how I can think of doing it... there's probably a much easier way though ;)
let u=a's unit vector and v=b's unit vector and z= required vector
so $u=\dfrac{2}{3}i - \dfrac{1}{3}j + \dfrac{2}{3}k$ and $v = \dfrac{4}{5}i + \dfrac{3}{5}k$

let the angle between u and v = 2x
therefore, the angle between u and z = x
and the angle between v and z = x
$|u|=|v|=|x|=1$

$\dfrac{u.v}{1} = \dfrac{14}{15} = sin(2x)$ $\dfrac{u.z}{1} = sin(x)$ $\dfrac{v.z}{1} = sin(x)$
Say that $z= ai + bj + ck$
Solving $sin^{-1}(\dfrac{14}{15})=2x$ and $\dfrac{2}{3}a - \dfrac{1}{3}b + \dfrac{2}{3}c = sin(x)$ and $\dfrac{4}{5}a + \dfrac{3}{5}c = sin(x)$ and $a^2+b^2+c^2=1$ given that
We have 4 variables with 4 equations so it can be solved... (i used a calculator)
Two results were obtained because x has both an acute and obtuse solution, we only want the acute one
The exact solutions were hideous, so to two decimal places,
$z=-0.03i + 0.21j + 0.98k$ the other solution i think was the colinear obtuse solution,
as i said, i'm sure there must be an easier way, but this was the best that I could come up with!

My only other thoughts would be to average the i, j and k components of vectors a and b and find a unit vector of the result.
Now that I think about it, that way should work and is much easier than my previous solution which I think is probably wrong, but it took ages to type so I'm leaving it there!
Title: Re: Perpendicular Vector
Post by: kamil9876 on April 25, 2010, 12:40:00 pm: let $u=OA, v=OB$

triangle OAB is isosceles (|u|=|v|), therefore the line segment that bisects angle AOB is is OM where M is the midpoint of AB.

Therefore OM=(u+v)/2
Title: Re: Perpendicular Vector
Post by: tram on April 25, 2010, 04:12:27 pm: just a question, would you be marked wrong if you did use the cross product to find the perpendicular vector? Like.... it's still a mathatically sound solution.
Title: Re: Perpendicular Vector
Post by: rainbows. on April 26, 2010, 05:08:53 pm: Hrmm sorry for the slow reply, thank you for the soultion !!
I'll ask the teacher soon if theres a more easier way of solving this :)

OA = 2i + 3j -k and OB = i - 2j + 3k
Find a unit vector parallel to BA.

This is a tricky question... anyone with the willingness to help :) ?
Title: Re: Perpendicular Vector
Post by: brightsky on April 26, 2010, 05:43:01 pm: $\vec{\mathbf{OA}} = 2i + 3j - k$

$\vec{\mathbf{OB}} = i - 2j + 3k$

Hence, $\vec{\mathbf{BA}} = \vec{\mathbf{BO}} + \vec{\mathbf{OA}} = -(i - 2j + 3k) + (2i + 3j - k) = -i + 2j - 3k + 2i + 3j - k = i + 5j - 4k$

Let $\mathbf{u} = ai + bj + ck$ be a vector parallel to $\vec{\mathbf{BA}}$ , where a, b, c are constants.

Using calculate the cross product $\mathbf{u} \times \vec{\mathbf{BA}}$ and set it to 0, then solve for a, b, c respectively.

Then use the result to find the unit vector $\hat{{\mathbf{u}}} = \frac{\mathbf{u}}{|\mathbf{u}|}$ .
Title: Re: Perpendicular Vector
Post by: rainbows. on April 26, 2010, 05:47:06 pm: "Using calculate the cross product " ?
Im sorry,stupid question, but using what? :/ :S
Title: Re: Perpendicular Vector
Post by: m@tty on April 26, 2010, 05:47:40 pm: $BA=-OB+OA=i+5j-4k$

$|BA|=\sqrt{1+25+16}=\sqrt{37}$

$\hat{BA}=\frac{1}{\sqrt{37}}\cdot (i+5j-4k)$

Or any vector of the form $\frac{1}{a\sqrt{37}}\cdot(ai+5aj-4ak), a \in \mathbb{R}\setminus\{0\}$

Will be parallel to BA, due to the ratio of components, and will be one unit in length due to being divided by its modulus.

I think...
Title: Re: Perpendicular Vector
Post by: rainbows. on April 26, 2010, 05:54:47 pm: Wait, wait, im confuse :S !

So the answer contain a constant now? o.o
Title: Re: Perpendicular Vector
Post by: m@tty on April 26, 2010, 06:05:09 pm: Wait, don't worry, the "a" cancels out :buck2: got confused...
Title: Re: Perpendicular Vector
Post by: tram on April 26, 2010, 09:06:19 pm: Quote from: rainbows. on April 26, 2010, 05:47:06 pm
"Using calculate the cross product " ?
Im sorry,stupid question, but using what? :/ :S

Lol don't worry, it is most definitely not a stupid question. Actually the question is aimed more at matty. Basically, the cross product is a way of finding a vector that is perpendicular to two other vectors, i.e. exactly what this question is asking. However, it is not a part of the spech(note spelling) course. It is taught in Uni maths, i was just interested if we were allowed to use it in spech(note spelling).
Title: Re: Perpendicular Vector
Post by: m@tty on April 26, 2010, 09:16:17 pm: Hmm, I don't know about Spech. In fact, what is Spech?

Though seriously, I don't know if we are allowed to use methods that are outside of the study design for Spesh.
Title: Re: Perpendicular Vector
Post by: tram on April 26, 2010, 09:20:02 pm: lol, spech stands for specialist maths matty. What does spesh stand for???;)

And yea.....i though as much......TBH i reackon if you get the right examiner, you'd get away with it, but some examiners would be like.....NO, THAT'S WRONG, beta not to take the chance
Title: Re: Perpendicular Vector
Post by: Mao on April 26, 2010, 09:45:58 pm: Quote from: theuncle on April 24, 2010, 05:01:33 pm
Quote from: rainbows. on April 24, 2010, 12:15:07 am
Find the unit vector which bisects the angel between a = 2i - j + 2k and b = 4i +3k. Hint: First find the unit vectors in the directions of the given vectors) >:S
Here's how I can think of doing it... there's probably a much easier way though ;)
let u=a's unit vector and v=b's unit vector and z= required vector
so $u=\dfrac{2}{3}i - \dfrac{1}{3}j + \dfrac{2}{3}k$ and $v = \dfrac{4}{5}i + \dfrac{3}{5}k$

let the angle between u and v = 2x
therefore, the angle between u and z = x
and the angle between v and z = x
$|u|=|v|=|x|=1$

$\dfrac{u.v}{1} = \dfrac{14}{15} = sin(2x)$ $\dfrac{u.z}{1} = sin(x)$ $\dfrac{v.z}{1} = sin(x)$
Say that $z= ai + bj + ck$
Solving $sin^{-1}(\dfrac{14}{15})=2x$ and $\dfrac{2}{3}a - \dfrac{1}{3}b + \dfrac{2}{3}c = sin(x)$ and $\dfrac{4}{5}a + \dfrac{3}{5}c = sin(x)$ and $a^2+b^2+c^2=1$ given that
We have 4 variables with 4 equations so it can be solved... (i used a calculator)
Two results were obtained because x has both an acute and obtuse solution, we only want the acute one
The exact solutions were hideous, so to two decimal places,
$z=-0.03i + 0.21j + 0.98k$ the other solution i think was the colinear obtuse solution,
as i said, i'm sure there must be an easier way, but this was the best that I could come up with!

My only other thoughts would be to average the i, j and k components of vectors a and b and find a unit vector of the result.
Now that I think about it, that way should work and is much easier than my previous solution which I think is probably wrong, but it took ages to type so I'm leaving it there!

If you draw it out, you'll see that $c = \hat{a} + \hat{b}$ bisects the angle between a and b. Thus, the unit vector of the angle bisector is $\hat{c} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$
Title: Re: Perpendicular Vector
Post by: AzureBlue on April 27, 2010, 09:33:14 pm: Quote from: tram on April 26, 2010, 09:20:02 pm
lol, spech stands for specialist maths matty. What does spesh stand for???;)
Ok then... spech = specialist maths and spesh = spestroscopy :)
Title: Re: Perpendicular Vector
Post by: tram on April 27, 2010, 10:22:52 pm: ^YESSSSS TOTALLY AGREED
Title: Re: Perpendicular Vector
Post by: Mao on April 28, 2010, 03:07:58 am: Stop this pointless crap on spesh vs spech¹, or else.

¹That is unless you find something even more inconsequential to get hyped up about, like 'does my arse look big in this?'. Yes it fucking does, you sound like a fucking child.
Title: Re: Perpendicular Vector
Post by: tram on April 28, 2010, 07:03:59 pm: point taken. I'll stop