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May 06, 2026, 03:45:39 pm

Author Topic: Perpendicular Vector  (Read 3375 times)  Share 

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VeryCrazyEdu.

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Perpendicular Vector
« on: April 21, 2010, 09:12:59 pm »
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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 :)

kamil9876

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Re: Perpendicular Vector
« Reply #1 on: April 21, 2010, 09:44:25 pm »
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Suppose your two vectors are and .

then by the dot product you want to solve:





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)
« Last Edit: April 21, 2010, 09:59:23 pm by kamil9876 »
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

VeryCrazyEdu.

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Re: Perpendicular Vector
« Reply #2 on: April 21, 2010, 09:56:15 pm »
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oh haha thats so easy compared to what i was thinking! Thanks heaps :)

rainbows.

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Re: Perpendicular Vector
« Reply #3 on: April 24, 2010, 12:15:07 am »
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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 :)

theuncle

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Re: Perpendicular Vector
« Reply #4 on: April 24, 2010, 05:01:33 pm »
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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 and

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


              
Say that
Solving and and and 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,
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!

kamil9876

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Re: Perpendicular Vector
« Reply #5 on: April 25, 2010, 12:40:00 pm »
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let

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
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

tram

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Re: Perpendicular Vector
« Reply #6 on: April 25, 2010, 04:12:27 pm »
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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.

rainbows.

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Re: Perpendicular Vector
« Reply #7 on: April 26, 2010, 05:08:53 pm »
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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 :) ?

brightsky

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Re: Perpendicular Vector
« Reply #8 on: April 26, 2010, 05:43:01 pm »
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Hence,

Let be a vector parallel to , where a, b, c are constants.

Using calculate the cross product and set it to 0, then solve for a, b, c respectively.

Then use the result to find the unit vector .

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rainbows.

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Re: Perpendicular Vector
« Reply #9 on: April 26, 2010, 05:47:06 pm »
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"Using calculate the cross product " ?
Im sorry,stupid question, but using what? :/ :S

m@tty

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Re: Perpendicular Vector
« Reply #10 on: April 26, 2010, 05:47:40 pm »
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Or any vector of the form

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...
« Last Edit: April 26, 2010, 06:06:03 pm by m@tty »
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rainbows.

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Re: Perpendicular Vector
« Reply #11 on: April 26, 2010, 05:54:47 pm »
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Wait, wait, im confuse :S !

So the answer contain a constant now? o.o

m@tty

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Re: Perpendicular Vector
« Reply #12 on: April 26, 2010, 06:05:09 pm »
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Wait, don't worry, the "a" cancels out :buck2: got confused...
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tram

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Re: Perpendicular Vector
« Reply #13 on: April 26, 2010, 09:06:19 pm »
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"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).

m@tty

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Re: Perpendicular Vector
« Reply #14 on: April 26, 2010, 09:16:17 pm »
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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.
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