How do I find a unit vector perpendicular to another vector?

My textbook only shows how to find the component of vector v perpendicular to vector w. But I don't have v and need to actually find it if that makes sense.
I know how to work out how to get a unit vector, that's fine.

I've basically been given 3 vectors a b c and have to find a unit vector that is perpendicular to b and to a - c. Another problem is I don't know if I have to find 1 unit vector perpendicular to those 2 given above, or 1 that is perpendicular to both simultaneously.

It's for an assignment, hence not posting the equations for the vectors a, b, c. (The equations are essentially arbitary - you guys can pick random 3D equations :smile:

)

Help much appreciated ^^

Reply 1

ghostwalker

Original post by PhysicsGal

...

I would expect it to be simultaneously.

If you've covered the vector product, then just taking the vector product of two vectors, gives a third vector that is perpendicular to both.

Reply 2

osmt1

You can work out a-c by taking each element in c from the corresponding element in a: so if a is 4i+5j+6k and c is i+2j+k, a-c = 3i+3j+5k
to find a vector perpendicular to both, you need to take the cross product. let's say b =2i+4j+6k. You can get the cross product by finding the determinant of
| i j k |
|3 3 5 | <- this is a-c written horizontally
|2 4 6 | <- this is b written horizontally
Then you find the unit version of the vector that gives you.
Hope this helps

(edited 10 years ago)

Reply 3

PhysicsGal

Original post by ghostwalker

I would expect it to be simultaneously.

If you've covered the vector product, then just taking the vector product of two vectors, gives a third vector that is perpendicular to both.

I used the cross product just now and realised that m x n = - (n x m).

So does this mean every pair of vectors has (atleast) 2 vectors perpendicular to it (which are both in opposite directions)?

Reply 4

PhysicsGal

Original post by osmt1

Thanks for your response, I did cross product the long winded way (although both seem long :P) where I do (a2b3 - a3b2)i etc...

Would I get 2 vectors perpendicular to the original pair? Since a x b = -b x a (or b x (-a)) ?

Reply 5

aznkid66

Algebraically, both methods are the same ^^

Mhmm, every pair of vectors that are not going in the same direction or going in opposite directions will have exactly 2 directions perpendicular to both, and thus exactly 2 directions for the cross.

I'd assume you'd only need to find 1 solution, not all possible solutions XD

Reply 6

PhysicsGal

Original post by aznkid66

Thank you

Yep, it's just asking for 1 solution, luckily! This is worth like 2 marks of a 25 mark assignment, that is worth about 5% of a maths unit..and still it's driving me crazy!

Thanks guys for your help :smile:

Reply 7

Hunarench95

Vector product is 0...

(a1B1 + a2B2 + a3B3)=0

Cos(x)=0 when x = 90 (perpendicular)

Reply 8

aznkid66

Original post by Hunarench95

Vector product is 0...

(a1B1 + a2B2 + a3B3)=0

Cos(x)=0 when x = 90 (perpendicular)

Are you doing dot product? If so, that'll only give you two equations--b . (x,y,z) and (a-c) . (x,y,z)--with variables x,y,z that, even if solved, would only resolve into the equation of a line. Of course, you can pick any ordered triple (x,y,z) on the line as the components of the vector, but the point is that attempting to use the dot product would be needlessly complicated!

Reply 9

PhysicsGal

Original post by ghostwalker

Original post by osmt1

Original post by aznkid66

Original post by Hunarench95

I have found the unit vector perpendicular to both the vectors in the OP. To check it was right, I tried to doing the dot product (twice, once with the unit vector dot each of the vectors in OP) and did not get 0 (zero) as a result :/

Where could I have gone wrong?

Reply 10

aznkid66

How will we know for certain if you don't show working? ^^;

First thing I would check would be the signs; make sure you did (a2b3 - a3b2)i - (a1b3 - b1a3)j + (a1b2-a2b1)k or something equivalent.

Also, check if the original cross (before you..."unitized" it) dots to 0.

Are you dotting it with b and a-c? Did you cross the correct vectors in the first place?

(edited 10 years ago)

Reply 11

PhysicsGal

Original post by aznkid66

Ahaha disaster averted, I did (-1)(-4) = - 5, instead of (-1)(-4) = 4 :colondollar:

I subtracted instead of multiplying :ahee:

Now it works out :colone:

Reply 12

Hunarench95

Original post by aznkid66

I agree that it would be complicated :wink:

but I'm reasonably sure that's the only way to tackle the question. Use dot product, to work out your gradient (making vectorA*VectorB = 0) without even necessarily using a scalar parameter, then after you have found your gradient, put a scalar parameter in front of the gradient, and write a coordinate of any point on the line.