[Mr Maths]

Find a unit vector ai + bj + ck which is perpendicular to the two vectors 2i + 2j -k and 4i + 2k

I understand for two vectors to be perpendicular their dot product must be 0, provided they are both non zero vectors.

So I worked out the dot product for the vector I am trying to find, and the two vectors given.

2a + 2b - c = 0 and 4a + 2c = 0

Is this right so far? I don't think there is a way to solve them like this

[James94]

Find the cross product of the two vectors, then divide each coefficient of i, j and k by a constant, such that it's magnitude is 1. If you don't know how to do this yet then I'll explain a little more, but you won't learn anything if I just give you the answer haha.

[mathz]

for this you need to remember there wont be a unique solution as if (x,y,z) is perp then so will k(x,y,z)

so if you could write

a=kb
b=b
c=lb

for some k,l then your vector would be b(k,1,l) and then choose b so its a unit vector.

[Mr Maths]

(Original post by James94)
Find the cross product of the two vectors, then divide each coefficient of i, j and k by a constant, such that it's magnitude is 1. If you don't know how to do this yet then I'll explain a little more, but you won't learn anything if I just give you the answer haha.

So you find the point of intersection of the two vectors given, is that what you mean by cross product? I know what a unit vector is, its a vector of magnitude 1. So you divide each component of the vector by the magnitude of the initial vector.

[WhiteGroupMaths]

Cross product is meant to help you arrive at a vector which is perpendicular to the original constituents.

ie a cross b produces a vector n; this vector n is perpendicular to both a and b.

Cross product is not solving for intersections whatsoever.

Hope this helps. Peace.

[Implication]

Just thought I'd add that the cross product is not required for Edexcel C4.


Also, remember that it isn't the exact values of what you've called a, b and c you are actually interested in - it's their relative values. You can make up any number and use it for a, b or c (it's usually easiest to use 1) and then solve your two simultaneous equations to find the other two.

[bkhan]

What would be the best way for revising for vectors. I find them the hardest in C4
Thanks

[TheDefiniteArticle]

(Original post by bkhan)
What would be the best way for revising for vectors. I find them the hardest in C4
Thanks

Do lots and lots and lots of questions - they don't deviate too much from the norm.

[Rational Paradox]

You can't find a specific solution as there are more variables than equations,but you can write each in terms of the others: e.g.

a + b + c = 0
2a + c = 0

so c = -2a, and b = a, so you have a vector (a, a, -2a), and then just let a be a constant such that the vector (a, a, -2a) has length is 1