Proving 3 Non-Coplanar Vectors with Position Vector r

[kidsmoker]

Hi there, I'm a bit stuck on this question:

" Given 3 non-coplanar vectors a, b and c convince yourself that the position vector r of any point in space may be represented by

r = λa + μb + γc

for some real numbers λ, μ and γ.

Show that

r.(bxc) = λa.(bxc) ,

r.(axb) = γa.(bxc) ,

r.(cxa) = μa.(bxc) . "


I understand how they get the first one - the cross product of b and c is perpendicular to both of them, so won't contain any b or c components. Hence when you do the dot product you'll be multiplying the b and c bits of r by zero so they disappear. However I don't get the other two...?

Please help!

 

[tiny-tim]

r = λa + μb + γc

for some real numbers λ, μ and γ.

Show that

r.(bxc) = λa.(bxc) ,

r.(axb) = γa.(bxc) ,

r.(cxa) = μa.(bxc) . "

I understand how they get the first one - the cross product of b and c is perpendicular to both of them, so won't contain any b or c components. Hence when you do the dot product you'll be multiplying the b and c bits of r by zero so they disappear. However I don't get the other two...?

HI kidsmoker!

No, you're kidding yourself … you don't understand how they got the first one.

They got it by dot-producting both sides of r = λa + μb + γ with (bxc).

Now try dot-producting both sides of r = λa + μb + γ with (axb) instead …

what do you get?

 

[kidsmoker]

They got it by dot-producting both sides of r = λa + μb + γ with (bxc).

Now try dot-producting both sides of r = λa + μb + γ with (axb) instead …

I thought that's what I was doing? axb will be perpendicular to both a and b so won't contain a or b components? So when you dot it with r surely the a and b parts of r will be multiplied by zero?

Btw did you mean to put r = λa + μb + γc rather than r = λa + μb + γ or am I just confused?

Thanks.