Proving Vectors are collinear under certain conditions

Paul616 · #1 $Old$ October 25th, 2009, 05:09 AM

Three points X, Y, Z have position vectors x,y,z. Show that X, Y and Z are collinear iff x ^ y + y ^ z + z ^ x = 0.

Here "^" denotes the vector cross product

Opalg · #2 $Old$ October 25th, 2009, 02:15 PM

Quote:

Originally Posted by Paul616 $View Post$

Three points X, Y, Z have position vectors x,y,z. Show that X, Y and Z are collinear iff x ^ y + y ^ z + z ^ x = 0.

Here "^" denotes the vector cross product

If X, Y and Z are collinear then $z=\lambda x + (1-\lambda)y$ for some scalar $\lambda$ . You can then verify that $y\times z + z\times x + x\times y=0$ (remembering that the cross product of a vector with itself is always 0).

For the converse, if $y\times z + z\times x + x\times y=0$ then $0 = x.(y\times z) + x.(z\times x) + x.(x\times y) = x.(y\times z)$ (since the other two terms are 0). So $y\times z$ is orthogonal to x. But it is also orthogonal to y and z. If x, y and z are linearly independent then $y\times z$ is orthogonal to the whole space and is therefore 0. But that would mean that y and z are not linearly independent. That contradiction shows that the three vectors must be linearly dependent.

So one of them, z say, is a linear combination of the others, $z = \lambda x + \mu y$ . Substitute that value for z into the equation $y\times z + z\times x + x\times y=0$ and you will find that $\mu = 1-\lambda$ , which is the condition for X, Y and Z to be collinear.