Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 1 / 8D / Solution

Two nonzero vectors are linearly independent when $ax+by=0$ implies $a=b=0$. Their span then has dimension two. If they are linearly dependent, one is a scalar multiple of the other, so their span is a nonzero line and has dimension one. Thus the requested dimensions are two and one, respectively.

The dot product and corresponding Euclidean norm areThe Cauchy-Schwarz inequality states $\boxed{|x\cdot y|\le \Vert x\Vert \Vert y\Vert }$. If $y=0$ this is immediate. Otherwise the nonnegativity of a squared norm, with $t=(x\cdot y)/\Vert y{\Vert }^2$, givesMultiplying by $\Vert y{\Vert }^2$ and taking square roots proves the inequality. Equality holds precisely when the vectors are linearly dependent. Expanding another squared norm and using the inequality givesand hence the triangle inequality $\boxed{\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert }$.

In ${\mathbb{R}}^3$, the two vectors lie in a common plane. If their directions make angle $\alpha$, the law of cosines gives $\Vert x-y{\Vert }^2=\Vert x{\Vert }^2+\Vert y{\Vert }^2-2\Vert x\Vert \Vert y\Vert \cos \alpha$. Comparing with the dot-product expansion yieldsThe sketch shows the same relation as a projection: multiply the length of $x$ by the signed component of $y$ in its direction. That component is negative for obtuse $\alpha$.

A unit vector is a vector of Euclidean norm one, equivalently $u\cdot u=1$. In particular the dot product of two unit vectors is the cosine of their mutual angle.