= 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 \b[two and one, respectively].
The <dot product> and corresponding <Euclidean norm> are
$$
x\cdot y=\sum_{j=1}^n x_jy_j,\qquad \|x\|=\sqrt{x\cdot x}=\left(\sum_{j=1}^n x_j^2\right)^{1/2}.
$$
The <Cauchy-Schwarz inequality> states $\boxed{|x\cdot y|\le\|x\|\,\|y\|}$. If $y=0$ this is immediate. Otherwise the nonnegativity of a squared norm, with $t=(x\cdot y)/\|y\|^2$, gives
$$
0\le\|x-ty\|^2=\|x\|^2-\frac{(x\cdot y)^2}{\|y\|^2}.
$$
Multiplying by $\|y\|^2$ and taking square roots proves the inequality. Equality holds precisely when the vectors are linearly dependent. Expanding another squared norm and using the inequality gives
$$
\|x+y\|^2=\|x\|^2+2x\cdot y+\|y\|^2\le(\|x\|+\|y\|)^2,
$$
and hence the <triangle inequality> $\boxed{\|x+y\|\le\|x\|+\|y\|}$.
In $\mathbb R^3$, the two vectors lie in a common plane. If their directions make angle $\alpha$, the <law of cosines> gives $\|x-y\|^2=\|x\|^2+\|y\|^2-2\|x\|\|y\|\cos\alpha$. Comparing with the dot-product expansion yields
$$
\boxed{x\cdot y=\|x\|\,\|y\|\cos\alpha.}
$$
The 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 $\alpha$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ia/paper-1-dot-product-projection.png]
{title=Dot product as a signed projection in the plane of two vectors}
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.
Back to article page
