On to 3 dimensions

Vectors in three dimensions behave just like vectors in the plane, except that they have three entries instead of two (and so are harder to draw). To the three coordinate axes are associated basic unit vectors $\bf i$, $\bf j$, and $\bf k$ of length $1$ in the direction of the $x$-axis, $y$-axis, and $z$-axis respectively. Then, the earlier figure relating the coordinates of a point $P(a,\,b,\,c)$ to the coordinate axes shows that the vector $a\,{\bf i} + b\,{\bf j} +c\,{\bf k}$ is equal to the displacement vector $\overrightarrow{OP}$. So a vector ${\bf v}$ in $3$-space can be represented by $${\bf v} \ = \ \langle\, a,\,b,\, c\,\rangle \ = \ a\,{\bf i} + b\,{\bf j} +c\,{\bf k} $$ The values of $a,\, b,$ and $c$ are called the components of $\bf v$. Addition, subtraction and scalar multiplication of vectors in 3-space then proceeds one variable at a time, just as before: $${\bf u} + {\bf v} = \langle a_1,\, b_1,\,c_1 \rangle + \langle a_2,\, b_2,\, c_2 \rangle = \langle a_1+a_2,\, b_1+b_2,\, c_1+c_2 \rangle\,,$$ $$\lambda {\bf w} = \lambda \langle c,\, d,\, e\rangle =\langle \lambda c,\, \lambda d,\, \lambda e \rangle\,.$$ Notice also that $${\bf i} = \langle 1,\,0,\,0\rangle, \qquad {\bf j} = \langle 0,\,1,\,0\rangle, \qquad {\bf k} = \langle 0,\,0,\,1\rangle, $$ $$\|a\,{\bf i} + b\,{\bf j} +c\,{\bf k}\| \ = \ \sqrt{a^2+b^2+c^2}\,.$$ By introducing coordinates into vectors, many algebraic, geometric and function-theoretic possibilities become available - that's the whole point!!