As it hath been foretold, now we'll generalize the theory of vectors in 2- and 3-dimensional space to \(m\)-dimensional space for any positive integer \(m\).

*Definition*. *Euclidean* \(m\)-space \(\mathbf R^m\) is the collection of \(m\)-dimensional *points*, that is \(m\)-tuples of numbers \(P(x_1,\dotsc,x_m)\). The numbers \(x_i\) are called the *coordinates* of the point. A *vector in \(\mathbf R^m\) (\(m\)-vector)* can also be described as an \(m\)-tuple \(\mathbf v=(v_1,\dotsc,v_m)\). For a vector, the numbers \(x_i\) are called its *components*. The *zero vector* in \(\mathbf R^m\) is \(\boldsymbol0=(0,\dotsc,0)\). Two vectors \(\mathbf v,\mathbf w\) are equal, if all their components are equal: \(v_i=w_i,\,1\le i\le m\).

*Remarks*. (1) For typographical reasons I had to switch to horizontal notation when writing vectors by components inline. I still find column vectors more intuitive.

In \(\mathbf R^m\) too vectors can be defined as arrows where two are equal when they are translates.

With \(m\)-dimensional vectors too we adopt the notation that if I say that \(\mathbf v\) is a vector in \(\mathbf R^m\), then that implies that \(v_1,\dotsc,v_m\) denote its components.

*Definition*. Let \(P(p_1,\dotsc,p_m),Q(q_1,\dotsc,q_m)\) be two points in \(\mathbf R^m\). Then their *distance* is \[
d(P,Q)=\sqrt{(p_1-q_1)^2+\dotsb+(p_m-q_m)^2}.
\] Let \(\mathbf v=(v_1,\dotsc,v_m)\) be a vector in \(\mathbf R^m\). Then its *length (magnitude, norm)* is \[
\|\mathbf v\|=\sqrt{v_1^2+\dotsb+v_m^2}.
\] A vector in \(\mathbf R^m\) is a *unit vector*, if it has length 1.

*Remark*. We say that \(\mathbf R^m\) is *Euclidean* \(m\)-space, because it is equipped with the above distance function. As another example (which of course is not covered in this class), special relativity theory is formulated in *Minkowski space*. That is \(\mathbf R^4\), but the distance function is different: \[
d(P(x_1,y_1,z_1,t_1),Q(x_2,y_2,z_2,t_2))=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2-(t_1-t_2)^2}.
\] Note that unlike Euclidean space, it is possible to get negative distances in Minkowski space. Vectors with negative length are called *timelike*, and vectors with positive length are called *spacelike*.

*Exercise*. Out of the following vectors, which is a unit vector? \[
\left(\begin{smallmatrix}2\\0\\-1\\-1\end{smallmatrix}\right),\,\left(\begin{smallmatrix}0\\1/3\\0\\1/3\\0\\1/3\end{smallmatrix}\right),\,\left(\begin{smallmatrix}2/3\\0\\2/3\\0\\1/3\end{smallmatrix}\right),\,\left(\begin{smallmatrix}-1/2\\0\\1/2\end{smallmatrix}\right).
\]

*Definition*. Let \(c\) be a scalar, and \(v\) an \(m\)-vector. Then the *scalar multiple* is \(c\mathbf v=(cv_1,\dotsc,cv_m)\). Two \(m\)-vectors \(\mathbf u,\mathbf v\) are *parallel (collinear)* if one is a scalar multiple of the other. Suppose that the \(\mathbf u,\mathbf v\) are parallel nonzero vectors. Then there exists a scalar \(c\) such that \(c\mathbf u=\mathbf v\). If \(c>0\) then we say *\(\mathbf u\) and \(\mathbf v\) have the same direction*, and if \(c<0\) then we say *they have the opposite direction*.

*Theorem*. Let \(c\) be a scalar, and \(v\) an \(m\)-vector. Then we have \[
\|c\mathbf v\|=|c|\cdot\|\mathbf v\|.
\]

*Exercise*. Find the unit vector with the opposite direction of \((9,-3,0,12)\).

*Definition*. Let \(\mathbf u,\mathbf v\) be \(m\)-vectors. Then their *vector sum* is \(\mathbf u+\mathbf v=(u_1+v_1,\dotsb,u_m+v_m)\). Their *vector difference* is \(\mathbf u-\mathbf v=(u_1-v_1,\dotsc,u_m-v_m)\). Their *dot product* is \(\mathbf u\cdot\mathbf v=u_1v_1+\dotsb+u_mv_m\). We say that \(\mathbf u\) and \(\mathbf v\) are *orthogonal*, if we have \(\mathbf u\cdot\mathbf v=0\).

*Remark*. As we have said before, if \(m\ne3\), then there is no cross product for \(m\)-vectors.

*Exercise* Find \(k\) so that the vectors \(2(1,3,0,2)-(0,-1,2,0)\) and \((a,2,0,1)\) are orthogonal.

*Exercise*. Let \(\ell\) be the line through \(P(0,3,-1,2)\) and \(Q(1,-3,0,1)\). Write down a two-point form equation, a point-parallel form equation, and parametric equations for it.

*Definition*. Let \(P\) be a point in \(\mathbf R^m\), and \(\mathbf n\) an \(m\)-vector. Then the *hyperplane through \(P\) with normal vector \(\mathbf n\)* is the collection of points \(Q(x_1,\dotsc,x_m)\) the coordinates of which satisfy the *point-normal equation* \[
\mathbf n\cdot(\mathbf x-\mathbf p)=0,
\] or the *standard form equation* \[
n_1x_1+\dotsb+n_mx_m=c,
\] where \(c=\mathbf n\cdot\mathbf p\).

*Exercise*. Find a normal vector for the hyperplane with standard form equation \[
3x_1+x_3-2x_5=15.
\] Find a point on this hyperplane.