Draw the vector \(\uvec{u} = (3,2)\) in the \(xy\)-plane, then draw a representation of the decomposition \(\uvec{u} = 3 \uvec{e}_1 + 2 \uvec{e}_2\text{,}\) where \(\uvec{e}_1\) and \(\uvec{e}_2\) are the standard basis vectors in \(\R^2\text{.}\)
We will universally refer to the algebraic formulas in Task 13.1.d as the norm of the vector \(\uvec{v}\text{,}\) and denote this concept by \(\unorm{v}\text{.}\) For two- and three-dimensional vectors the norm of a vector computes its geometric length. In higher dimensions where we cannot visualize the geometry directly, we turn this around and instead define the geometric length of a vector to be the numeric result of computing the vector’s norm.
In this activity, make sure you can answer the questions for all dimensions, and make sure you can justify your answer using the formula for norm from Discovery 13.2, not just geometrically.
Plot points \(P(1,3)\) and \(Q(4,-1)\) in the \(xy\)-plane. Now draw in the vectors \(\uvec{u}\) and \(\uvec{v}\) that correspond to \(\abray{OP}\) and \(\abray{OQ}\text{.}\) Complete the triangle by drawing a vector between \(P\) and \(Q\text{.}\) Do you remember how to express this vector as a combination of \(\uvec{u}\) and \(\uvec{v}\text{?}\) Now compute the distance between \(P\) and \(Q\) by computing the norm of this third vector.
In the \(xy\)-plane, what is the angle between \(\uvec{e}_1\) and \(\uvec{e}_2\text{?}\) … between \(\uvec{e}_1\) and \(\uvec{u} = (1,1)\text{?}\) … between \(\uvec{e_1}\) and \(2\uvec{e}_1\text{?}\) … between \(\uvec{e_1}\) and \(-\uvec{e}_2\text{?}\) … between \(\uvec{e}_1\) and \(\uvec{v} = (1,-1)\text{?}\) … between \(\uvec{e_1}\) and \(-\uvec{e}_1\text{?}\)
Consider the vectors in the diagram to be two-dimensional. There is a version of Pythagoras that applies here even though \(\theta\) may not be a right angle, called the law of cosines:
\begin{equation}
a^2 + b^2 - c^2 = 2 a b \cos\theta \text{.}\tag{✶}
\end{equation}
A diagram to assist in exploring the law of cosines via vector geometry. Three directed line segments are arranged in a triangular configuration. One directed line segment representing a vector labelled \(\uvec{u} \) extends steeply upwards and slightly rightwards from a point in the lower-left corner to a point at the top of the diagram. A second directed line segment representing a vector labelled \(\uvec{v} \) extends rightwards and slightly upwards from the same initial point as \(\uvec{u} \) in the lower-left corner to a point at the right edge of the diagram. A third directed line segment representing an unnamed vector compeletes the triangle, extending from the terminal point of \(\uvec{v} \) to the terminal point of \(\uvec{u} \text{.}\) Finally, the angle formed by \(\uvec{u} \) and \(\uvec{v} \) in the lower-left corner is labelled \(\theta \text{.}\)
Applied to our diagram, \(a \) represents the length of \(\uvec{u} \text{,}\)\(b \) represents the length of \(\uvec{v} \text{,}\) and \(c \) represents the length of the “hypotenuse” \(\uvec{w} \text{.}\) (If \(\theta\)is\(\degree{90}\text{,}\) the right-hand side of this equality equals zero and this law “collapses” to the same equality as Pythagoras.)
The formula in the components of vectors \(\uvec{u} \) and \(\uvec{v} \) that fills in the last blank of Task c of Discovery 13.7 is a simple but important one. It is called the Euclidean inner product or standard inner product (or just simply the dot product) of \(\uvec{u}\) and \(\uvec{v}\text{,}\) and written
The importance of this formula is that it is a bridge between geometry and algebra. Replacing the left-hand side of the law of cosines (✶) with our final, simplified expression from the previous discovery activity and then dividing both sides by \(2 a b \text{,}\) we obtain
Let’s extend the computational pattern from Discovery 13.7. In the two-dimensional case in Task a below, you should just enter the formula for the dot product that you discovered in Task c of Discovery 13.7 (but without the “\(2 \) times”). In the subsequent tasks in higher dimensions, use the pattern from the two-dimensional case to create a similar higher-dimensional formula.
Suppose \(\uvec{u}\) and \(\uvec{v}\) are \(n\)-dimensional column vectors and \(A\) is an \(n\times n\) matrix. Use what you discovered in Task a to fill in the blank:
