Recall that for \(\uvec{u} = (u_1,u_2,\dotsc,u_n)\) and \(\uvec{v} = (v_1,v_2,\dotsc,v_n)\text{,}\) the dot product of \(\uvec{u}\) and \(\uvec{v}\) is defined by the formula given below on the left. It is an important formula because if \(\theta\) is the angle between two nonzero vectors \(\uvec{u}\) and \(\uvec{v}\text{,}\) then \(\theta\) satisfies both \(0 \le \theta \le \pi\) and the formula given below on the right.
Based on the provided portion of the cosine graph, what can you say about \(\udotprod{u}{v} \) in the case that \(\theta \) is acute? … obtuse? … right?
A partial graph of \(y = \cos\theta \) on axes labelled \(\theta \) on the horizontal and \(y \) on the vertical. A half-period of the graph is displayed, beginning at the \(y \)-intercept at point \((0, 1) \text{,}\) continuing down through the \(\theta \)-intercept at point \((\pi / 2, 0) \text{,}\) and ending at the point \((\pi, -1) \text{.}\)
Extending the concept of perpendicular to higher dimensions, vectors \(\uvec{u}\) and \(\uvec{v}\) are called orthogonal if \(\udotprod{u}{v} = 0\text{.}\)
Can you guess a vector \(\uvec{v} = (v_1,v_2)\) that is orthogonal to \(\uvec{u} = (1,-3)\) in the plane? Make sure your guess satisfies the definition of orthogonal: you need \(\udotprod{u}{v} = 0\text{.}\)
Draw a diagram of your vectors \(\uvec{u}\) and \(\uvec{v}\text{,}\) both with initial points at the origin. On your diagram, how can you modify your intial guess \(\uvec{v}\) geometrically while still maintaining orthogonality with \(\uvec{u}\text{?}\)
Turn the pattern of your guess from Task a into a general pattern for vectors in the plane: if \(\uvec{u} = (a,b)\text{,}\) then an example of a vector orthogonal to \(\uvec{u}\) is
Draw the vector \(\uvec{a} = (3,1)\) in the \(xy\)-plane with its tail at the origin. Now imagine you were to also draw in every possible scalar multiple of \(\uvec{a}\) (positive, negative, zero, fractional, whatever). What geometric shape would these scalar multiples of \(\uvec{a}\) trace out? Draw this shape on your diagram.
Plot the point \(Q(4,4)\) on your diagram. On the line defined by \(\uvec{a}\) that you drew in the first part of this activity, draw in the point that you think is closest to \(Q\text{.}\) Label this point \(P\text{.}\) Now draw \(\abray{PQ}\text{,}\) and label this vector as \(\uvec{n}\text{.}\)
Vector \(\abray{OP}\) is parallel to \(\uvec{a}\text{,}\) so \(\abray{OP}\) is a scalar multiple of \(\uvec{a}\text{.}\) Our goal is to determine the scalar \(k\) so that the head of \(k\uvec{a}\) lies at \(P\text{.}\) Complete the triangle in your diagram by drawing in the vector \(\uvec{u} = \abray{OQ}\text{.}\)
Substitute your expression for \(\uvec{n}\) from Task c into your equation for \(\udotprod{n}{a}\) from Task b, and then solve for \(k\) as a formula in \(\uvec{u}\) and \(\uvec{a}\text{.}\)
(where in the brackets you should fill in a formula in the variable letters \(\uvec{u}\) and \(\uvec{a}\text{,}\)without using their actual numerical components, that describes how to compute \(k\uvec{a}\) from \(\uvec{u}\) and \(\uvec{a}\)).
The vector \(k\uvec{a}\) in Discovery 13.3 is called the orthogonal projection of \(\uvec{u}\) onto \(\uvec{a}\), and we write \(\uproj{u}{a}\) to mean this vector. It is also sometimes called the vector component of \(\uvec{u}\) parallel to \(\uvec{a}\). The vector \(\uvec{n} = \uvec{u}-\uproj{u}{a}\) is called the vector component of \(\uvec{u}\) orthogonal to \(\uvec{a}\).
Suppose \(\uvec{u}\) is orthogonal to \(\uvec{a}\text{.}\) What is \(\uproj{u}{a}\text{?}\) What is the component of \(\uvec{u}\) orthogonal to \(\uvec{a}\text{?}\)
If \(\ell\) is the line through the origin and parallel to a vector \(\uvec{a}\text{,}\) and \(\uvec{u}\) is some other vector, then our construction in Discovery 13.3 guarantees that \(\proj_{\uvec{a}} \uvec{u}\) represents the closest point on \(\ell\) to the terminal point of \(\uvec{u}\text{.}\)
The distance between a point and a line is defined as the shortest (so, perpendicular) distance between the two. Use the orthogonal projection to come up with a procedure to determine the distance between the line \(\ell\colon\,y=x/2\) and the point \(Q(2,4)\text{.}\)
Recall that a point \((x,y)\) lies on the line if and only if its coordinates satisfy the given equation. Let’s consider such a point as the terminal point of the vector \(\uvec{x} = (x,y)\) with its initial point at the origin. Does the left-hand side of the equation for the line look like the formula for some quantity related to \(\uvec{x}\) and some other vector? Perhaps some quantity that we’ve been exploring in detail recently?
In light of the first part of this activity, what does the right-hand side of the equation for the line say about the relationship between a vector \(\uvec{x}=(x,y)\) that lies along the line and the other special vector you identified in Task a?
Draw the line and label the point \(P(1,2)\text{.}\) Choose another arbitrary point on the line and label it \(Q(x,y)\text{.}\) Draw the vector \(\uvec{v} = \abray{PQ}\) along the line. Express the components of \(\uvec{v}\) as formulas in \(x\) and \(y\text{.}\)
Draw the vector \(\uvec{n} = (2,3)\) (from the coefficients in the line equation, just as in Discovery 13.6) with its tail at \(P\text{.}\) What do you notice about the relationship between this normal vector and the vector \(\uvec{v}\) parallel to the line? Express this relationship in terms of the dot product, and then expand out this dot product.
The same sort of analysis can be carried out for a plane in space determined by algebraic equation \(a x + b y + c z = d\text{.}\) The coefficients form a normal vector \(\uvec{n} = (a,b,c)\) that is perpendicular to the plane (that is, orthogonal to every vector that is parallel to the plane), and given some specific point \(\uvec{x}_0 = (x_0,y_0,z_0)\) that lies on the plane, the plane can be described by the point-normal form \(\dotprod{\uvec{n}}{(\uvec{x}-\uvec{x}_0)} = 0\text{.}\)
Consider the planes \(\Pi_1\text{,}\)\(\Pi_2\text{,}\) and \(\Pi_3\) described algebraically below.
\begin{align*}
\Pi_1 \amp \colon x - y + 2 z = 2 \amp
\Pi_2 \amp \colon 2 x - 2 y + 4 z = 7 \amp
\Pi_3 \amp \colon x - y + 3 z = 2
\end{align*}
Use the concept of normal vector to justify the claim that \(\Pi_1\) and \(\Pi_2\) are parallel, but that \(\Pi_3\) is not parallel to either of \(\Pi_1\) or \(\Pi_2\text{.}\)
Orthogonal projection onto a plane in space is a little more complicated, and is likely something you would learn about in a second course in linear algebra. But it’s possible to use a different strategy to determine the distance between a point and a plane by using the fact that a plane has one unique normal “direction.”
Come up with a procedure using vectors to determine the distance between parallel planes. Do not assume that either of the planes passes through the origin.
A three-dimensional diagram illustrating a procedure for determining the distance \(d\) between a point \(Q\) and a plane \(\Pi\) in three-dimensional space. A parallelogram with a shaded-in interior is drawn. The interior of this parallelogram should be imagined as if it is a two-dimensional, solid, rectangular surface suspended within a three-dimensional space (similar to a tabletop “suspended” above the floor in a room), but viewed at an angle from above. This surface is labelled as representing a portion of the plane \(\Pi\text{.}\) The point \(Q\) is plotted above this surface, external to it, and a dashed line segment is drawn from it down to meet the surface at a right angle. The point where the dashed line meets the surface is also plotted.
Another point is plotted on the surface and labelled \(R\text{.}\) The directed line segment \(\abray{RQ}\) is drawn and labelled as representing a vector \(\uvec{u}\text{.}\) Another directed line segment is drawn with its initial point at \(R\text{,}\) rising up out of the shaded surface at a right angle, and is labelled as representing a vector \(\uvec{n}\text{.}\) A longer, directed line segment with initial point at \(R\text{,}\) representing an unnamed vector, is drawn in parallel with \(\uvec{n}\text{,}\) rising to the same height above the shaded surface as \(Q\text{.}\) A dashed line segment is drawn from the terminal point of this vector to \(Q\text{,}\) and a parallel dashed line segment is drawn along the shaded surface from \(R\) to the point “underneath” \(Q\text{.}\) Together, these two dashed line segments, the dashed line segment from \(Q\) down to the plane, and the vector parallel to \(\uvec{n}\) up to the same height as \(Q\) form rectangle sitting on and rising vertically up from the shaded surface.
Finally, an arrow with a dotted line for the shaft points suggestively from the shaft of vector \(\uvec{u}\) to the shaft of the longer, unlabelled vector that is parallel to \(\uvec{n}\text{.}\)
