Recall that for and , the dot product of and is defined by the formula given below on the left. It is an important formula because if is the angle between two nonzero vectors and , then satisfies both and the formula given below on the right.
Discovery13.1.
Based on the graph of on domain provided below, what can you say about in the case that is acute? … obtuse? … right?
Graph of .
Extending the concept of perpendicular to higher dimensions, vectors and are called orthogonal if .
Discovery13.2.
(a)
Can you guess a vector that is orthogonal to in the plane? Make sure your guess satisfies the definition of orthogonal: you need .
(b)
What relationship to your initial guess will other vectors in the plane that are orthogonal to have?
Hint.
Draw a diagram of your vectors and , both with initial points at the origin. On your diagram, how can you modify your intial guess geometrically while still maintaining orthogonality with ?
(c)
Turn the pattern of your guess from Task a into a general pattern for vectors in the plane: if , then an example of a vector orthogonal to is
.
Discovery13.3.
(a)
Draw the vector in the -plane with its tail at the origin. Now imagine you were to also draw in every possible scalar multiple of (positive, negative, zero, fractional, etc.). What geometric shape would these scalar multiples of trace out? Draw this shape on your diagram.
(b)
Plot the point on your diagram. On the line defined by that you drew in the first part of this activity, draw in the point that you think is closest to . Label this point . Now draw , and label this vector as .
What is the relationship between and the line? What is the value of ?
(c)
Vector is parallel to , so is a scalar multiple of . Our goal is to determine the scalar so that the head of lies at . Complete the triangle in your diagram by drawing in the vector .
Then express as a combination of and .
(d)
Substitute your expression for from Task c into your equation for from Task b, and then solve for as a formula in and .
Now complete the general formula:
(where in the brackets you should fill in a formula in the variable letters and ,without using their actual numerical components, that describes how to compute from and ).
The vector in Discovery 13.3 is called the orthogonal projection of onto , and we write to mean this vector. It is also sometimes called the vector component of parallel to . The vector is called the vector component of orthogonal to .
The same problem can be solved in higher dimensions by the same formula for .
Discovery13.4.
(a)
Suppose is orthogonal to . What is ? What is the component of orthogonal to ?
(b)
Answer the same two questions in the case that is parallel to .
Discovery13.5.
If is the line through the origin and parallel to a vector , and is some other vector, then our construction in Discovery 13.3 guarantees that represents the closest point on to the terminal point of .
The distance between a point and a line is defined as the shortest (i.e. perpendicular) distance between the two. Use the orthogonal projection to come up with a procedure to determine the distance between the line and the point .
Discovery13.6.
The homogeneous linear equation defines a line through the origin in (i.e. the -plane).
(a)
Recall that a point 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 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 and some other vector? Perhaps some quantity that we’ve been exploring in detail recently?
(b)
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 that lies along the line and the other special vector you identified in the previous part?
Discovery13.7.
The nonhomogeneous linear equation defines a line through the point in .
(a)
Draw the line and label the point . Choose another arbitrary point on the line and label it . Draw the vector along the line. Express the components of as formulas in and .
(b)
Draw the vector (from the coefficients in the line equation, just as in Discovery 13.6) with its tail at . What do you notice about the relationship between this normal vector and the vector 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 . The coefficients form a normal vector that is perpendicular to the plane (i.e. orthogonal to every vector that is parallel to the plane), and given some specific point that lies on the plane, the plane can be described by the point-normal form .
Discovery13.8.
Consider the planes ,, and described algebraically below.
Use the concept of normal vector to justify the claim that and are parallel, but that is not parallel to either of or .
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.”
Discovery13.9.
(a)
Using the diagram below as inspiration, come up with a procedure to determine the distance between a point and a plane .
Hint.
Determine a vector that represents an equivalent distance, and then will be the norm of this vector.
(b)
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.
Hint.
Find a way to reduce this problem to the problem in the first part of this activity.
Diagram illustrating procedure to determine the distance between a point and a plane in space.