Section2.3Lines and planes in \(\mathbb{R}^2\) and \(\mathbb{R}^3\)
In this section we apply the tools we developed in the previous two sections to the study of lines and planes. We start by describing two different ways of describing lines in \(\mathbb{R}^2\text{,}\) and then see that those methods generalize differently to lines and planes in \(\mathbb{R}^3\text{.}\)
Subsection2.3.1Lines in \(\mathbb{R}^2\)
In previous mathematics courses you have seen that most lines in \(\mathbb{R}^2\) can be expressed by equations of the form \(y = ax+b\text{,}\) where \(a\) is the slope of the line and \(b\) is the \(y\)-intercept of the line. One downside to this kind of equation is that not every line can be expressed in this way; specifically, vertical lines do not have equations of this form. By slightly rearranging we get another form that addresses this problem.
Definition2.3.1.
The general form equation of a line \(L\) is an equation of the form
\begin{equation*}
ax+by=c\text{,}
\end{equation*}
where \(a\text{,}\)\(b\text{,}\) and \(c\) are constants, such that the points on \(L\) are exactly the points \((x, y)\) that satisfy the equation.
Let \(L\) be the line in \(\mathbb{R}^2\) that passes through the points \((1, 3)\) and \((-1, 7)\text{.}\) The equation of this line in point-intercept form is \(y=-2x+5\text{.}\) Rearranging this we get the general form equation \(2x+y=5\text{.}\)
Notice that we could have rearranged the equation a bit differently, and obtained \(-2x-y=-5\text{.}\) This is also a general form equation for the same line \(L\text{,}\) because the solutions to this equation are the same as the solutions to the equation \(2x+y=5\text{.}\) In fact, if we multiply both sides of our equation by any non-zero constant we will get another general form equation for the same line. Thus, for instance, \(8x+4y=20\) is another general equation for \(L\text{.}\)
We now turn to interpreting the general form equation using vectors. Given a general equation \(ax+by=c\text{,}\) let \(\vec{n} = \begin{bmatrix}a\\b\end{bmatrix}\) and \(\vec{v} = \begin{bmatrix}x\\y\end{bmatrix}\text{.}\) Then we see that the general equation \(ax+by=c\) is equivalent to \(\vec{n} \cdot \vec{v} = c\text{.}\) In order to explain what the number \(c\) means, it is useful to draw a picture and consider an example.
We once again consider the line \(L\) with general form equation \(2x+y=5\text{.}\) Let \(\vec{n} = \begin{bmatrix}2\\1\end{bmatrix}\text{,}\)\(\vec{v} = \begin{bmatrix}x\\y\end{bmatrix}\text{,}\) and \(\vec{p} = \begin{bmatrix}1\\3\end{bmatrix}\text{.}\)
Figure2.3.5.The line \(2x+y=5\) with the vectors \(\vec{n}, \vec{v}, \vec{p}\text{.}\)
We can see from the image that \(\vec{n}\) is orthogonal to the direction of the line \(L\text{.}\) Moreover, if \((x, y)\) is a point on \(L\) then \(\vec{v} - \vec{p}\) is a vector in the direction of \(L\text{.}\) Thus \(\vec{v}-\vec{p} \perp \vec{n}\text{,}\) meaning \((\vec{v} - \vec{p})\cdot \vec{n} = 0\text{.}\) Expanding this and rearranging we get \(\vec{n}\cdot\vec{v} = \vec{n}\cdot\vec{p}\text{.}\) The left side expands to \(2x+y\text{,}\) and the right side expands to \(5\text{.}\)
As an exercise, take a moment to verify that if you take \(\vec{p}\) to be the vector to another point on the line (instead of \((1, 3)\)) then it is still true that \(\vec{p}\cdot\vec{n} = 5\text{.}\)
The phenomenon of the example works for every line: Given \(ax+by=c\text{,}\) the vector \(\vec{n} = \begin{bmatrix}a\\b\end{bmatrix}\) is orthogonal to the direction of the line, and if \(\vec{p}\) is the vector from the origin to any fixed point on the line, then \(\vec{n}\cdot\vec{p} = c\text{.}\)
Definition2.3.6.
Let \(L\) be a line in \(\mathbb{R}^2\text{.}\) Let \(\vec{n}\) be a vector orthogonal to the direction of the line, and let \(\vec{p}\) be the vector from the origin to any fixed point on the line. Let \(\vec{v} = \begin{bmatrix}x\\y\end{bmatrix}\text{.}\) The normal form equation of the line \(L\) is the equation
The vector \(\vec{n}\) is called the normal vector to the line.
It is important to notice that the general form equation and the normal form equation are fundamentally the same equation, just represented in different ways. In particular, the normal vector to the line \(ax+by=c\) is \(\vec{n} = \begin{bmatrix}a\\b\end{bmatrix}\text{,}\) and the number \(c\) can be found by taking the dot product of the normal vector with any point on the line.
We now turn to a different way of describing a line in \(\mathbb{R}^2\text{.}\) Instead of describing a line by giving an equation that the points on the line must satisfy, we instead describe how to create the line by starting at a particular point and then following the direction of the line.
Definition2.3.7.
Let \(L\) be a line in \(\mathbb{R}^2\text{.}\) Let \(\vec{d}\) be a vector such that if \(P\) is any point on \(L\) then starting from \(P\) and following any multiple of \(\vec{d}\) results in another point on \(L\) (such a vector is called a direction vector for the line). Let \(\vec{p}\) be a vector which, in standard position, ends on \(L\text{.}\) Let \(\vec{v} = \begin{bmatrix}x\\y\end{bmatrix}\text{.}\) The vector form equation of \(L\) is the equation
The variable \(t\) appearing in the vector equation is a parameter. Given any value of \(t\text{,}\) we can plug that value in to \(\vec{p}+t\vec{d}\) to get the coordinates of a point on the line. Conversely, given any point \(\vec{q}\) on the line, there is a unique value of \(t\) such that \(\vec{q} = \vec{p}+t\vec{d}\text{.}\) We may use any variable, not just \(t\text{,}\) as the name of the parameter.
We continue again with the line \(L\) passing through \(P = (1, 3)\) and \(Q = (-1, 7)\text{.}\) A direction vector for this line is \(\vec{d} = \vec{PQ} = \begin{bmatrix}-2\\4\end{bmatrix}\text{.}\) We could take any multiple of this vector as a direction vector instead, so let us choose to use \(\vec{d} = \begin{bmatrix}1\\-2\end{bmatrix}\text{.}\) A vector form equation of \(L\) is:
If we choose \(t=3\text{,}\) we find that \(\begin{bmatrix}1\\3\end{bmatrix} + 3\begin{bmatrix}1\\-2\end{bmatrix} = \begin{bmatrix}4\\-3\end{bmatrix}\text{,}\) so we find that \((4, -3)\) is a point on \(L\text{.}\)
We already know that \((-1, 7)\) is a point on the line. Therefore there should be a number \(t\) such that \(\begin{bmatrix}-1\\7\end{bmatrix} = \begin{bmatrix}1\\3\end{bmatrix} + t\begin{bmatrix}1\\-2\end{bmatrix}\text{.}\) By expanding this vector equation into a system of two linear equations we can find that \(t=-2\text{.}\)
The vector equation we found is not the only one. We can choose any point on the line, so another perfectly good vector equation for this line is:
Given a vector equation \(\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}p_1\\p_2\end{bmatrix} + t\begin{bmatrix}d_1\\d_2\end{bmatrix}\) we can expand it into two equations:
\begin{gather*}
x = p_1+td_1\\
y = p_2+td_2
\end{gather*}
These equations are called a parametric form equation of the line. Since the parametric form and the vector form are so closely related, we will usually not distinguish between them.
Subsection2.3.2Lines in \(\mathbb{R}^3\)
In \(\mathbb{R}^2\) the only "flat" objects are straight lines. When we move up to \(\mathbb{R}^3\) we have two different kinds of "flat" objects, namely lines and planes. We start with lines, and will explore planes in the next section.
It is easiest to start with the vector form equations. A line in \(\mathbb{R}^3\) can be described by specifying a point on the line and a direction vector for the line, just like we did in \(\mathbb{R}^2\text{.}\)
Definition2.3.9.
Let \(L\) be a line in \(\mathbb{R}^3\text{.}\) Let \(\vec{v} = \begin{bmatrix}x\\y\\z\end{bmatrix}\text{,}\) let \(\vec{p}\) be a vector to a point on \(L\text{,}\) and let \(\vec{d}\) be a direction vector for the line. A vector form equation of the line \(L\) is the equation
Let \(L\) be the line in \(\mathbb{R}^3\) passing through \(P = (1, 2, 3)\) and \(Q = (-1, 0, 1)\text{.}\) A direction vector for \(L\) is \(\vec{d} = \vec{PQ} = \begin{bmatrix}-2\\-2\\-2\end{bmatrix}\text{.}\) We can use any multiple of this as a direction vector, so we choose to use \(\vec{d} = \begin{bmatrix}1\\1\\1\end{bmatrix}\text{.}\) A vector equation for \(L\) is thus:
As in \(\mathbb{R}^2\text{,}\) expanding out the vector equation into three linear equations gives the parametric form.
Our next goal is to describe the general form and normal form equations for a line. This turns out to be more complicated in \(\mathbb{R}^3\) than it is in \(\mathbb{R}^2\text{,}\) because in \(\mathbb{R}^3\) there are infinitely many different directions that are orthogonal to any fixed line. At any particular point the vectors that are orthogonal to a line form a plane, so to give a normal form equation for the line we will need to give enough normal vectors to describe this plane. Intuitively, you might expect that planes are ``two-dimensional", and hence that two (non-parallel) vectors should be sufficient. This is true (but it will be some time before we are ready to make it precise).
Definition2.3.11.
The normal equations for a line in \(\mathbb{R}^3\) are a system of two equations,
Here \(\vec{v} = \begin{bmatrix}x\\y\\z\end{bmatrix}\text{,}\)\(\vec{p}\) is a vector to any chosen point on the line, and \(\vec{n_1}\) and \(\vec{n_2}\) are non-parallel vectors each of which is orthogonal to the line.
If we expand the dot products we obtain a system of two linear equations, which is referred to as the general form equations of the line.
Let us find normal and general equations for the line from the previous example, which is the line passing through \(P = (1, 2, 3)\) and \(Q = (-1, 0, 1)\text{.}\) A direction vector for this line is \(\vec{d} = \begin{bmatrix}1\\1\\1\end{bmatrix}\text{.}\) We need to find two normal vectors, that is, two non-parallel, non-zero vectors \(\vec{n_1}\) and \(\vec{n_2}\) such that \(\vec{n_1}\cdot\vec{d} = 0 = \vec{n_2}\cdot\vec{d}\text{.}\) There are several ways we could do this:
By inspection, i.e., by guess-and-check.
By inspection for \(\vec{n_1}\text{,}\) and then using a cross product to find \(\vec{n_2}\text{.}\)
By setting up a system of equations.
The first two methods both rely on some guesswork, and the second one uses a technique (the cross product) that will not be part of this course, so we illustrate the third (and most systematic) method.
Let \(\vec{n_1} = \begin{bmatrix}a\\b\\c\end{bmatrix}\text{.}\) The equation \(\vec{n_1}\cdot\vec{d} = 0\) in this example expands to \(a+b+c=0\text{.}\) If we rearrange this a bit, we get \(a = -b-c\text{,}\) so that:
We now see that we can choose any values of \(b\) and \(c\) that we like, and we will be able to get a normal vector for the line. To ensure that we get two non-parallel normal vectors we will get \(\vec{n_1}\) by choosing \(b=1, c=0\) and we will get \(\vec{n_2}\) by choosing \(b=0, c=1\text{.}\) We thus get:
Now we need \(\vec{p}\text{,}\) for which we can take any point on the line. For this example we will choose \(\vec{p} = \begin{bmatrix}1\\2\\3\end{bmatrix}\text{,}\) but any other point on the line would also be fine. We are now prepared to write down normal form equations of our line:
To turn this into general form equations we simply expand the dot products. We calculate \(\vec{n_1}\cdot\vec{p} = 1\) and \(\vec{n_2}\cdot\vec{p} = 2\text{.}\) The general form equations are:
In this example we will reverse what we did in example Example 2.3.12. Suppose that we are told that a line has general equations
\begin{gather*}
-x+y=1 \\
-x+z=2
\end{gather*}
and we wish to find a vector equation.
We notice that the general equations are a system of two linear equations in three variables. If we set these up in an augmented matrix and row reduce, we have:
This is already a vector equation, but if we want to make it look more like the forms we have seen before we could choose to introduce a parameter \(t\text{,}\) set \(z=t\text{,}\) and obtain:
It is worth noticing that we obtained the same direction vector that we found in Example 2.3.12. This time our equation used the point \(\begin{bmatrix}-2\\-1\\0\end{bmatrix}\text{,}\) but since the vector equation can use any point on the line this is not a problem.
In summary, a line in \(\mathbb{R}^3\) is described by a vector equation with one direction vector, or by two normal or general form equations, and we have seen how to translate between these forms.
Subsection2.3.3Planes in \(\mathbb{R}^3\)
We now turn to planes in \(\mathbb{R}^3\text{.}\) While a line in \(\mathbb{R}^3\) can be described by either one direction vector or two normal vectors, describing a plane requires two direction vectors or one normal vector. You might notice that for both lines and planes the number of direction vectors plus the number of normal vectors is \(3\text{,}\) and we are working in \(\mathbb{R}^3\text{.}\) This is certainly not an accident, but we will have to wait until Chapter 6 for a full explanation.
Definition2.3.14.
The normal equation of a plane in \(\mathbb{R}^3\) is an equation of the form
Here \(\vec{v} = \begin{bmatrix}x\\y\\z\end{bmatrix}\text{,}\)\(\vec{p}\) is a vector to any chosen point on the plane, and \(\vec{n}\) is a vector orthogonal to the plane.
If we expand the dot product in the normal equation of a plane we obtain the general equation of the plane.
Definition2.3.15.
The vector equation of a plane in \(\mathbb{R}^3\) is an equation of the form
Here \(\vec{v} = \begin{bmatrix}x\\y\\z\end{bmatrix}\text{,}\)\(\vec{p}\) is a vector to any chosen point on the plane, and \(\vec{d_1}\) and \(\vec{d_2}\) are non-parallel direction vectors for the plane.
The three equations obtained by looking at the \(x, y, z\) components of the vector equation are the parametric equations for the plane.
Consider the plane passing through the points \(P = (1, 2, 1)\text{,}\)\(Q = (0, 0, 1)\text{,}\) and \(R = (-1, 0, -1)\text{.}\) Two direction vectors for this plane are \(\vec{QP} = \begin{bmatrix}1\\2\\0\end{bmatrix}\) and \(\vec{QR} = \begin{bmatrix}-1\\0\\-2\end{bmatrix}\text{.}\) Thus a vector equation for the plane is
As with the other examples in this chapter, this is not the only possible vector equation for this plane.
To obtain a normal equation we need to find a normal vector for the plane, that is, a vector that is orthogonal to both of the two direction vectors we have. If you know about the cross product you can take \(\vec{n} = \vec{QP}\times\vec{QR}\text{.}\) Alternatively, if \(\vec{n} = \begin{bmatrix}a\\b\\c\end{bmatrix}\) then we need \(0 = \vec{n}\cdot\vec{QP} = a+2b\) and \(0 = \vec{n}\cdot\vec{QR} = -a-2c\text{,}\) yielding a system of equations we can solve. Regardless of which method we use to find \(\vec{n}\text{,}\) the only possible choices for \(\vec{n}\) are multiples of \(\vec{n} = \begin{bmatrix}2\\-1\\-1\end{bmatrix}\text{.}\) We then choose \(\vec{p} = \begin{bmatrix}1\\2\\1\end{bmatrix}\) (but any other point on the plane would also be fine), to get the normal equation
You can verify that this equation is correct by checking that the points \(P, Q, R\) all satisfy the equation.
As an illustration of the techniques we have developed so far, we end this section with an example of finding the shortest distance from a point to a plane. We will solve the next example in two ways, because the two techniques are sometimes useful in different circumstances.
The key observation is that the shortest distance from \(B\) to our plane will be made at a point \(C\) on the plane where the vector \(\vec{BC}\) is normal to the plane. We begin by finding any point on the plane, that is, any point that satisfies the equation \(x-y-2z=1\text{;}\) the point \(A = (1, 0, 0)\) will do. Now consider the vector \(\vec{AB} = \begin{bmatrix}0\\1\\1\end{bmatrix}\text{.}\) This vector goes from the plane to \(B\text{,}\) but it isn't normal to the plane. What we're interested in is the component of \(\vec{AB}\) along the direction normal to the plane. That is, the distance we want is \(\norm{\operatorname{proj}_{\vec{n}}(\vec{AB})}\text{.}\)
The only ingredient we haven't found yet is the normal vector \(\vec{n}\) to the plane. Finding \(\vec{n}\) is exactly the same as it was for finding the normal vector to a line in \(\mathbb{R}^2\text{;}\) we just read the coefficients from the general equation. Thus our normal vector is \(\vec{n} = \begin{bmatrix}1\\-1\\-2\end{bmatrix}\text{.}\)
Finally, we do the required calculation. The shortest distance from the plane to \(B = (1, 1, 1)\) is:
Therefore the shortest distance from \(B = (1, 1, 1)\) to a point on the plane \(x-y-2z=1\) is \(\sqrt{3/2}\text{.}\)
If we wanted to find the point on the plane where this distance occured, we want to follow the direction of \(\vec{n}\) from \(B\) for a distance of \(\sqrt{3/2}\text{.}\) First, let
be the unit vector in the direction of \(\vec{n}\text{.}\) We will be starting at \(B\text{,}\) but we aren't yet sure if we want to travel along \(\sqrt{3/2}\vec{m}\) or \(-\sqrt{3/2}\vec{m}\text{.}\) The easiest solution is to do both, and consider the points \(C_1 = B + \sqrt{3/2}\vec{m} = (3/2, 1/2, 0)\) and \(C_2 = B-\sqrt{3/2}\vec{m} = (1/2, 3/2, 2)\text{.}\) We see that only the point \(C_1\) satisfies the plane's equation \(x-y-2z=1\text{,}\) so that's the point we're looking for.
The point on the plane \(x-y-2z=1\) closest to \(B = (1, 1, 1)\) is therefore \(C = (3/2, 1/2, 0)\text{.}\)
We now give a second solution to the same problem, which uses a different approach. As described in the first solution, we know that we are looking for a point on the plane that can be reached from \(B\) by following the vector normal to the plane. We also know from the general equation of the plane that the normal vector is \(\vec{n} = \begin{bmatrix}1\\-1\\-2\end{bmatrix}\text{.}\) We are therefore looking for a point that satisfies \(x-y-2z=1\) and lies on the line given in vector form as \(\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}+t\begin{bmatrix}1\\-1\\-2\end{bmatrix}\text{.}\) That is, we want all of the following equations to be satisfied:
and the only solution to this equation is \(t = 1/2\text{.}\) The point we are looking for is therefore
\begin{equation*}
C = B + \frac{1}{2}\vec{n} = (3/2, 1/2, 0),
\end{equation*}
which is the same answer we found in the first solution. If we wanted to know what the distance is, we could compute \(\norm{\vec{BC}} = \sqrt{3/2}\text{,}\) which is also the same answer we found in the first solution.
Exercises2.3.4Exercises
1.
Find the vector equation for the line through \((-7,6,0)\) and \((-1,1,4)\text{.}\) Then, find the parametric equations for this line.
To check that we did not make a mistake: for \(t=0\text{,}\) we are indeed at the point \((-7,6,0)\text{,}\) and for \(t=1\text{,}\) we are at the point
Our line can be described by \(\vec{x} = \vect{OP} + t\vect{PQ}\) since this makes sure that, for \(t=0\text{,}\) our line passes through \(P\text{,}\) and for \(t=1\text{,}\) it passes through \(Q\text{.}\) The vector equation is therefore given by
Our line can be described by \(\vec{x} = \vect{OP} + t\vect{PQ}\) since this makes sure that, for \(t=0\text{,}\) our line passes through \(P\text{,}\) and for \(t=1\text{,}\) it passes through \(Q\text{.}\) The vector equation is therefore given by
The line passing through \(P=(1,0,-3)\) and parallel to the line with parametric equations \(x_1 = -1 +2t\text{,}\)\(x_2 = 2-t\text{,}\) and \(x_3= 3+3t\text{.}\)
First, let us find a directional vector for the line whose parametric equations are given. For \(t=0\text{,}\) that line passes through the point \(Q=(-1,2,3)\) and for \(t=1\text{,}\) it passes through \(R=(1,1,6)\text{.}\) Therefore, the vector
First, let us find a directional vector for the line whose parametric equations are given. For \(t=0\text{,}\) that line passes through the point \(Q=(2,1,0)\) and for \(t=1\text{,}\) it passes through \(R=(1,1,1)\text{.}\) Therefore, the vector
The lines through \(P=(1,0,1)\) that meet the line \(\ell\) whose vector equation is given by \(\mathbf{x} = \begin{bmatrix} 1 \\ 2\\ 0 \end{bmatrix} +t \begin{bmatrix} 2 \\ -1\\ 2 \end{bmatrix} \) at points with distance 3 from \(Q=(1,2,0)\text{.}\)
This means, we need \(9t^2=9\text{,}\) or in other words, \(t=\pm 1\text{.}\) Therefore, there are two points \(R_+, R_-\) on \(\ell\) which have distance \(3\) from \(Q\text{,}\) namely the points that satisfy
Find the normal, general, and vector form of each of the following planes. Hint.
In order to determine the normal form of a plane, you need both a single point \(P\) on the plane and a normal vector \(\vec{n}\) to the plane. See Definition 2.3.14.
Passing through \(A=(2,1,3)\text{,}\)\(B=(1,1,3)\text{,}\) and \(C=(1,2,-3)\text{.}\)
We begin by finding two direction vectors for this plane, namely \(\vec{d_1} = \vec{AB} = \begin{bmatrix}-1\\0\\0\end{bmatrix}\) and \(\vec{d_2} = \vec{AC} = \begin{bmatrix}-1\\1\\-6\end{bmatrix}\text{.}\) We can then use any point on the plane to give a vector equation, such as this one:
To find the normal form we need a normal vector, that is, a vector \(\vec{n}\) such that \(\vec{n}\cdot\vec{d_1} = 0\) and \(\vec{n}\cdot\vec{d_2} = 0\text{.}\) If you are comfortable with the cross product, one solution is to take \(\vec{n} = \vec{d_1} \times \vec{d_2}\text{.}\) We are not teaching the cross product in this course, so we will find \(\vec{n}\) another way. Suppose that \(\vec{n} = \begin{bmatrix}a\\b\\c\end{bmatrix}\text{.}\) Then we need:
\begin{equation*}
0 = \vec{n}\cdot\vec{d_1} = \begin{bmatrix}a\\b\\c\end{bmatrix} \cdot\begin{bmatrix}-1\\0\\0\end{bmatrix} = -a
\end{equation*}
We have here a system of two linear equations in three variables, and any solution to the system will give us a normal vector. Solving the system using our usual techniques, we find that \(a=0, b=6, c=1\) is a solution, so we take \(\vec{n} = \begin{bmatrix}0\\6\\1\end{bmatrix}\text{.}\) Then if we use the point \(A\) to define our normal equation, we get
We follow the same strategy as in the previous example. As direction vectors we take \(\vec{d_1} = \vec{AB} = \begin{bmatrix}-1\\1\\-5\end{bmatrix}\) and \(\vec{d_2} = \vec{BC} = \begin{bmatrix}4\\7\\-12\end{bmatrix}\text{.}\) Then, using the point \(B\) this time as our chosen point on the plane, we get a vector equation:
One solution to this system is \(a=23, b=-32, c=-11\text{,}\) so we take \(\vec{n} = \begin{bmatrix}23\\-32\\-11\end{bmatrix}\text{.}\) Using the point \(B\text{,}\) our normal form equation is:
Two planes are parallel if and only if they have the same normal vectors. Therefore, we should first find a normal vector to the plane whose general equation is given: If we chose
so \(\vec{n}\) is indeed a normal vector. With this at hand, any point \(X=(x_1,x_2,x_3)\) on the plane whose normal form we are after can be described by \(\vec{n}\cdot\vect{PX}=0\text{,}\) or more explicitly:
For the vector form, we need two non-parallel vectors \(\vec{u}\text{,}\)\(\vec{v}\) which are orthogonal to \(\vec{n}\text{,}\) i.e. they need to satisfy
Two planes are parallel if and only if they have the same normal vectors. Therefore, we should first find a normal vector to the plane whose general equation is given: If we chose
so \(\vec{n}\) is indeed a normal vector. With this at hand, any point \(X=(x_1,x_2,x_3)\) on the plane whose normal form we are after can be described by \(\vec{n}\cdot\vect{PX}=0\text{,}\) or more explicitly:
For the vector form, we need two non-parallel vectors \(\vec{u}\text{,}\)\(\vec{v}\) which are orthogonal to \(\vec{n}\text{,}\) i.e. they need to satisfy
Since the plane contains the line, it in particular contains the point \(Q=(0,0,2)\) (choose \(t=0\) in the line equation). Since \(P\) is assumed to be in the plane, we conclude that the line that goes through \(Q=(0,0,2)\) and \(P\text{,}\) whose vector form is given by
is also contained in the plane. Since this second line's directional vector is not parallel to the first line's directional vector, we conclude that the plane is spanned by those two directional vectors and thus has vector form
We can find a normal vector \(\vec{n}\) by letting \(\vec{n} = \begin{bmatrix}a\\b\\c\end{bmatrix}\) and solving the system \(\vec{n} \cdot \vec{d_1} = 0 = \vec{n}\cdot\vec{d_2}\text{.}\) We find that a solution is
Since the plane contains the line, it in particular contains the point \(Q=(3,-1,2)\) (choose \(t=0\) in the line equation). Since \(P\) is assumed to be in the plane, we conclude that the line that goes through \(Q\) and \(P\text{,}\) whose vector form is given by
is also contained in the plane. Since this second line's directional vector is not parallel to the first line's directional vector, we conclude that the plane is spanned by those two directional vectors and thus has vector form
We can find a normal vector \(\vec{n}\) by setting \(\vec{n}\cdot\vec{d_1} = \vec{n}\cdot\vec{d_2} =0\) and solving the resulting system. We find that one solution is
First, we have to find out in which point the two lines meet. (If they do not meet, then they are skew lines since they are clearly not parallel. In this case, they are not in the same plane.) In other words, we are looking for two real numbers \(s, t\) such that
\begin{equation*}
1 + s = t;\;
-1 + s = -t;\;
2 + s = 2.
\end{equation*}
The last equation yields \(s=0\) and both other equations then yield \(t = 1\text{.}\) We conclude that the two lines meet in the point \((1,-1,2)\) because
Since the two lines are not parallel, their directional vectors span the plane that contains them, so we arrive at the following vector equation of the plane,
First, we have to find out in which point the two lines meet. (If they do not meet, then they are skew lines since they are clearly not parallel. In this case, they are not in the same plane.) In other words, we are looking for two real numbers \(s, t\) such that
\begin{equation*}
3 + s = 2t;\;
1 - s = -2+t;\;
3s = 5 - t.
\end{equation*}
The second equation says \(t=3-s\text{,}\) which we plug into the other equations to get \(s = 1\text{.}\) We conclude that the two lines meet in the point \((4,0,3)\) because
Since the two lines are not parallel, their directional vectors span the plane that contains them, so we arrive at the following vector equation of the plane,
So if we plug both of these into (2.3.1) and cancel the summand of \(\norm{\vect{OX}}^2\) on both sides, then we see that \(\norm{\vect{PX}}= \norm{\vect{QX}}\) is equivalent to
which is its normal form. Lastly, for the vector form, we need three points in the plane which do not lie on a line. Let us choose \(R_1= (0, -3 , -\frac{1}{8} )\text{,}\)\(R_2= (\frac{3}{2}, 0, 1)\text{,}\) and \(R_3= (6,\frac{1}{4} ,0)\) which all satisfy the general equation. Now, the plane's vector equation is given by
which is its normal form. Lastly, for the vector form, we need three points in the plane which do not lie on a line. Let us choose \(R_1= (3,0,0)\text{,}\)\(R_2= (0, -3, 0)\text{,}\) and \(R_3= (0,0,-3)\) which all satisfy the general equation. Now, the plane's vector equation is given by
\begin{align*}
\vec{x}
\amp =
\vect{OR_1}
+
t
\vect{R_1 R_2}
+
s
\vect{R_1 R_3}\\
\amp =
\begin{bmatrix}
3\\0\\0
\end{bmatrix}
+
t
\begin{bmatrix}
-3\\-3\\0
\end{bmatrix}
+
s
\begin{bmatrix}
-3\\0\\-3
\end{bmatrix}
\end{align*}
where \(\vec{a}\) is a point on the line and \(\mathbf{d}\) is a direction vector of the line. We now only have to plug in the values for the two situations:
so that \(Q= (\tfrac{29}{7}, \tfrac{2}{7}, \tfrac{11}{7})\text{.}\)
\(P=(1,-1,3)\text{,}\)\(\ell \colon \vec{x} = \begin{bmatrix} 1 \\ 0\\ -1 \end{bmatrix} +t\begin{bmatrix} 3 \\ 1\\ 4 \end{bmatrix} \) For the given \(\ell\text{,}\) we have
where \(\vec{n}\) is normal to the plane, \(\vec{a}\) is the position vector of any point on the plane, \(\vec{p}:=\vect{OP}\text{,}\) and \(\operatorname{proj}\) is as defined in Hint 2.2.5.8.1.>
Find a plane \(\mathcal{P}\) to which \(\ell_1\) (and therefore also \(\ell_2\)) is orthogonal. In other words, the directional vector of \(\ell_1\) is normal to \(\mathcal{P}\text{.}\) Find the points \(Q_1, Q_2\) which \(\mathcal{P}\) has in common with \(\ell_1\) resp. \(\ell_2\text{.}\) What do you know about the line from \(Q_1\) to \(Q_2\text{?}\)
We follow the idea of Hint 2.3.4.6.1 and make one additional remark: The line from \(Q_1\) to \(Q_2\) is, by construction, contained in the plane since the two points are in the plane. In particular, said line is orthogonal to \(\ell_1\) and \(\ell_2\) and hence describes the direct path from \(\ell_1\) to \(\ell_2\text{,}\) meaning
We now follow this recipe for the two given pairs of lines.
We choose the point \(Q_1= (2 , -1, 3 ) \) on \(\ell_1\) and the declare the directional vector \(\vec{n}=\begin{bmatrix} 1 \\ -1\\ 4 \end{bmatrix} \) of \(\ell_1\) to be normal to the plane \(\mathcal{P}\text{.}\) The normal form of \(\mathcal{P}\) is then given by \(\vec{n} \cdot \vec{x} = \vec{n} \cdot \vect{OQ_1}\text{,}\) so its general form is
Next, we are looking for the intersection of \(\mathcal{P}\) with \(\ell_2\text{.}\) In other words, we are looking for \(t\) such that \(\begin{bmatrix} 1 \\ 0\\ 1 \end{bmatrix} +t\begin{bmatrix} 1 \\ -1\\ 4 \end{bmatrix}\) satisfies the general equation of \(\mathcal{P}\text{:}\)
We choose the point \(Q_1= (3 , 0, 2 ) \) on \(\ell_1\) and the declare the directional vector \(\vec{n}=\begin{bmatrix} 3 \\ 1\\ 0 \end{bmatrix} \) of \(\ell_1\)to be normal to the plane \(\mathcal{P}\text{.}\) The normal form of \(\mathcal{P}\) is then given by \(\vec{n} \cdot \vec{x} = \vec{n} \cdot \vect{OQ_1}\text{,}\) so its general form is
Next, we are looking for the intersection of \(\mathcal{P}\) with \(\ell_2\text{.}\) In other words, we are looking for \(t\) such that \(\begin{bmatrix} -1 \\ 2\\ 2 \end{bmatrix} +t\begin{bmatrix} 3 \\ 1\\ 0 \end{bmatrix}\) satisfies the general equation of \(\mathcal{P}\text{:}\)
Find a normal vector \(\vec{n}_1\) resp. \(\vec{n}_2\) for each of the planes. Using the Cosine Formula 2.2.5.5.1, compute the angle \(\theta=\sphericalangle (\vec{n}_1, \vec{n}_2)\) between these vectors. Then,
Find a normal vector \(\vec{n}\) for \(\mathcal{P}\) and a directional vector \(\vec{d}\) of \(\ell\text{.}\) Compute \(\theta=\sphericalangle (\vec{n}, \vec{d})\) using the Cosine Formula 2.2.5.5.1. Then
The angle \(\theta\) between this vector and the directional vector \(\vec{d} = \begin{bmatrix} 1 \\ 1\\ 4 \end{bmatrix}\) of the line can be computed using the Cosine Formula 2.2.5.5.1:
