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Section 4.3 Lines, planes, and generalizations

Subsection 4.3.1 Review: scalar multiplication in \(\R^2\)

One of the most attractive features of scalar multiplication is that it may be interpreted geometrically in \(\R^2\text{.}\) Take the vector \(\vec v=(1,2)\text{,}\) and consider the following points in \(\R^2\text{:}\)

  1. \(\displaystyle \vec v=(1,2)\)

  2. \(\displaystyle 2\vec v=(2,4)\)

  3. \(\displaystyle 3\vec v=(3,6)\)

  4. \(\displaystyle \frac12\vec v=(\frac12,1)\)

  5. \(\displaystyle -1\vec v=(-1,-2)\)

  6. \(\displaystyle -2\vec v=(-2,-4)\)

  7. \(\displaystyle -3\vec v=(-3,-6)\)

These are all scalar multiples of \(\vec v\text{,}\) that is, each one is of the form \(t\vec v\) for some number \(t\text{.}\) Here is an accurate plot of these points in \(\R^2\text{:}\)

Figure 4.3.1.

Notice that all of the points lie on a straight line that passes through \(\vec 0=(0,0)\text{.}\) In addition, if we start from \((0,0)\) and follow the line through the first quadrant, we pass through \(\vec 0\text{,}\) \(\frac12\vec v\text{,}\) \(\vec v\text{,}\) \(2\vec v\) and \(3\vec v\) in that order. It would seem that as \(t\) starts at \(0\) and increases through positive values, the points \(t\vec v\) starts at \(\vec0\) and moves along the line in the first quadrant. Similarly, as \(t\) starts at \(0\) and decreases through negative values, the points \(t\vec v\) starts at \(\vec0\) and moves along the line in the third quadrant.

A linearity test.

Suppose we wish to determine whether or not two vectors \(\vec v\) and \(\vec w\) are collinear with \(\vec 0\text{.}\) There are two cases, depending on whether or not \(\vec 0\) lies between \(\vec v\) and \(\vec w\text{:}\)

Figure 4.3.2.

In the right-hand figure, the angle \(\theta\) between \(\vec v\) and \(\vec w\) (at \(\vec0\)) is \(0\) and so \(\cos\theta=1\text{,}\) while in the left-hand figure, the angle \(\theta\) between \(\vec v\) and \(\vec w\) is \(\pi\) and so \(\cos\theta=-1.\) Hence \(\vec v\) and \(\vec w\) are collinear with \(\vec 0\) if and only if \(|\cos\theta|=1\text{.}\)

However, we know from Proposition 4.2.14 that

\begin{equation*} |\vec x\cdot\vec y|= \|\vec x\| \|\vec y\|\, |\cos \theta| \end{equation*}

and so we now have another equivalent condition for linearity:

\begin{equation*} |\vec x\cdot\vec y|= \|\vec x\| \|\vec y\| \end{equation*}

Notice that this condition says that the Cauchy-Schwarz inequality given in Theorem 4.2.17). is actually equality.

First, suppose that \(\vec w\) is on the line joining \(\vec v\) and \(\vec 0\text{.}\) Here is the picture:

Figure 4.3.4.

Notice that the red lines indicate that \(\tan(\theta)=\frac ba\) and the green lines indicate that \(\tan(\theta)=\frac yx\text{.}\)This implies that \(\frac ba=\frac yx\text{.}\) Now let \(t=\frac xa\text{.}\) We then have \(x=ta\) and \(y=\frac{bx}a=\frac{bta}{a}=bt\text{,}\) and so \(\vec w=(x,y)=(ta,tb)=t(a,b)=t\vec v\text{.}\)

Next we start with \(\vec v\) and consider \(\vec w=t\vec v\text{.}\) We want to show that \(\vec w\) is on the line joining \(\vec0\) and \(\vec v\text{.}\) To verify this with the linearity test, we evaluate:

\begin{equation*} |\vec v\cdot \vec w|=|\vec v\cdot t\vec v|=|t| |\vec v\cdot \vec v|=|t|\|\vec v\|^2\\ \|\vec v\| \|\vec w\|=\|\vec v\| \|t\vec v\|=|t| \|\vec v\| \|\vec v\|=|t|\|\vec v\|^2 \end{equation*}

Hence \(\vec w\) is on the line joining \(\vec0\) and \(\vec v\text{,}\) as desired.

Subsection 4.3.2 Review: addition in \(\R^2\)

A construction: completing the parallelogram.

Start with two points \((a,b)\) and \((c,d)\text{,}\) both not \(\vec0\text{.}\) Further, assume that the three points \((a,b)\text{,}\) \((c,d)\) and \(\vec0\) do not line on a single line.

Figure 4.3.5.

Draw the line through \(\vec0\) and \((a,b)\) (red in the figure) and a line through \(\vec0\) and \((c,d)\) (green in the figure). At \((a,b)\) draw a line parallel to the one through \(\vec0\) and \((c,d)\) (green in the figure) and, similarly, one through \((c,d)\) parallel to the line through \(\vec0\) and \((a,b)\text{.}\) Call the point where these two new lines meet \((x,y)\text{.}\) The four points \(\vec0\text{,}\) \((a,b)\text{,}\) \((c,d)\) and \((x,y)\) are the four vertices of a parallelogram. Starting with three points \(\vec0\text{,}\) \((a,b)\) and \((c,d)\text{,}\) the construction or the fourth point \((x,y)\) is called completing the parallelogram.

Next, we want to find the values of \(x\) and \(y\text{.}\) We drop perpendiculars from \((a,b)\text{,}\) \((c,d)\) and \((x,y)\) to the \(x\)-axis.

Figure 4.3.6.

Notice that the two yellow triangles are the same size (that is, congruent). This means that the distance from \(x\) to \(c\) on the \(x\)-axis is the same as that from \(a\) to \(0\text{.}\) In other words, \(x-c=a-0\text{,}\) or \(x=a+c\text{.}\) Using similar perpendiculars to the \(y\)-axis, we get \(y=b+d\text{.}\) We conclude that

\begin{equation*} (x,y)=(a+c,b+d)=(a,b)+(c,d) \end{equation*}

This gives a geometric method for constructing the sum of two vectors. The figure constructed above has both points in the first quadrant, and using points in other quadrants changes the figure. If there are points in the first and second quadrant, the figure becomes

Figure 4.3.7.

In this case, \(a=x-c\) and once again

\begin{equation*} (x,y)=(a+c,b+d)=(a,b)+(c,d) \end{equation*}

Variations of the same argument work when the points are in other quadrants.

Subsection 4.3.3 Directed vectors in \(\R^2\) and \(\R^n\)

We have looked at vectors as \(n\)-tuples, that is, \(\vec x=(x_1,x_2,\ldots,x_n)\text{.}\) There is another way to look at vectors, namely, as objects having both length and direction. We visualize this as an arrow joining two points with the length being the length of the arrow and the direction that where the arrow points. Here are two such vectors in \(\R^2\text{.}\)

Figure 4.3.9.

The point \(P_1\) is the tail of \(\overrightarrow{P_1Q_1}\) while \(Q_1\) is the head. Apparently the vectors \(\overrightarrow{P_1Q_1}\) and \(\overrightarrow{P_2Q_2}\) have the same direction and length, and so they should be the same vector. We next give the criterion that determines whether or not \(\overrightarrow{P_1Q_1}=\overrightarrow{P_2Q_2}\) in general.

Figure 4.3.10.

Starting with \(\overrightarrow{P_1Q_1}\text{,}\) construct \(\overrightarrow{\vec0 X}\) that is parallel to it and has the same length. This constructs the yellow parallelogram in the figure above. The parallelogram rule then implies \(P_1+X=Q_1\text{,}\) or \(X=Q_1-P_1\text{.}\) Repeating the construction with \(\overrightarrow{P_2Q_2}\text{,}\) we get the same vector \(\overrightarrow{\vec0 }\) if and only if \(\overrightarrow{P_1Q_1}\) and \(\overrightarrow{P_2Q_2}\) have the same length and direction, which means \(X=Q_2-P_2\text{.}\) While the picture is in \(\R^2\text{,}\) but the idea is easily extended to the same picture in \(\R^3\) by considering the plane containing \(\vec0\text{,}\) \(\vec x\) and \(\vec y\text{.}\) Indeed, the definition in \(\R^n\) follows the same pattern: A vector \(\overrightarrow{PQ}\) is visualized as an arrow with tail and head at \(P\) and \(Q\) in \(\R^n\text{.}\)

Definition 4.3.11. Equality of directed vectors.
\begin{equation*} \overrightarrow{P_1Q_1}=\overrightarrow{P_2Q_2} \textrm{ if and only if } Q_1-P_1=Q_2-P_2. \end{equation*}

This says, roughly speaking, that head minus tail of the first vector equals head minus tail of the second vector. We can call this the head minus tail rule.

Definition 4.3.12. Direction number of a vector.

The direction number of the vector \(\overrightarrow{PQ}\) is \(Q-P\)

Subsection 4.3.4 Scalar multiplication of directed vectors

By definition, for any directed vector \(\overrightarrow{PQ}\text{,}\) we have \(\overrightarrow{PQ}=\overrightarrow{\vec0(Q-P)}\text{.}\) Scalar multiples of \(\overrightarrow{PQ}\) are defined so that it matches our usual definition for \(n\)-tuples and also satisfies the parallelogram rule:

Figure 4.3.13.

The geometric interpretation of scalar multiplication is now clear. Multiplying a vector \(\overrightarrow{PQ}\) by a positive scalar \(r\) keeps the tail and direction fixed and stretches the arrow by a factor of \(r\text{.}\) When \(r\) is negative, the arrow is stretched by a factor of \(-r\) and the direction is reversed.

Subsection 4.3.5 Addition of directed vectors

We next give a geometric interpretation for the addition of directed vectors. We wish to use the same pattern as we did with scalar multiplication: be consistent with the geometric interpretation of the parallelogram rule for \(n\)-tuples.

We start with two vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{RS}\text{.}\) Using the tail minus head rule, these are the same as the vectors \(\vec x=\overrightarrow{\vec0(Q-P)}\) and \(\vec y=\overrightarrow{\vec0(S-R)}\text{.}\) Now \(\vec x\) and \(\vec y\) can be viewed as \(n\)-tuples, and so the sum gives \(\vec x+\vec y=Q-P+S-R\text{.}\)

Figure 4.3.14.

On the other hand, we can take the two vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{RS}\) and find the vector equal to \(\overrightarrow{RS}\) but with tail at \(Q\text{.}\) By the head minus tail rule, the head of this vector is at \(Q+S-R\) (since \((Q+S-R)-Q=S-R\)). Now consider the vector with tail at \(P\) and head at \(Q+S-R\text{.}\) This is the third side of the triangle. The head minus tail rule of evaluation gives \(Q+S-R-P\text{,}\) which is exactly what we computed for the sum. This gives us a wonderful geometric interpretation of the sum of two vectors.

Figure 4.3.15.

Subsection 4.3.6 Lines in \(\R^2\) and \(\R^n\)

There are different equations whose solutions are lines in \(\R^2\text{.}\) If a line \(L\) passes through the points \((x_0,y_0)\) and \((x_1,y_1)\text{,}\) has slope \(m\) and intersects the \(y\)-axis at \((0,b)\text{,}\) then the point \((x,y)\) is on \(L\) if and only if it satisfies equations of the following form:

  • The point-slope form:

    \begin{equation*} \frac{y-y_0}{x-x_0}=m \end{equation*}

  • The slope-intercept form:

    \begin{equation*} y=mx+b \end{equation*}

  • The point-point form:

    \begin{equation*} \frac{y-y_0}{x-x_0}=\frac{y_1-y_0}{x_1-x_0} \textrm{ (if } x_1\not=x_0\textrm{)} \end{equation*}

  • The symmetric form: For this example, let \(a=x_1-x_0\not=0\) and \(b=y_1-y_0\not=0\text{.}\) Then

    \begin{equation*} \frac{x-x_0}a=\frac{y-y_0}b \end{equation*}

  • The parametric form:

    \begin{equation*} (x,y)=(x_0,y_0) + t\bigl((x_1,y_1)-(x_0,y_0)\bigr)=(1-t)(x_0,y_0)+t(x_1,y_1) \end{equation*}

    which may be visualized as

    Figure 4.3.16.
Definition 4.3.17. General equation of a line in \(\R^2\).

A line \(L\) is the set of points \((x,y)\) satisfying the equation

\begin{equation*} ax+by+c=0 \end{equation*}

where \(a\) and \(b\) are not both \(0\text{.}\)

Figure 4.3.19.

Notice that a line in \(\R^2\) is the solution of a single linear equation in two unknowns.

Lines in \(\R^n\) mirror the symmetric and parametric definitions for \(\R^2\text{.}\)

Definition 4.3.20. Lines in \(\R^n\).

If \(\vec x\) and \(\vec y\) are two vectors in \(\R^n\text{,}\) then the line \(L\) containing \(\vec x\) and \(\vec y\) consists of all vectors \(\vec z\) such that

\begin{equation*} \vec z=(1-t)\vec x+t\vec y \end{equation*}

for some real number \(t\text{.}\)

The following figure gives the geometric interpretation on the number \(t\text{.}\) Obviously \(t=0\) gives \(\vec x\) and \(t=1\) gives \(\vec y\text{.}\) As \(t\) goes from \(0\) to \(1\text{,}\) the vector \(\vec z\) moves from \(\vec x\) to \(\vec y\text{.}\) In fact, when \(t=\frac12\text{,}\) the point \(\vec z\) is halfway between \(\vec x\) and \(\vec y\text{.}\) As \(t\) becomes larger than \(1\text{,}\) the vector \(\vec z\) passes \(\vec y\) and moves further down the line.

Figure 4.3.21.

There is a symmetric form for lines in \(\R^n\text{.}\) If \(\vec x=(x_1,x_2,\ldots,x_n)\text{,}\) \(\vec y=(y_1,y_2,\ldots,y_n)\) and \(\vec z=(z_1,z_2,\ldots,z_n)\text{,}\) then \(\vec z\) is on the line joining \(\vec x\) and \(\vec y\) if and only if

\begin{equation*} z_i=(1-t)x_i+ty_i \end{equation*}

for \(i=1,2,\ldots,n\text{,}\) and in this case

\begin{align*} t \amp=\frac{z_1-x_1}{y_1-x_1} = \frac{z_2-x_2}{y_2-x_2} =\cdots = \frac{z_n-x_n}{y_n-x_n}\\ \amp=\frac{z_1-x_1}{a_1} = \frac{z_2-x_2}{a_2} =\cdots = \frac{z_n-x_n}{a_n} \end{align*}

The vector \((a_1,a_2,\ldots,a_n)\) is the direction number of the line.

Determining if a point is on a line.

From our last result, this is easy. The equation \(\vec z=(1-t)\vec x+t\vec y\) gives us \(n\) linear equations in the single variable \(t\text{.}\) If this system of linear equations is consistent, then the point is on the line.

Let \(\vec x=(1,2,-1,3)\) and \(\vec y=(1,-2,1,3)\text{.}\) Is \(\vec z=(1,-2,-1,3)\) on the line joining \(\vec x\) and \(\vec y\text{?}\) The equation \(\vec z=(1-t)\vec x+t\vec y\) gives four equations, one for each coordinate:

\begin{align*} 1(1-t)+1t\amp=1 \\ 2(1-t)+(-2)t\amp=-2 \amp\amp\to t=1\\ (-1)(1-t)+1t\amp=-1 \amp\amp\to t=0\\ 3(1-t)+3t\amp=3 \end{align*}

Now we have a contradiction. The second equation implies \(t=1\) while the third implies \(t=0\text{,}\) and so the point is not on the line.

On the other hand, is \(\vec z=(1,-6,3,3)\text{.}\) then \(\vec z=(1-t)\vec x+t\vec y\) gives four equations,

\begin{align*} 1(1-t)+1t\amp=1 \amp\amp\to 1=1\\ 2(1-t)+(-2)t\amp=-6 \amp\amp\to t=2\\ (-1)(1-t)+1t\amp=3 \amp\amp\to t=2\\ 3(1-t)+3t\amp=3 \amp\amp\to 3=3 \end{align*}

Since all four equation are valid when \(t=2\) the vector \(\vec z\) is on the line. It is easy to verify that \(\vec z=-\vec x+2\vec y\text{.}\)