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Section 5.4 Equations of lines in \(\R^3\)

We want to find the equation of a line \(L\text{,}\) that is, an equation using the vector \((x,y,z)\) so that \((x,y,z)\) is on \(L\) if and only if \((x,y,z)\) satisfies the equation.

We start with the case of a line \(L\) that passes through the origin \(\vec0=(0,0,0)\text{.}\) Let \(\vec n=(a,b,c)\) be any nonzero point on \(L\text{.}\) In this case it is easy to get the equation we want.

Figure 5.4.1.

Any point \((x,y,z)\) is on \(L\) if and only if \((x,y,z)=t(a,b,c)\) for some real number \(t\text{.}\) In other words, if we have a given vector \(\vec x=(x_0,y_0,z_0)\) then \(\vec x\) is on \(L\) if and only if there is a solution \(t\) to the system of equations

\begin{equation*} ta=x_0\\ tb=y_0\\ tc=z_0 \end{equation*}

The vector \(\vec n=(a,b,c)\) is an indication of the direction of the line. For this reason, it is called the direction vector of the line. The components of this vector, that is, \(a\text{,}\) \(b\) and \(c\text{,}\) are called the direction numbers. Direction vectors for lines in \(\R^3\) are analogous to the slope of a line in \(\R^2\text{.}\)

We next consider lines as before, but with the more general assumption that it passes through \((x_0,y_0,z_0)\text{.}\) We wish to keep the concept of the direction vector. The parallelogram rule is the tool that leads the way:

Figure 5.4.3.

We start with the line \(L\) and consider the line, \(L'\) through \(\vec0\) and with the same direction vector as (that is, parallel to) \(L\text{.}\) The parallelogram rules shows us that \((x,y,z)\) is on \(L'\) if and only if \((x,y,z)+(x_0,y_0,z_0)\) is on \(L\text{.}\) Since every point on \(L'\) is of the form \(t(a,b,c)\) for some real number \(t\text{,}\) we conclude that every point on \(L\) is of the form \(t(a,b,c)+(x_0,y_0,z_0)\text{.}\)

Finally, we consider a line \(L\) passing through two points \((x_0,y_0,z_0)\) and \((x_1,y_1,z_1)\text{.}\) Let \(L'\) be the line parallel to \(L\) passing through \(\vec0\text{.}\) Then \(\vec n=(x_1,y_1,z_1)-(x_0,y_0,z_0)\) is a nonzero point on \(L'\text{,}\) from which follows that every point on \(L'\) is of the form \(t\left((x_1,y_1,z_1)-(x_0,y_0,z_0)\right) =t(x_1,y_1,z_1)-t(x_0,y_0,z_0)\text{.}\) This in turn implies that every point on \(L\) is of the form \(t(x_1,y_1,z_1)-t(x_0,y_0,z_0)+(x_0,y_0,z_0)=t(x_1,y_1,z_1)+(1-t)(x_0,y_0,z_0)\text{.}\)

Figure 5.4.5.