Passing through a point and parallel to a vector

Theorem 6.1 (vector form, point + direction vector). Suppose a line $L$ passes through a fixed point $A$ whose position vector is $\bar{a}$, and $L$ is parallel to a given vector $\bar{b}$. Let $P$ be any other point on $L$, with position vector $\bar{r}$. Since $P$ lies on $L$ and $L$ is parallel to $\bar{b}$, the vector $\overline{AP}$ must itself be parallel to $\bar{b}$, so it can be written as a scalar multiple of $\bar{b}$: $\overline{AP} = \lambda \bar{b}$ for some real number $\lambda$. But by the triangle law of vector addition, $\overline{AP} = \overline{OP} - \overline{OA} = \bar{r} - \bar{a}$. Combining the two gives $\bar{r} - \bar{a} = \lambda \bar{b}$, i.e. $\bar{r} = \bar{a} + \lambda \bar{b}.$ This is the vector equation of the line. The real number $\lambda$ is called a parameter: as $\lambda$ runs over all real numbers, the point $P$ traces out every point of the line exactly once, so there is a one-to-one correspondence between the points of $L$ and the values of $\lambda$ — this is why the equation is called the parametric form of the vector equation of a line. A quick activity worth doing mentally is to substitute a few different values of $\lambda$ (say $0$, $1$, $-1$) into $\bar{r} = \bar{a} + \lambda\bar{b}$ to see three different position vectors, all lying on the same line. The same geometric fact — that $\bar{r} - \bar{a}$ is parallel to $\bar{b}$ — can also be written without a parameter at all, using the property that two parallel (non-zero) vectors have zero cross product: $(\bar{r} - \bar{a}) \times \bar{b} = \bar{0}$. This is called the non-parametric form of the vector equation, and it is completely equivalent to the parametric form; it simply says directly that the vector from $A$ to any point of the line is always parallel to $\bar{b}$, without introducing $\lambda$ explicitly.

Theorem 6.2 (Cartesian form, point + direction ratios). The same line can be described using coordinates instead of vectors. Suppose $L$ passes through $A(x_1, y_1, z_1)$ and has direction ratios $a, b, c$ (meaning a vector parallel to $L$ has components proportional to $a, b, c$). Let $P(x, y, z)$ be any other point on $L$. The direction ratios of the segment $AP$ are then $x - x_1,\ y - y_1,\ z - z_1$. Since $AP$ lies along $L$, these direction ratios must be proportional to the line's own direction ratios $a, b, c$ — that is exactly what it means for two sets of direction ratios to represent the same direction. Writing this proportionality as a chain of equal ratios gives $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c},$ which are the Cartesian equations of the line. Unlike a line in a plane, a line in space genuinely needs two independent equations (this chain of ratios is really two equations joined together) — a single equation in $x, y, z$ describes a plane, not a line, so it is impossible to pin down a line in space with just one Cartesian equation.

A few standing conventions are worth stating explicitly, since they recur throughout the chapter. If $\bar{b} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$ is a vector parallel to a line, then $a_1, b_1, c_1$ are direction ratios of that line, and conversely any set of direction ratios $a_1, b_1, c_1$ gives a parallel vector $\bar{b} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$ — the vector and Cartesian descriptions are simply two faces of the same underlying direction. When the equal ratios above are all set equal to the parameter $\lambda$ itself, i.e. $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda$, the result is called the symmetric form of the Cartesian equations. Separating $\lambda$ out of each ratio instead gives $x = x_1 + \lambda a,\ y = y_1 + \lambda b,\ z = z_1 + \lambda c$, called the parametric form of the Cartesian equations; the coordinates of the general point on the line are therefore $(x_1 + \lambda a,\ y_1 + \lambda b,\ z_1 + \lambda c)$, and exactly as with the vector form, each real value of $\lambda$ gives one point of the line and each point of the line corresponds to exactly one value of $\lambda$. The symmetric form is only valid when none of $a, b, c$ is zero, since each one appears as a denominator; if even one of the three direction ratios is zero, the equations must be written in the parametric form instead. Finally, if we use direction cosines $l, m, n$ (a unit vector along the line) in place of general direction ratios $a, b, c$, the general point becomes $(x_1 + \lambda l,\ y_1 + \lambda m,\ z_1 + \lambda n)$, and a short computation shows why direction cosines are especially convenient: $AP^2 = (\lambda l)^2 + (\lambda m)^2 + (\lambda n)^2 = \lambda^2\{l^2 + m^2 + n^2\} = \lambda^2,$ using $l^2+m^2+n^2=1$ for direction cosines. Hence $AP = |\lambda|$, so with direction cosines the parameter $\lambda$ itself directly measures the distance of the point from the base point $A$ (with sign showing which side of $A$ it lies on).

Worked examples.

Ex.(1). We must check whether the point with position vector $4\hat{i} - 11\hat{j} + 2\hat{k}$ lies on the line $\bar{r} = (6\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(2\hat{i} + 7\hat{j} + 3\hat{k})$. Setting $\bar{r} = 4\hat{i} - 11\hat{j} + 2\hat{k}$ in the line's equation and matching components gives three separate equations: $6 + 2\lambda = 4$, $-4 + 7\lambda = -11$, and $5 + 3\lambda = 2$. Solving each on its own gives $\lambda = -1$ every single time, so all three coordinate equations are satisfied by one common value of $\lambda$; therefore the given point does lie on the line. An equally valid alternative is to use the non-parametric idea directly: writing $\bar{a} = 6\hat{i} - 4\hat{j} + 5\hat{k}$ and $\bar{b} = 2\hat{i} + 7\hat{j} + 3\hat{k}$, a point $\bar{r}$ lies on the line exactly when $\bar{r} - \bar{a}$ is a scalar multiple of $\bar{b}$. Here $\bar{r} - \bar{a} = (4\hat{i} - 11\hat{j} + 2\hat{k}) - (6\hat{i} - 4\hat{j} + 5\hat{k}) = -2\hat{i} - 7\hat{j} - 3\hat{k} = -(2\hat{i} + 7\hat{j} + 3\hat{k}) = -1\cdot\bar{b}$, which is indeed a scalar multiple of $\bar{b}$, confirming the point lies on the line.

Ex.(2). Find the vector equation of the line through the point with position vector $4\hat{i} - \hat{j} + 2\hat{k}$, parallel to $-2\hat{i} - \hat{j} + \hat{k}$. This is a direct application of Theorem 6.1 with $\bar{a} = 4\hat{i} - \hat{j} + 2\hat{k}$ and $\bar{b} = -2\hat{i} - \hat{j} + \hat{k}$, giving $\bar{r} = (4\hat{i} - \hat{j} + 2\hat{k}) + \lambda(-2\hat{i} - \hat{j} + \hat{k})$.

Ex.(3). Find the vector equation of the line through $2\hat{i} + \hat{j} - 3\hat{k}$, perpendicular to both $\bar{b} = \hat{i} + \hat{j} + \hat{k}$ and $\bar{c} = \hat{i} + 2\hat{j} - \hat{k}$. Here we are not handed the required direction vector directly — instead we must build one. The cross product $\bar{b} \times \bar{c}$ is always perpendicular to both $\bar{b}$ and $\bar{c}$, so it must be parallel to the line we want (any line perpendicular to two given, non-parallel vectors must run along their cross product). Computing the determinant: $\bar{b}\times\bar{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\1&2&-1\end{vmatrix} = \hat{i}(1\cdot(-1) - 1\cdot2) - \hat{j}(1\cdot(-1) - 1\cdot1) + \hat{k}(1\cdot2 - 1\cdot1) = -3\hat{i} + 2\hat{j} + \hat{k}.$ So the required line passes through $2\hat{i} + \hat{j} - 3\hat{k}$ and is parallel to $-3\hat{i} + 2\hat{j} + \hat{k}$, giving $\bar{r} = (2\hat{i} + \hat{j} - 3\hat{k}) + \lambda(-3\hat{i} + 2\hat{j} + \hat{k})$. …