Equation of a Line, Parametric & Cartesian

Starting from $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ with $\mathbf{r} = \langle x, y, z \rangle$, $\mathbf{a} = \langle a_1, a_2, a_3 \rangle$, $\mathbf{b} = \langle b_1, b_2, b_3 \rangle$, matching components gives the parametric form:

If every $b_i \neq 0$, solve each equation for $\lambda$: $\lambda = \dfrac{x-a_1}{b_1} = \dfrac{y-a_2}{b_2} = \dfrac{z-a_3}{b_3}$. Dropping the $\lambda =$ gives the symmetric Cartesian form:

Worked through the hook with $\mathbf{a} = \langle 1, 2, -1 \rangle$, $\mathbf{b} = \langle 3, -1, 2 \rangle$:

• Parametric: $x = 1 + 3\lambda$, $y = 2 - \lambda$, $z = -1 + 2\lambda$.
• Cartesian: $\dfrac{x-1}{3} = \dfrac{y-2}{-1} = \dfrac{z+1}{2}$.

Degenerate case. If, say, $\mathbf{b} = \langle 0, 2, 5 \rangle$, then $x$ is constant ($x = a_1$) and the Cartesian form becomes $x = a_1$, $\dfrac{y - a_2}{2} = \dfrac{z - a_3}{5}$.

Parametric: $x = a_1 + \lambda b_1$, $y = a_2 + \lambda b_2$, $z = a_3 + \lambda b_3$ · Cartesian: $\dfrac{x-a_1}{b_1} = \dfrac{y-a_2}{b_2} = \dfrac{z-a_3}{b_3}$ (only when every $b_i \neq 0$) · $b_i = 0$ $\Rightarrow$ replace that fraction with $x_i = a_i$ · Each fraction equals $\lambda$, that is why they are all equal

Pause, copy the parametric form ($x=a_1+\lambda b_1$ etc.), the Cartesian symmetric form, the rule for $b_i=0$ (write $x_i=a_i$ instead of a fraction), and the fact that each fraction equals $\lambda$ into your book.

We just saw the chain vector → parametric → Cartesian: set $x = a_1+\lambda b_1$, $y = a_2+\lambda b_2$, $z = a_3+\lambda b_3$, then eliminate $\lambda$ to get $(x-a_1)/b_1 = (y-a_2)/b_2 = (z-a_3)/b_3$ (when no $b_i = 0$). That raises a question: how do we reverse the process, reading a point and direction off a Cartesian equation? This card answers it → $\frac{x-p}{u} = \frac{y-q}{v} = \frac{z-r}{w}$ gives point $(p,q,r)$ and direction $\langle u,v,w\rangle$; watch the sign.

Given a symmetric Cartesian equation $\dfrac{x - p}{u} = \dfrac{y - q}{v} = \dfrac{z - r}{w}$, the line passes through $(p, q, r)$ in direction $\langle u, v, w \rangle$. So the equivalent vector form is:

Reading carefully: the numerators give $\mathbf{a}$ (after flipping signs as needed: $x - p$ means $p$, not $-p$). The denominators give $\mathbf{b}$.

Degenerate Cartesian form. If the Cartesian appears as "$x = a_1$, $\dfrac{y - a_2}{b_2} = \dfrac{z - a_3}{b_3}$" then $b_1 = 0$ and the line is parallel to the $yz$-plane (its $x$-coordinate never changes). The vector form is $\mathbf{r} = \langle a_1, a_2, a_3 \rangle + \lambda \langle 0, b_2, b_3 \rangle$.

Point-on-line via Cartesian. A point $(x_0, y_0, z_0)$ lies on $\dfrac{x-a_1}{b_1} = \dfrac{y-a_2}{b_2} = \dfrac{z-a_3}{b_3}$ iff all three fractions evaluate to the same number. This is often quicker than solving the parametric system.

Cartesian $\dfrac{x-p}{u} = \dfrac{y-q}{v} = \dfrac{z-r}{w}$ $\Rightarrow$ point $(p,q,r)$, direction $\langle u, v, w \rangle$ · Watch sign: $x + 3 = x - (-3)$ so $p = -3$ · If one denominator is zero, that coordinate is constant (write it as $x_i = a_i$) · Point on line test: evaluate all three fractions; equal $\Rightarrow$ on the line

Pause, copy the Cartesian → vector reading rule (denominators = direction, constants after $-$ = point), the sign trap ($x+3 \Rightarrow p=-3$), and the point-on-line test (all three fractions equal) into your book.