Equation of a Line in 3D — Vector Form

Let $A$ be a known point on the line with position vector $\mathbf{a} = \overrightarrow{OA}$, and let $\mathbf{b}$ be a non-zero vector parallel to the line. If $R$ is a general point on the line with position vector $\mathbf{r} = \overrightarrow{OR}$, then $\overrightarrow{AR}$ is parallel to $\mathbf{b}$, so $\overrightarrow{AR} = \lambda\mathbf{b}$ for some scalar $\lambda$. Then:

Through two points $A$ and $B$: take $\mathbf{a} = \overrightarrow{OA}$ and direction $\mathbf{b} = \overrightarrow{AB} = \mathbf{B} - \mathbf{A}$. Then $\lambda = 0$ gives $A$ and $\lambda = 1$ gives $B$. Worked through the hook with $A(1,2,3)$ and $B(4,0,5)$:

• $\mathbf{a} = \langle 1, 2, 3 \rangle$, $\mathbf{b} = \mathbf{B} - \mathbf{A} = \langle 3, -2, 2 \rangle$.
• Vector equation: $\mathbf{r} = \langle 1, 2, 3 \rangle + \lambda \langle 3, -2, 2 \rangle$.
• Check: $\lambda = 0 \Rightarrow \mathbf{r} = \langle 1, 2, 3 \rangle = A$. $\lambda = 1 \Rightarrow \mathbf{r} = \langle 4, 0, 5 \rangle = B$. ✓
• Equally valid: $\mathbf{r} = \langle 4, 0, 5 \rangle + \mu \langle -3, 2, -2 \rangle$ — same line, different $\mathbf{a}$ and $\mathbf{b}$.

Vector equation of a line: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$, $\lambda \in \mathbb{R}$ · $\mathbf{a}$ = position vector of any point on the line; $\mathbf{b}$ = direction vector (any non-zero parallel vector) · Through $A$ and $B$: $\mathbf{b} = \mathbf{B} - \mathbf{A}$, so $\lambda = 0$ gives $A$ and $\lambda = 1$ gives $B$

Pause — copy the vector line equation $\mathbf{r} = \mathbf{a}+\lambda\mathbf{b}$, the roles of $\mathbf{a}$ (point) and $\mathbf{b}$ (direction), and the two-point construction $\mathbf{b} = \mathbf{B}-\mathbf{A}$ into your book.

We just saw the vector equation of a line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$, where $\mathbf{a}$ is any point on the line and $\mathbf{b}$ is the direction vector, with $\lambda = 0$ and $\lambda = 1$ giving the two defining points. That raises a question: two lines with proportional direction vectors might be the same line or distinct parallel lines — how do we tell? This card answers it → check whether any point of one line satisfies the equation of the other; consistent $\lambda$ means same line, no solution means parallel and distinct.

The vector equation of a line is not unique. Two equations $\mathbf{r} = \mathbf{a}_1 + \lambda\mathbf{b}_1$ and $\mathbf{r} = \mathbf{a}_2 + \mu\mathbf{b}_2$ describe the same line iff:

• $\mathbf{b}_1$ and $\mathbf{b}_2$ are parallel: $\mathbf{b}_1 = k\mathbf{b}_2$ for some non-zero $k$, AND
• The point with position vector $\mathbf{a}_2$ lies on the first line (or equivalently $\mathbf{a}_1$ on the second).

If the direction vectors are parallel but neither line contains the other's point, the lines are parallel but distinct. If the direction vectors are not parallel, the lines either intersect or are skew (the latter only in 3D).

Testing if a point $P$ is on a line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$: solve $\mathbf{p} = \mathbf{a} + \lambda\mathbf{b}$ component by component. If all three components give the same $\lambda$, the point lies on the line; otherwise it doesn't.

Lines parallel iff $\mathbf{b}_1 = k\mathbf{b}_2$ for some scalar $k \neq 0$ · Same line: directions parallel AND a point of one lies on the other · Parallel but distinct: directions parallel BUT no shared point · Point-on-line test: solve $\mathbf{p} = \mathbf{a} + \lambda\mathbf{b}$; need consistent $\lambda$ in all three components

Pause — copy the parallel test ($\mathbf{b}_1 = k\mathbf{b}_2$), the same-line test (directions parallel AND a point of one lies on the other), and the point-on-line method (solve $\mathbf{p} = \mathbf{a}+\lambda\mathbf{b}$ for consistent $\lambda$) into your book.