# Intuition behind equation of line in Vector form

I'm new to Vector and 3D Geometry and today I was taught the equation of a line passing through two points $$A(\vec{a})$$ and $$B(\vec{b})$$ as $$\vec{r}=\vec{a} + λ(\vec{b}-\vec{a})$$

It is difficult for me to imagine how this works. Why do we have a $$λ$$ here? What I can think of is that $$\vec{r}=\vec{a} + λ(\vec{b}-\vec{a}$$) means that if we draw a line passing through $$A$$ and $$B$$ and we plug in different values of $$λ$$, we will get the position vector of different points that lie on that line. Am i right in thinking so? So this is a bit different from the stuff we use in $$2D$$ geometry or other forms like expressing the line in terms of $$x,y,z$$?

• Yes you are right in your thinking. $\lambda=0$ corresponds to $A(\vec{a})$ and if you move parallel to a vector $\vec{b}$ you will get the line $\vec{r}=\vec{a}+\lambda \vec{b}$ Oct 20 at 15:29
• That equation isn't quite right. Oct 20 at 15:30
• @PM2Ring oops, my bad, I did correct it now.
– Vega
Oct 20 at 16:21
• That's much better. :) Oct 20 at 16:22
• FWIW, another useful way to write this is $$\vec{r}=(1-\lambda)\vec{a} + \lambda\vec{b}$$ Oct 20 at 16:35

Here is a visual explanation, for vectors in $$\mathbb{R}^2$$. The vectors $$\vec{a}$$ and $$\vec{b}$$ fix two points in space. Consider the difference $$\vec{b}-\vec{a}$$ shown by the red arrows. Then, we move a certain number of lengths of $$\vec{b}-\vec{a}$$ (given in the picture by $$\lambda=5$$) along the direction of $$\vec{b}-\vec{a}$$. The resulting vector $$\vec{a}+\lambda(\vec{b}-\vec{a})=\vec{a}+5(\vec{b}-\vec{a})$$ has its tip on a line that clearly goes through the tips of $$\vec{a}$$ and $$\vec{b}$$ and is parallel to $$(\vec{b}-\vec{a})$$.

In general, $$\lambda$$ can be anything, so there can be any number of $$\vec{b}-\vec{a}$$'s, including fractional amounts, added to $$\vec{a}$$. This traces out the entire line. • +1, Awesome visual representation. So you basically have $\vec{a}$ and then $\vec{b}-\vec{a}$ is a vector parallel to the line joining $A$ and $B$ and this multiplied by $λ$ gives different vectors along the line which when added with $\vec{a}$ by triangular law of addition gives the position vector of different points on the line. Did I get it correct?
– Vega
Oct 20 at 16:26
• @Vega I updated me answer to correspond to your updated post. Oct 20 at 16:31
• And yes the original (typo) basically corresponds to the equation of a line passing through A and parallel to a vector $\vec{b}$
– Vega
Oct 20 at 16:37
• You got it correct. Oct 20 at 16:38

Remember, Euclidean vectors are not points; $$\begin{pmatrix} 1 \cr 1 \cr 1\end{pmatrix}$$ is neither the point $$(1,1,1)$$ nor affixed between the origin and $$(1,1,1);$$ rather it is "portable" and represents a particular translation in space.

The parametric equation$$\vec{r}=\vec{a} + λ\vec{u}$$ of a line is actually very intuitive: it is saying that to land on any point $$(x,y,z)$$ on the line, start at the origin and move directly to point $$A$$ (after all, $$\vec{a}$$ is just $$A$$'s position vector), then move along $$\vec{u}$$ by a $$some$$ (negative/zero/positive) multiple of its length.

Then, the position vector $$\vec{r}=\begin{pmatrix} x \cr y \cr z\end{pmatrix}$$ of that point is given by the vector sum of these two movements/translations.

There is one parameter҂, $$\lambda,$$ because a line has one degree of freedom҂: the real-valued multiple that I mentioned above is the value of $$\lambda$$ that lands us on that particular point (and which gives its particular position vector $$\vec{r}$$). As $$\lambda$$ varies over $$\mathbb R,$$ every single point on that line will have been covered.

҂ Lines (including curved lines) in $$3$$-D space are nonetheless $$1$$-dimensional objects in the sense that when constrained to the line, a single number (parameter) is sufficient to fully specify the position of each point. This is why their parametric equations (whether in $$2$$-D space or $$3$$-D space) contain only a single parameter. Planes, on the other hand, have two parameters in their parametric equation $$\vec{r}=\vec{a} + λ\vec{u}+\gamma\vec{v}$$.
• @RyanG Yeah, either way, in both questions, the concept was the same, which I understood
– Vega
Oct 20 at 17:18

Rather than looking at a line lets examine a line segment beginning at the point represented by a position vector $$p$$ and in the direction of some other vector $$x$$ that tells us the orientation and length of the segment. Geometrically this means we simple add $$p + x$$ to find where the line segment ends, but what if I want to label every position on the line segment? Well, looking at the half way point along the segment we see that it's at position $$p + \frac{1}{2}x$$ and a quarter is $$p + \frac{1}{4}x$$. Examining other points on the segment we see that as long as we pick some scalar multiple of $$x$$ between zero and one it represents a point on the segment. This gives us the geometric intuition as to why we can now parameterize the line segment as $$p + \lambda x$$ for some real number $$0 \leq \lambda \leq 1$$. Now that we have the segment parameterized all we have to do to get the entire line is let $$\lambda$$ take on any real value. Note that the dimension of the space was irrelevant here as everything was happening in the plane containing $$p$$ and $$x$$ so the two dimensional case is sufficient for geometric purposes.