Parametrizing a triangle in $\mathbb{R}^3$?

Question

I have the points $a = (1,2,3),b = (1,1,1), c = (3,1,2).$
I know this is the region S such that
$S = a + \lambda (b - a) + \mu (c-a)$ for some $\lambda,\mu \in \mathbb{R}$.
How would I find the restrictions on $\lambda, \mu$?

Answer 1

We rewrite an element of $S$ as $$ a+\lambda(b-a)+\mu(c-a)=(1-\lambda-\mu)a+\lambda b+\mu c. $$ Since $S$ is the triangle generated by $a,b,c$, we have $\lambda,\mu, \lambda+\mu\in [0,1]$. Indeed, the triangle generated by $a,b,c$ is given by \begin{eqnarray*} S&=&\{\alpha a+\beta b+\gamma c:\alpha+\beta+\gamma=1,\alpha,\beta,\gamma\in[0,1]\}\\ &=&\{(1-\beta-\gamma) a+\beta b+\gamma c:1-\beta-\gamma,\beta,\gamma\in[0,1]\}\\ &=&\{(1-\beta-\gamma) a+\beta b+\gamma c:\beta+\gamma,\beta,\gamma\in[0,1]\}. \end{eqnarray*}

Answer 2

You can start with a segment.

If you have two points $A$ and $B$, take the vector $\vec {AB}$ and then any vector $$\vec{AP}=\alpha \vec{AB}$$

with $0\le \alpha\le 1$ have a starting point at $A$ (obviously) and a ending point between $A$ and $B$ (because $0\le \alpha\le 1$). So,

$$P-A=\alpha(B-A)\to P=(1-\alpha)A+\alpha B$$

Now let's see what happen with a triangle. See the picture:

Taking to account what we just said, we have that

$$D=(1-\alpha)B+\alpha C$$

and any point $P$ in $AD$ is given by:

$$P=(1-\beta)A+\beta D$$

so,

$$P=(1-\beta)A+\beta [(1-\alpha)B+\alpha C]$$

$$P=(1-\beta)A+[\beta (1-\alpha)]B+\beta\alpha C$$

we then can call

$\lambda_1=1-\beta$

$\lambda_2=\beta (1-\alpha)$

$\lambda_3=\beta \alpha$

and see that $\lambda_1,\lambda_2,\lambda_3\in [0,1]$ and $\lambda_1+\lambda_2+\lambda_3=1$

$$P=\lambda_1A+\lambda_2 B+\lambda_3 C$$

Speaking of your case

$$S = a + \lambda (b - a) + \mu (c-a)=(1-\lambda-\mu)a+\lambda b+\mu c$$

Can you find out the conditions for $\lambda$ and $\mu$?