0
$\begingroup$

Thanks for your help, this is related to this question

The calculation is: $\lambda A+(1−\lambda)B$ for $\lambda \in[0,1]$

With $A = (0,0)$ and $B = (2,4)$

I am no mathematician, so if you could show the steps to the solution, I would be most grateful.

Cheers, Rudy

  • 2
    Do you know how to add vectors? Do you know how to multiply by scalars? What is the sticking point? – 2017-01-06
  • 0
    $$P=\lambda A + (1-\lambda) B = \big(\lambda \cdot 0 + (1-\lambda)\cdot 2, \lambda \cdot 0 + (1-\lambda)\cdot 4\big) \;=\; \big(2 - 2 \lambda, 4-4\lambda\big)\,$$ When $\lambda=0\,$ $\,P\equiv B\,$ and when $\lambda =1\,$ $\,P \equiv A$. For $0 \lt \lambda \lt 1\,$ $\,P$ is the point on segment $AB$ which divides it in the ratio $(1-\lambda) : \lambda\,$. – 2017-01-06
  • 0
    So very far out of my depth here. I do not unfortunately know, but will do some research to try and understand. – 2017-01-06

2 Answers 2

0

You can parametrize the segment joining $(0, 0)$ to $(2, 4)$ as $\{(t, 2t) : 0 \leq t \leq 2\}$. Hope this helps!

  • 0
    Thank you Ben, appreciated. – 2017-01-06
0

This is a parameterization of the line segment joining $B$ and $A$. The expression can be rearranged into $B+\lambda(A-B)$, which might be a bit easier to interpret: $(A-B)$ is a direction vector pointing from $B$ to $A$, so you could think of this as describing the motion of a particle that starts at $B$ when $\lambda=0$ and moves at a constant speed toward $A$ until reaching it at $\lambda=1$. If you let $\lambda$ range over all of $\mathbb R$ instead, you’ll get a parameterization of the line through $A$ and $B$.

Another way to think of the expression $\lambda A+(1-\lambda)B$ is as a weighted average of $A$ and $B$, with the weight corresponding to the proportion of the distance between them. This might be a bit more obvious if you write it as $${\lambda A+(1-\lambda)B\over\lambda+(1-\lambda)}.$$ If you take $\lambda=1/2$, for instance, the expression becomes $(A+B)/2$, which you might recognize as a formula for calculating the midpoint of $A$ and $B$.

For your specific example, we have $$\lambda A+(1-\lambda)B=\lambda(0,0)+(1-\lambda)(2,4)=(2-2\lambda,4-4\lambda).$$

  • 0
    Thank you amd, appreciated. – 2017-01-09