1. ## Further Manipulation: Vectors

How was:
$\b{r} = \left(3\b{i}+1\b{j}-\frac72\b{k}\right)+s \left(8\b{i}-4\b{j}+5\b{k}\right)$

Manipulated to:
$\b{r} = \left(7\b{i}-1\b{j}-1\b{k}\right)+\lambda \left(8\b{i}-4\b{j}+5\b{k}\right)$

2. Originally Posted by Air
How was:
$\b{r} = \left(3\b{i}+1\b{j}-\frac72\b{k}\right)+s \left(8\b{i}-4\b{j}+5\b{k}\right)$

Manipulated to:
$\b{r} = \left(7\b{i}-1\b{j}-1\b{k}\right)+\lambda \left(8\b{i}-4\b{j}+5\b{k}\right)$
Make the replacement $s = \lambda + \frac{1}{2}$.

Since both $s$ and $\lambda$ are any real number, you get the same $\vec{r}$ using either one.