I've recently been reading up on vectors, specifically the vector equation of a straight line.
I'm stumped on this question:
Find the vector that is parallel to the line $r = 2i – j + λ(i + 2j)$ and is of magnitude $5\sqrt{5}$.
The answer stated is $5i + 10j$, but I can't arrive there. Here is my method:
$$
r = (2 + λ)i + (-1 + 2λ)j\\
|r| = \sqrt{(2λ)^2 + (-2λ)^2}\\
5√5 = \sqrt{4λ^2 + 4λ^2}\\
125 = 8λ^2\\
λ = \sqrt{\frac{125}{8}}\
$$
$$
r = \left(2 + \sqrt{\frac{125}{8}}\right)i + \left(-1 + 2\sqrt{\frac{125}{8}}\right)j\\
r = \left(2 + \frac{5√5}{4}\right)i + \left(-1 + \frac{10√5}{4}\right)j\\
r = \left(\frac{8 + 5√5}{4}\right)i + \left(\frac{-4 + 10√5}{4}\right)j
$$
Where am I going wrong?
Best Answer
The parametric form of a line can be written like this: $$\mathbf{r(\lambda)}=\mathbf{u}+\lambda\mathbf{v}$$ The line is the set of all points (or position vectors, if you prefer) with this form. You can think of it as describing the motion of an object through space: $\mathbf{r}(\lambda)$ is the location at time $\lambda$.
You can think of $\mathbf{r}(0)=\mathbf{u}$ as the starting point, or the location of the moving object at time $\lambda=0$. At time $\lambda=1$, the object is at $\mathbf{r}(1)=\mathbf{u}+\mathbf{v}$. At time 2, the object is at $\mathbf{r}(2)=\mathbf{u}+2\mathbf{v}$.
It is pretty easy to see that $\mathbf{v}$ can be seen as the velocity and $\mathbf{u}$ some sort of offset. (In fact, it is not wrong to think of velocity as the derivative of position with respect to time $\lambda$ here, just as in one dimensional calculus. In this case, $\mathbf{r}'(\lambda)=\mathbf{v}$, as you'd expect.)
So, $\mathbf{v}$ is the velocity. As with any vector, a velocity vector has two components: the direction and the magnitude (called the speed in this case).
Now we can rephrase the question: all you need to do is find a vector that is pointing in the same direction as the velocity vector, and with a magnitude of $5\sqrt{5}$. To do this, the main thing is to find the velocity vector. In other words, we have to put your original position function in the form $\mathbf{u}+\lambda\mathbf{v}$.
I'll leave this for you to finish, except to note that $\mathbf{u}$ and $\mathbf{v}$ are both going to be linear combinations of the basis vectors $\mathbf{i}$ and $\mathbf{j}$.