Q:

Mark the points $\frac{3}{4}v + \frac{1}{4}w$ from $\frac{1}{2}v + \frac{1}{2}w$

Recently, I do self-learning "Linear Algebra" by using this book "Introduction to Linear Algebra, 3rd Edition" by Gilbert Strang with his lecture on MIT Opencourseware. I am having problem with one of his problem in his book. The following is the problem description.

The figure shows $\frac{1}{2}v + \frac{1}{2}w$. Mark the points
  $\frac{3}{4}v + \frac{1}{4}w$ and $\frac{1}{4}v + \frac{1}{4}w$ and $v + w$.  

For $\frac{1}{4}v + \frac{1}{4}w$ and $v + w$ are very easy. The first one is half of vector u and the second one is double of vector u. The problem is $\frac{3}{4}v + \frac{1}{4}w$.

Here is my attempt:
First, I double the vector u, so $2u = v + w$. Then, I can construct vector v
by connecting from vector w to vector 2u.

Then, I plot $\frac{1}{4}w$ on the vector w and $\frac{3}{4}v$ on the vector v. Connecting those two point and I get $\frac{3}{4}v + \frac{1}{4}w$. See the picture below.

Questions

Do I have the correct reasoning?
Is there any better way to plot $\frac{3}{4}v + \frac{1}{4}w$?

A:

Seems O.K., as far as the vectors go. However, since the problem says "mark the points", it may be that you're expected to draw a vector in a way that it originates at $(0,0)$. You can achieve this by translating your solution.
As for the better way, I'd say no. You obviously need $v$, and you did manage to get it quite easily.

