I was studying Gilbert Strang's linear algebra book (An Introduction to Linear Algebra (4e)) and came across a problem that I couldn't understand. I've noticed someone's asked the same question here but the only answer didn't address the question.

The "figure" that I'm referring to in the title is this:

enter image description here

Locate $\frac{1}{3}\vec{u} + \frac{1}{3}\vec{v} + \frac{1}{3}\vec{w}$ and $\frac{1}{2}\vec{u} + \frac{1}{2}\vec{w}$ on the figure. Challenge problem: Under what restrictions on $c$, $d$, $e$ will the combinations $c\vec{u} + d\vec{v} + e\vec{w}$ fill in the dashed triangle?

It's not hard finding answers to problems, but even with the answer I cannot grasp the solution. The answer is that "$\frac{1}{3}\vec{u} + \frac{1}{3}\vec{v} + \frac{1}{3}\vec{w}$ is the center point of the dashed triangle," "$\frac{1}{2}\vec{u} + \frac{1}{2}\vec{w}$ is the point between $\vec{u}$ and $\vec{w}$."

I can somewhat understand how the second answer came to be, but I'm really having difficulty understanding how the sum of vector thirds leads to the center point of the triangle.

Would anybody kind enough to explain?

Thank you.

  • 1
    $\begingroup$ You can try at least two things: 1) think of $u/3+v/3+w/3=u/3+2/3(v/2+w/2)$; 2) the scalars $c,d,e$ must satisfy an equation so that the linear combination lies on the plane; that equation will in fact be of the type $\alpha c+\beta d+\gamma e=1$ with $\alpha,\beta,\gamma$ known numbers. $\endgroup$ – Chrystomath Feb 6 at 8:58
  • $\begingroup$ Hello, thanks for the comment. May I ask, why do those three scalars sum up to $1$? $\endgroup$ – Seankala Feb 6 at 13:37

It depends on what you mean by "the centre" of a triangle. There are thousands of points which people call a "centre" of a triangle. In this case, the "centre" being referred to is the Centroid.

The most common characterisation of the centroid is the common points of intersection between the lines from each vertex to the midpoint of its opposite side. The midpoint of $u$ and $v$ is $\frac{u}{2} + \frac{v}{2}$, so the line through $w$ and $\frac{u}{2} + \frac{v}{2}$ is given by $$x(t) = w + t\left(\frac{u}{2} + \frac{v}{2} - w\right).$$ Note that $$x\left(\frac{2}{3}\right) = \frac{u}{3} + \frac{v}{3} + \frac{w}{3}.$$ Also note that this expression is completely symmetric, so doing the same with, say, $v$ and $\frac{u}{2} + \frac{w}{2}$ will produce a line with the same point on it. Therefore, all three possible lines will all pass through this one point.

  • $\begingroup$ Hello thanks for the answer. You've cleared up the issue quite nicely. It may seem like a bit of a dumb question, but how did you come up with the equation for the line? $\endgroup$ – Seankala Feb 6 at 15:01
  • 2
    $\begingroup$ @Seankala, if $S$ is the starting point of a line segment and $T$ is the target (endpoint), moving from $S$ to $T$ means starting at $S$ and moving in the direction $T-S$. Every point on the line segment can be expressed as $S+\lambda(T-S)$ with a $\lambda \in [0,1]$, the starting point $S$ has $\lambda=0$ and the target point $T$ has $\lambda=1$. In this special case, the line segment starts at one of the corners ($w$) and ends on the midpoint of the opposite side ($\frac u2+\frac v2$). $\endgroup$ – Wolfgang Kais Feb 6 at 15:22
  • $\begingroup$ Hello @WolfgangKais thanks for the answer, it cleared up a lot. I was reviewing this question today and noticed that the result seems to differ depending on what we choose as the starting point. Is this correct? $\endgroup$ – Seankala Feb 11 at 9:17
  • 1
    $\begingroup$ @Seankala The center of the triangle does not depend on the corner you start with, check this out using $\lambda=2/3$ und you will allways get that same amazing result. :-) $\endgroup$ – Wolfgang Kais Feb 11 at 10:13

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