Let P and Q be two points and let R be the point on PQ whose distance from P is twice its distance from Q.Verify that $w= \frac{1}{3}u + \frac{2}{3}v$

Question

Let $P$ and $Q$ be two points in space and let $R$ be the point on $PQ$ whose distance from $P$ is twice its distance from $Q$. If $U=OP$, $v=OQ$ and $W=OR$. Verify that $w= \frac{1}{3}u + \frac{2}{3}v$

I tried the following:

• $P=(p_1,p_2)$
• $Q=(q_1,q_2)$
• $R=(r_1,r_2)$
• Im assuming O as the origin, but im not sure.

$w= \frac{1}{3}u + \frac{2}{3}v$

$\implies 3\sqrt{r_1^2 +r_2^2}=\sqrt{p_1^2 +p_2^2} +2\sqrt{q_1^2 +q_2^2}$

Also, I know that:

$||QR||+||RP||=||QP||$ and $||QP||=3||QR||$

So, from this equality I tried to build what should be verified. However, I don't come up with a result that works for me.

Answer 1

$$\vec{OP} = \textbf u, \vec{OQ} = \textbf v, \vec{OQ} = \textbf w$$

$ \textbf w = \vec{OR} = \vec{OP} + \vec{PR} = \textbf u + \frac{2}{3}\vec{PQ} \tag1$

$$\vec{OP} + \vec{PQ} = \vec{OQ} \implies \vec{PQ} = \vec{OQ} - \vec{OP} = \textbf v - \textbf u$$

So substituting in $(1)$,

$$ \textbf w = \frac{1}{3} \textbf u + \frac{2}{3} \textbf v $$

Answer 2

$A, B\in \mathbb {R}^n$, if $C=nA+(1-n)B$ then $\|C-A\|=\|(n-1)A-(n-1)B\|=|n-1|\|A-B\|$. The statement is an if and only if but it should be clear, $C\equiv R$, $P\equiv A$, $Q\equiv B$ and $n=1/3$.

Answer 3

If you want to use your initial Cartesian coordinate approach:

Let $\vec{u} = OP = (u_x, u_y)$, $\vec{v} = OQ = (v_x, v_y)$, and $\vec{w} = OR = (w_x, w_y)$.

Let's assume $\vec{w} = \frac{1}{3} \vec{u} + \frac{2}{3} \vec{v}$, i.e. $$\left\lbrace \begin{aligned} w_x &= \frac{1}{3} u_x + \frac{2}{3} v_x = \frac{u_x + 2 v_x}{3} \\ w_y &= \frac{1}{3} u_y + \frac{2}{3} v_y = \frac{u_y + 2 v_y}{3} \\ \end{aligned} \right.$$ The distance between $P$ and $R$ is $$\begin{aligned} \lVert R - P \rVert & = \lVert OR - OP \rVert = \lVert \vec{w} - \vec{u} \rVert \\ ~ & = \sqrt{(w_x - u_x)^2 + (w_y - u_y)^2} \\ ~ & = \sqrt{\left(\frac{u_x + 2 v_x - 3 u_x}{3}\right)^2 + \left(\frac{u_y + 2 v_y - 3 u_y}{3}\right)^2} \\ ~ & = \sqrt{\left(\frac{2 v_x - 2 u_x}{3}\right)^2 + \left(\frac{2 v_y - 2 u_y}{3}\right)^2} \\ ~ & = \sqrt{\frac{2^2}{3^2}\biggl( (v_x - u_x)^2 + (v_y - u_y)^2 \biggr)} \\ ~ & = \frac{2}{3}\sqrt{(v_x - u_x)^2 + (v_y - u_y)^2} \\ ~ & = \frac{2}{3}\sqrt{(u_x - v_x)^2 + (u_y - v_y)^2} \\ ~ & = \frac{2}{3} \lVert \vec{v} - \vec{u} \rVert = \frac{2}{3} \lVert PQ \rVert \end{aligned}$$ Similarly, the distance between $Q$ and $R$ is $$\begin{aligned} \lVert R - Q \rVert & = \lVert OR - OQ \rVert = \lVert \vec{w} - \vec{v} \rVert \\ ~ & = \sqrt{(w_x - v_x)^2 + (w_y - v_y)^2} \\ ~ & = \sqrt{\left(\frac{u_x + 2 v_x - 3 v_x}{3}\right)^2 + \left(\frac{u_y + 2 v_y - 3 v_y}{3}\right)^2} \\ ~ & = \sqrt{\left(\frac{u_x - v_x}{3}\right)^2 + \left(\frac{u_y - v_y}{3}\right)^2} \\ ~ & = \sqrt{\frac{1}{3^2} \biggl( ( u_x - v_x )^2 + (u_y - v_y)^2 \biggr) } \\ ~ & = \frac{1}{3}\sqrt{(u_x - v_x)^2 + (u_y - v_y)^2} \\ ~ & = \frac{1}{3} \lVert \vec{v} - \vec{u} \rVert = \frac{1}{3} \lVert PQ \rVert \\ \end{aligned}$$ i.e. $\lVert R P \rVert = 2 \lVert R Q \rVert$.

(In case you wonder about the notation, I've used upper case letters $O$, $P$, $Q$, and $R$ for position vectors, and $\vec{u}$, $\vec{v}$, and $\vec{w}$ for displacement vectors. Furthermore, displacement vector $OP$ corresponds to the difference in their position vectors, $P - O$; thus $PQ = Q - P = (OQ) - (OP) = (Q - O) - (P - O) = Q - P$, and $O$ is actually irrelevant. However, in vector algebra, $O$ should refer to origin, unless explicitly defined otherwise. I personally do not prefer to differentiate between position and displacement vectors – they're both vectors with the exact same rules – but the way the question is stated, I thought keeping the distinction might help understand the problem and the solution.)