MHB Finding the Position Vector of Point $B$ on a Line Segment Between $P$ and $Q$

[Guest2]

Let $P = (1,2,-1)$ and $Q = (3,1,0)$. A point $B$ lies on the line segment between $P$ and $Q$. Given that its distance from $P$ is twice its distance from $Q$. Find the position vector of $B$.

 

[Deveno]

The equation for a line segment is very similar to that for a line, it is:

$(x,y,z) = tQ + (1-t)P,\ 0 \leq t \leq 1$ (this one goes FROM $P$ TO $Q$, as $t$ goes from 0 to 1, if we switch $P$ and $Q$ it goes the other way).

In our case, this is $(2t+1,2-t,t-1)$ with $0 \leq t \leq 1$.

Suppose we choose $0 < t_0 < 1$. Let's use the standard distance formula, to calculate the distance of the point $B$ corresponding to $t_0$.

We have $d(P,B) = \sqrt{(1-(2t_0+1))^2 + (2-(2-t_0))^2 + (-1-(t_0-1))^2}$

$= \sqrt{4t_0^2 + t_0^2 + t_0^2} = \sqrt{6}t_0$

and $d(Q,B) = \sqrt{(3-(2t_0+1))^2 + (1-(2-t_0))^2 + (0-(t_0 - 1))^2}$

$= \sqrt{(2-2t_0)^2 + (-1+t_0)^2 + (t_0-1)^2} = \sqrt{6}(1-t_0)$ (we use $1-t_0$ since we want the positive square root).

Can you continue?

(Intuitively, how far along the interval from 0 to 1 do you have to be, to where how far you have to go is half of how far you've come? Use this to check your answer).

 

[Guest2]

The equation for a line segment is very similar to that for a line, it is:

$(x,y,z) = tQ + (1-t)P,\ 0 \leq t \leq 1$ (this one goes FROM $P$ TO $Q$, as $t$ goes from 0 to 1, if we switch $P$ and $Q$ it goes the other way).

In our case, this is $(2t+1,2-t,t-1)$ with $0 \leq t \leq 1$.

Suppose we choose $0 < t_0 < 1$. Let's use the standard distance formula, to calculate the distance of the point $B$ corresponding to $t_0$.

We have $d(P,B) = \sqrt{(1-(2t_0+1))^2 + (2-(2-t_0))^2 + (-1-(t_0-1))^2}$

$= \sqrt{4t_0^2 + t_0^2 + t_0^2} = \sqrt{6}t_0$

and $d(Q,B) = \sqrt{(3-(2t_0+1))^2 + (1-(2-t_0))^2 + (0-(t_0 - 1))^2}$

$= \sqrt{(2-2t_0)^2 + (-1+t_0)^2 + (t_0-1)^2} = \sqrt{6}(1-t_0)$ (we use $1-t_0$ since we want the positive square root).

Can you continue?

(Intuitively, how far along the interval from 0 to 1 do you have to be, to where how far you have to go is half of how far you've come? Use this to check your answer).

Thank you!

Since it's distance from $P$ is twice its distance from $Q$, it's

$\sqrt{6}t_0 = 2 \sqrt{6}(1-t_0) \implies t_0 = \frac{2}{3}$, so $\vec{B} = \left(\frac{7}{3}, \frac{4}{3}, -\frac{1}{3}\right)$