- #1

Guest2

- 193

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- MHB
- Thread starter Guest2
- Start date

In summary, the position vector of point $B$ lies on the line segment between $P$ and $Q$, given by the equation $(x,y,z) = tQ + (1-t)P,\ 0 \leq t \leq 1$, and can be found using the distance formula by setting the distance from $P$ equal to twice the distance from $Q$. By choosing a value of $t$ that satisfies this equation, we can find the position vector of $B$, which is $\left(\frac{7}{3}, \frac{4}{3}, -\frac{1}{3}\right)$.

- #1

Guest2

- 193

- 0

Mathematics news on Phys.org

- #2

Deveno

Science Advisor

Gold Member

MHB

- 2,726

- 6

$(x,y,z) = tQ + (1-t)P,\ 0 \leq t \leq 1$ (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).

- #3

Guest2

- 193

- 0

Deveno said:

$(x,y,z) = tQ + (1-t)P,\ 0 \leq t \leq 1$ (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 $\vec{B} = \left(\frac{7}{3}, \frac{4}{3}, -\frac{1}{3}\right)$

To find the position vector of point B on a line segment between P and Q, you can use the formula: *Position vector of B = Position vector of P + (Distance from P to B) * (Unit vector in the direction of PQ)*. This formula takes into account the distance from P to B and the direction of the segment from P to Q.

A position vector is a mathematical tool used to describe the location of a point in a coordinate system. It consists of a magnitude (length) and a direction, and is usually represented by an arrow pointing from the origin to the point.

Yes, you can find the position vector using the coordinates of P and Q. First, you will need to determine the distance from P to B and the unit vector in the direction of PQ. Then, you can use the formula mentioned in the first question to calculate the position vector of B.

The purpose of finding the position vector is to accurately describe the location of a point on a line segment. This can be useful in many applications, such as in physics and engineering, where precise measurements and calculations are necessary.

Yes, the position vector of B can still be calculated even if the line segment is not straight. However, the formula mentioned in the first question may need to be modified depending on the shape and direction of the line segment. For example, if the segment is curved, you may need to use calculus to determine the unit vector in the direction of PQ.

- Replies
- 9

- Views
- 1K

- Replies
- 1

- Views
- 1K

- Replies
- 4

- Views
- 1K

- Replies
- 22

- Views
- 668

- Replies
- 3

- Views
- 1K

- Replies
- 1

- Views
- 1K

- Replies
- 1

- Views
- 2K

- Replies
- 5

- Views
- 2K

- Replies
- 2

- Views
- 2K

- Replies
- 20

- Views
- 1K

Share: