Simple Linear Algrebra find a point

In summary, the problem is to find the point R, which is 1/4 of the way from Q to P given the points P(-4,-5,0) and Q(5,5,-3). The direction vector PQ is calculated as [-9,-10,3] and the correct way to find R is to add 1/4 of the direction vector PQ to the point Q, resulting in the coordinates (2.75, 2.5, -2.25) for R. It is important to note that the direction vector should be calculated from Q to P, not from P to Q.
  • #1
krisrai
15
0

Homework Statement


Given the two points P(-4,-5,0) and Q(5,5,-3) find the point R that is 1/4 of the way from Q to P.


Homework Equations



PQ = Point Q- Point P

The Attempt at a Solution


I got my direction vector as
PQ= [9,10,-3]

but i don't know how to find R

All i know is that the answer is supopsed to be:
R(2.75,2.5,-2.25)

Im wanted to add 1/4 of my direction vector PQ to my point q but this doesn't give me the right answer
 
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  • #2
Your vector is from P to Q. You want a vector from Q to P (so, QP=[-9,-10,3]).
Then, R = Q + 1/4*QP.
 
  • #3
You are on your way to the right answer!
When you calculate your direction vector [tex] \vec{PQ} [/tex] you actually get a vector which starts at point (0.0) and ends at the vector coordinates you get which in your case are (9,10,-3)
But your point Q is not (0,0), isn't it?
 

1. What is simple linear algebra and how is it used to find a point?

Simple linear algebra is a branch of mathematics that deals with the study of linear equations and their properties. It is used to find a point by using the concept of coordinates and equations to solve for the unknown variables.

2. What is a point in linear algebra?

A point in linear algebra is a specific location in a coordinate system that is identified by its position along the x and y axes. It is represented by an ordered pair (x,y) where x represents the horizontal position and y represents the vertical position.

3. How do you find a point using simple linear algebra?

To find a point using simple linear algebra, you need to have a linear equation in the form of y = mx + b, where m is the slope and b is the y-intercept. Plug in the known values for x and y into the equation and solve for the unknown variable to determine the point's coordinates.

4. Can simple linear algebra be used to find a point in higher dimensions?

Yes, simple linear algebra can be extended to higher dimensions, such as three-dimensional space. In this case, a point would be represented by an ordered triple (x,y,z) where x, y, and z represent the coordinates along the x, y, and z axes, respectively.

5. Why is simple linear algebra important for finding points?

Simple linear algebra is important for finding points because it provides a mathematical framework for solving linear equations and determining the coordinates of a point. This is especially useful in fields such as physics, engineering, and computer graphics where precise measurements and calculations are necessary.

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