Linear Algebra Plane Projection

In summary: Then use your formula.In summary, to find the projection of QP onto the plane containing the line l:(x,y,z)= t(6,4,2)+ (3,-4,2) and the point Q(5, -7, 7), the normal vector of the plane must be calculated first. To find this, two independent vectors in the plane are needed, which can be found by taking the cross product of the vector (6,4,2) and the difference between the given point Q and (3,-4,2). Using the dot product, the normal vector is found to be (1,-1,-1). Then, the projection of QP onto the plane can be calculated using the
  • #1
lina29
85
0

Homework Statement


Let 2 be the plane containing the line, l:(x,y,z)= t(6,4,2)+ (3,-4,2) and the point Q(5, -7, 7). Let P be the point (-6, -12,5)
a) Find the projection of QP onto 2.


Homework Equations


I know the projection eqn would be (( plane 2 dot QP)/ magnitude on QP) * components of QP


The Attempt at a Solution



I got the vector QP to be (-11, -5,-2). I'm just confused on have to find the eqn of plane 2
 
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  • #2
Hi lina29! :smile:


I don't understand your relevant equation.
What do you mean by "plane 2"?
And what would the "magnitude on QP" be?



This is what I would do.

First find a normal vector n to plane 2, and then use the projection formula:
[tex]\text{projection of QP on plane 2} = QP - {(QP · n)n \over ||n||^2}[/tex]

To find the normal vector n, first you need 2 independent vectors in plane 2.
You can calculate n for instance by taking their cross product.
 
  • #3
I meant plane 2 as the plane containing the line, l:(x,y,z)= t(6,4,2)+ (3,-4,2). The magnitude of QP would be the sqrt(150). How would I find 2 indpt vectors in plane 2?
 
  • #4
lina29 said:
I meant plane 2 as the plane containing the line, l:(x,y,z)= t(6,4,2)+ (3,-4,2). The magnitude of QP would be the sqrt(150). How would I find 2 indpt vectors in plane 2?

Well, you used the text "plane 2" in your formula in a place where it should be a vector.
But how is a plane a vector?
Is it possible that your relevant equation is for a line instead of a plane?


The first vector is the variable part in your line expression: (6,4,2).
The second vector is the difference between the 2 points in the plane: Q and (3,-4,2).
 
  • #5
I used plane 2 since that's what I initally thought it would be. Obviously, a plane isn't a vector I just wasn't thinking of that when I wrote the equation :). So I should do the dot product of QP and the first vector for projection?
 
  • #6
lina29 said:
I used plane 2 since that's what I initally thought it would be. Obviously, a plane isn't a vector I just wasn't thinking of that when I wrote the equation :). So I should do the dot product of QP and the first vector for projection?

No, not the first vector, but the normal vector (n) which you have yet to calculate.

TBH, I do not know which equations and concepts you know, and which you don't.
Are you familiar with the cross product (as opposed to the dot product)?
Or do you know perhaps how to define a plane by a Cartesian equation?
 
  • #7
So how would I go about calculating the normal vector?
 
  • #8
Let's start with the 2 vectors that are in the plane:

The first vector is (6,4,2).
The second vector is Q(5, -7, 7) - (3,-4,2) = (2, -3, 5).

You need a vector that is perpendicular to both these vectors.
That is the vector that we will call "n".
Being perpendicular means that the dot product is zero.

So let's see if we can find a vector perpendicular to (6,4,2) shall we?
Would (1,-1,-1) do the job?
 
  • #9
yes.
 
  • #10
Soooooo...? :rolleyes:
 
  • #11
lol so I would use that vector for the projection?
 
  • #12
lina29 said:
lol so I would use that vector for the projection?

Well, is it perpendicular to (2, -3, 5) as well?
 
  • #13
I got (-29/3, -19/3, -10/3) which was counted right. Thank you!
I'm also asked to find the distance between P and l. Using the formula
|ax0+by0+cz0+d|/ sqrt(a2+b2+c2).Where a,b,c,d are the variables of the line, and x,y,z are variables of the point P. Would the line I used be 6a+4b+2c=0
 
  • #14
lina29 said:
I got (-29/3, -19/3, -10/3) which was counted right. Thank you!
I'm also asked to find the distance between P and l. Using the formula
|ax0+by0+cz0+d|/ sqrt(a2+b2+c2).Where a,b,c,d are the variables of the line, and x,y,z are variables of the point P. Would the line I used be 6a+4b+2c=0

Oh, okay.

Ah well, the line to be used has (a,b,c)=(6,4,2).
What you wrote is actually a Cartesian plane equation with normal vector (6,4,2) that intersects the origin.
 
  • #15
so how would I find the formula?
 
  • #16
lina29 said:
so how would I find the formula?

Pick a point (x,y,z) on the line and fill it in, in ax+by+cz+d=0.
From this you can deduce d.
(Actually you get a plane that is perpendicular to the line and intersects it at the point (x,y,z).)

Then you need to fill in (x0,y0,z0) which is P - (x,y,z).
 

What is a plane projection in linear algebra?

A plane projection in linear algebra is a method of representing a three-dimensional object on a two-dimensional plane. It involves projecting each point of the object onto the plane along lines that are parallel to a chosen direction.

What is the purpose of performing a plane projection in linear algebra?

The purpose of performing a plane projection in linear algebra is to simplify the representation of a three-dimensional object in a way that is easier to analyze and manipulate. It can also help with visualizing and understanding the geometry of the object.

What are the key concepts involved in plane projection in linear algebra?

The key concepts involved in plane projection in linear algebra include vectors, matrices, and transformations. Vectors are used to represent points and directions in three-dimensional space, while matrices and transformations are used to perform the projection onto the plane.

What are some real-world applications of plane projection in linear algebra?

Plane projection in linear algebra has many real-world applications, including computer graphics, engineering and architectural design, and geographical mapping. It is also used in physics and other sciences to model and analyze three-dimensional systems.

What are the limitations of plane projection in linear algebra?

One of the main limitations of plane projection in linear algebra is that it can result in distortions and inaccuracies when projecting complex three-dimensional objects onto a two-dimensional plane. Additionally, it may not be suitable for representing objects with curved or irregular surfaces.

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