Welcome to our community

Be a part of something great, join today!

Finding the projection of a vector.

Jundoe

New member
Nov 23, 2013
10
I would like to verify this problem from an introductory to Linear Algebra course.

It goes as follows:

Let L be the line with parametric equations x=2+3t, y=1-2t, z=-2+t, and let v=(3,2,2). Find vectors w1 and w2 such that v=w1+w2, and such that w1 is parallel to L and w2 is perpendicular to L.
This is how I proceeded:

From the given parametric equations I constructed the vectors:
line L: a=(3, -2, 1) and b=(2,1,-2).

To find w1, I know that w1= kL

And to find k: (v.L)/||L||2
And w2 is just a matter of: w2=v-w1

The issue I am facing is, which vector do I chose for the L?
I have found 2 vectors from the parametric equations.
Should I simply take the difference? a-b= (1,-3,3)?

Thank You.

[edit.]

If I stick to my L line being equal to (1,-3,3), due to the fact that b is my position vector, then:

w1 = 3/19(1,-3,3)
w2 = (3,2,2) - 3/19(1,-3,3)

Am I completely off?
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
I would like to verify this problem from an introductory to Linear Algebra course.

It goes as follows:

Let L be the line with parametric equations x=2+3t, y=1-2t, z=-2+t, and let v=(3,2,2). Find vectors w1 and w2 such that v=w1+w2, and such that w1 is parallel to L andw2 is perpendicular to L.
This is how I proceeded:

From the given parametric equations I constructed the vectors:
line L: a=(3, -2, 1) and b=(2,1,-2).

To find w1, I know that w1= kL

And to find k: (v.L)/||L||2
And w2 is just a matter of: w2=v-w1

The issue I am facing is, which vector do I chose for the L?
I have found 2 vectors from the parametric equations.
Should I simply take the difference? a-b= (1,-3,3)?

Thank You.
Hi Jundoe, and welcome to MHB!

You have got the equation of the line L as $(x,y,z) = \mathbf{b} + \mathbf{a}t.$ In that equation, $\mathbf{b}$ (the constant) is a point on the line, and $\mathbf{a}$ (the coefficient of $t$) gives the direction of the line. So you want to take $\mathbf{a}$ as the parameter for $\mathbf{w}_1$, because you want $\mathbf{w}_1$ to point in the same direction as L.
 

Jundoe

New member
Nov 23, 2013
10
Hi Jundoe, and welcome to MHB!

You have got the equation of the line L as $(x,y,z) = \mathbf{b} + \mathbf{a}t.$ In that equation, $\mathbf{b}$ (the constant) is a point on the line, and $\mathbf{a}$ (the coefficient of $t$) gives the direction of the line. So you want to take $\mathbf{a}$ as the parameter for $\mathbf{w}_1$, because you want $\mathbf{w}_1$ to point in the same direction as L.
Thank you for replying! So in other words, the position of the vector–its origin–is irrelevant in this matter?

So, ignoring the point. I would have the following:

w1= 1/2(3,-2,1)
w2= (3,2,2)-1/2(3,-2,1)= (3/2, 3, 3/2)

Hope I didn't do any careless mistakes, does that seem about right?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Thank you for replying! So in other words, the position of the vector–its origin–is irrelevant in this matter?

So, ignoring the point. I would have the following:

w1= 1/2(3,-2,1)
w2= (3,2,2)-1/2(3,-2,1)= (3/2, 3, 3/2)

Hope I didn't do any careless mistakes, does that seem about right?
Correct! (Yes)