# Finding the projection of a vector.

#### Jundoe

##### New member
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
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
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
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! 