# Vector / Scalar Projection

• April 3rd 2009, 08:59 PM
Russ
Vector / Scalar Projection
This question has me majorly confused. I've been going quite well so far at Calculus (that's the name of the subject) but this question has me completely stumped. Complete answers aren't necessary since I want to get this done myself (i don't know what the rules are about giving out answers) but I'd really appreciate a couple of prods in the right direction.

Here we go:

Let u = (2,1,-3) and v = (1,4,2) be vectors in R3. Calculate and draw sketches to explain:

(a) the scalar projection of u on v
(b) the scalar projection of v on u
(c) the vector projection of u in the direction of v
(d) the vector projection of v in the direction perpendicular to uOkay. I started with (a) and went like this:

scalar projection is v . u where v is a unit vector.
After calculating that dot product as: (v / ||v||) . u
I was surprised to have it equal to 0, which means that the two vectors are perpendicular. Can somebody explain whether that is correct or not and what it means for solving the rest of the questions?

Thanks :)
Also, how can I include maths symbols in posts?
• April 3rd 2009, 09:12 PM
mr fantastic
Quote:

Originally Posted by Russ
This question has me majorly confused. I've been going quite well so far at Calculus (that's the name of the subject) but this question has me completely stumped. Complete answers aren't necessary since I want to get this done myself (i don't know what the rules are about giving out answers) but I'd really appreciate a couple of prods in the right direction.

Here we go:

Let u = (2,1,-3) and v = (1,4,2) be vectors in R3. Calculate and draw sketches to explain:

(a) the scalar projection of u on v
(b) the scalar projection of v on u
(c) the vector projection of u in the direction of v
(d) the vector projection of v in the direction perpendicular to uOkay. I started with (a) and went like this:

scalar projection is v . u where v is a unit vector.
After calculating that dot product as: (v / ||v||) . u
I was surprised to have it equal to 0, which means that the two vectors are perpendicular. Can somebody explain whether that is correct or not and what it means for solving the rest of the questions?

Thanks :)
Also, how can I include maths symbols in posts?

u and v are perpendicular.

Go to this subforum: http://www.mathhelpforum.com/math-help/latex-help/
• April 3rd 2009, 09:15 PM
Russ
Ah excellent. Good to know that they're perpendicular and I'm not just failing lol.
Can you point me in the right direction to find the scalar projections as a result? Is it just 0?

I will check that forum for the math equation things.
• April 3rd 2009, 09:20 PM
mr fantastic
Quote:

Originally Posted by Russ
Ah excellent. Good to know that they're perpendicular and I'm not just failing lol.
Can you point me in the right direction to find the scalar projections as a result? Is it just 0?
[snip]

u.v = 0 therefore ....
• April 3rd 2009, 09:29 PM
Russ
Hmm. Okay, give me a sec here.

The dot product is 0, so they are perpendicular, so cos ( $\theta$) must be equal to $\pi$/2

Is one of them a zero-vector in that case?
• April 3rd 2009, 09:35 PM
mr fantastic
Quote:

Originally Posted by Russ
Hmm. Okay, give me a sec here.

The dot product is 0, so they are perpendicular, so cos ( $\theta$) must be equal to $\pi$/2

Is one of them a zero-vector in that case?

You substitute u.v = 0 into the formulae for the things you've been asked to find.
• April 3rd 2009, 09:35 PM
Russ
Okay I went and checked my formulae and...

the scalar projection of u on to v is equal to: u $\cos (\theta )$

Now, we established that the angle between them is $\pi/2$ because they're perpendicular, so we need to find the length of u, which I can do.

:)
• April 3rd 2009, 09:46 PM
Russ
Okay.

projvu = ||u|| . $\cos ( \theta )$
= ||u|| . 0
= 0
Correct?
if the scalar projection of u onto v isn't 0, then I have no idea how to do this lol.

Hoping for somebody to confirm or deny so I can move onto the next parts.

EDIT: that Latex thing isn't working to produce the correct symbol. it just italicizes the input. Can somebody use code tags to show me the right input?
• April 4th 2009, 04:58 AM
LaTeX
Hello Russ
Quote:

Originally Posted by Russ
Okay.

projvu = ||u|| . $\cos ( \theta )$
= ||u|| . 0
= 0
Correct?
if the scalar projection of u onto v isn't 0, then I have no idea how to do this lol.

Hoping for somebody to confirm or deny so I can move onto the next parts.

EDIT: that Latex thing isn't working to produce the correct symbol. it just italicizes the input. Can somebody use code tags to show me the right input?

Mr F has fixed the LaTeX for you. If you want to see the amended code (or the code for any other posting that has a LaTeX image in it) just click on the item, and the code will pop up in a separate little window. It's really neat!

I also find the LaTeX/Mathematics - Wikibooks, collection of open-content textbooks page really useful.

• April 4th 2009, 11:37 AM
earboth
Quote:

Originally Posted by Russ
...
if the scalar projection of u onto v isn't 0, then I have no idea how to do this lol.

Hoping for somebody to confirm or deny so I can move onto the next parts.

...

1. The dot product of two vectors is defined by:

$\vec u \cdot \vec v = |\vec u| \cdot |\vec v| \cdot \cos(\theta)$

2. If you use brackets the meaning of the RHS of the definition is changing slightly:

$|\vec u| \cdot {\color{red}\bold{(}}|\vec v| \cdot \cos(\theta) {\color{red}\bold{)}}$

Here the vector $\vec v$ is projected onto the vector $\vec u$ . The value of the product correspond with the area of a rectangle.

$|\vec v| \cdot {\color{red}\bold{(}}|\vec u| \cdot \cos(\theta) {\color{red}\bold{)}}$

Here the vector $\vec u$ is projected onto the vector $\vec v$ . The value of the product correspond with the area of a rectangle.

Both rectangles have equal areas but different shapes.
• April 4th 2009, 03:47 PM
Russ
Those rectangles just confused me, but my answers were:

a) 0
b) 0
c) 0
d) v (the original vector)

Hoping that's correct, I don't think this is going to be the question chosen for marking thankfully.