# Projection of Vectors

• August 27th 2009, 06:43 PM
skeske1234
Projection of Vectors
The rectangular prism is measured 1x2x3 as shown, find the the projection of vector u onto vector v.

i'm having trouble finding the points of u and v

my guess:
u(1,0,3)
v(1,2,3)
• August 27th 2009, 06:55 PM
Calculus26
The formula for the projection of u on v is

proju on v = (u*v/||v||^2) v

what you are using for u and v is incorrect

u = (3,0,1) v = (3,2,-1)
• August 28th 2009, 03:58 AM
enjam
The projection of u along v is simply:
u . v/|v|
So you have:
(3i + 0j + k) . (3i + 2j - k) / (3^2 + 2^2 + (-1)^2)^0.5
= (3x3) + (0x2) + (1x-1) / (14)^0.5
= 8 / sqrt 14
• August 28th 2009, 06:26 AM
skeske1234
How did you get the x value to be 3? isn't the x value the width/depth of the rectangular prism, thus 1? and the height 3 = z value and the length 2=y value?
• August 28th 2009, 07:50 AM
Plato
Quote:

Originally Posted by skeske1234
How did you get the x value to be 3? isn't the x value the width/depth of the rectangular prism, thus 1? and the height 3 = z value and the length 2=y value?

You just learn the rule for projections.

$\text{proj}_v u = \frac{{u \cdot v}}
{{\left\| v \right\|^2 }}v = \frac{8}
{{14}}\left\langle {3,2, - 1} \right\rangle$
• August 28th 2009, 02:22 PM
Calculus26
Quote:

The projection of u along v is simply:
u . v/|v|
So you have:
(3i + 0j + k) . (3i + 2j - k) / (3^2 + 2^2 + (-1)^2)^0.5
= (3x3) + (0x2) + (1x-1) / (14)^0.5
= 8 / sqrt 14

The projection is a vector --what you have here is the magnitude of the projection
• August 28th 2009, 06:19 PM
skeske1234
Plato, I mean how did you get the point (3,2,-1)? like the vector points of u and v, in order to plug in formula
• August 28th 2009, 06:28 PM
Calculus26
At firs tglance it appears as though u goes from(0,0,1) to (3,0,1)

so u should b3 3i not 3i + k as I suggested

and v goes from (0,0,1) to (3,2,0) v = 3 i + 2j -k