Thread: 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)

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)

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

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?

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.

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

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

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