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Home Q & A Board Science-Related Homework Help Physics
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nrcrafton nrcrafton
wrote...
10 years ago
What is the orthogonal projection of v(6,-1,-7) onto the line L through (-5,-4,6) and the origin?

i tried finding the projection of L on u (L being (-5,-4,6)), then subtracting v by that projection to find the answer, but thats wrong
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wrote...
#1
10 years ago
Let
R=-5i-4j+6k be the position vector of (-5,-4,6), then
the unit vector of R,
r=(-5i-4j+6k)/sqrt(77)
The dot product
v.r=(-30+4-42)/sqrt(77)=>
v.r=-68/sqrt(77) is the orthogonal projection of v on L.
wrote...
#2
10 years ago Edited: 10 years ago, bio_man
See Gram-smith process .-

You may write vector v like the sum on v proyection on u plus orthogonal proyection on u .-
this is , V= Vu + Vn  , then
Vn= V- Vu

Vdot u = IVI IuI cos T

Vu = IVI cos T  < u >       (< u > unit vector on u direction )
Vu = (V dot u) /IuI   u/IuI
Vu = (Vdot u) /IuI^2  u
Vdot u = -30+4 -42 = -68
IuI^2 = u dot u = 25+16+36=77
Vu = (-68/77) ( -5,-4,6)
Vn=(6,-1,-7) -( -68/77) (-5,-4,6)
Vn= (6,-1,-7) + (-4.42,-3.53,5.30)
Vn=  (1.58, -4.53, -1.70)

Check , Vn dot u = 1.58*(-5) + (-4.53) *(-4)  +(-1.70) (6) = 0   ok /
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