Steve S. answered 03/12/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
“Determine wether u and v are orthogonal, parallel or neither, where: u = <1,2,-3> and v = <-4/3,-8/3,4>.
I am thinking they’re parallel.”
The Dot Product
u • v = (1)(-4/3) + (2)(-8/3) + (-3)(4), and
I am thinking they’re parallel.”
The Dot Product
u • v = (1)(-4/3) + (2)(-8/3) + (-3)(4), and
u • v = √(1+4+9)√(16/9 + 64/9 + 16) cos(θ),
where θ is the angle between the vectors.
cos(θ) = ((1)(-4/3) + (2)(-8/3) + (-3)(4))
where θ is the angle between the vectors.
cos(θ) = ((1)(-4/3) + (2)(-8/3) + (-3)(4))
/(√(1+4+9)√(16/9 + 64/9 + 16))
= (-4/3 -16/3 - 36/3)/√((14)(80/9 + 144/9))
= (-56/3)/√((14)(80/9 + 144/9))
= -56/√((14)(80 + 144))
= -56/56
= -1
θ = 180° so the vectors are parallel.
==
“1)Find the vector z = -2u + (1/2)v - 3w,
= (-4/3 -16/3 - 36/3)/√((14)(80/9 + 144/9))
= (-56/3)/√((14)(80/9 + 144/9))
= -56/√((14)(80 + 144))
= -56/56
= -1
θ = 180° so the vectors are parallel.
==
“1)Find the vector z = -2u + (1/2)v - 3w,
given u = <-1,3,2>, v = <6,-2,-2> and w = <5,0,-5>.
So i am thinking that i have to first find the internal and terminal point to find the component”
u = <-1,3,2>
v = <6,-2,-2>
w = <5,0,-5>
z = -2u + 1/2v - 3w
-2u = <2,-6,-4>
(1/2)v = <3,-1,-1>
-3w = <-15,0,15>
z = <(2+3-15,-6-1+0,-4-1+15>
z = <-10,-7,10>
So i am thinking that i have to first find the internal and terminal point to find the component”
u = <-1,3,2>
v = <6,-2,-2>
w = <5,0,-5>
z = -2u + 1/2v - 3w
-2u = <2,-6,-4>
(1/2)v = <3,-1,-1>
-3w = <-15,0,15>
z = <(2+3-15,-6-1+0,-4-1+15>
z = <-10,-7,10>