Asked by angles
Find two unit vectors that make an angle of 60° with v = ‹3, 4›. Give your answers correct to three decimal places.
I'm not sure what to do here; I believe that I use to definition of a dot product (or maybe cross product) with relation to the trig function (cos for dot, sin for cross) but for whatever reason I can't get my answer to come out.
I'm not sure what to do here; I believe that I use to definition of a dot product (or maybe cross product) with relation to the trig function (cos for dot, sin for cross) but for whatever reason I can't get my answer to come out.
Answers
Answered by
Reiny
let such a vector be (a,b)
then (a,b)⋅(3,4) = |(a,b)||(3,4)|cos60°
3a + 4b = 5√(a^2 + b^2)(1/2)
6a+8b = 5√(a^2+b^2)
let a= 1 , any non-zero number will do
6+8b = 5√(1+b^2)
square both sides
36+96b +64 b^2 = 25 + 25b^2
39b^2 + 96b + 11 = 0
b = (-96 ± √7500)/48
= appr. -2.341 or -0.1205
so two vectors could be (1, -2.341) or (1, -.1205)
but you want a unit vector, so using the second point,
|(1,-.1205)| = 1.00723
giving us a unit vector of (1/1.00723 , -.1205/1.00723)
or appr. (.9928 , -.1196)
check:
(3,4).(.9928,-.1196) = 2.5
|(3,4)||(.9928,-.1196)cos60 = 2.5
reverse the direction of the above vector to get
(-.9928, .1196) for a second vector, or
let a = some other value and repeat the above steps
then (a,b)⋅(3,4) = |(a,b)||(3,4)|cos60°
3a + 4b = 5√(a^2 + b^2)(1/2)
6a+8b = 5√(a^2+b^2)
let a= 1 , any non-zero number will do
6+8b = 5√(1+b^2)
square both sides
36+96b +64 b^2 = 25 + 25b^2
39b^2 + 96b + 11 = 0
b = (-96 ± √7500)/48
= appr. -2.341 or -0.1205
so two vectors could be (1, -2.341) or (1, -.1205)
but you want a unit vector, so using the second point,
|(1,-.1205)| = 1.00723
giving us a unit vector of (1/1.00723 , -.1205/1.00723)
or appr. (.9928 , -.1196)
check:
(3,4).(.9928,-.1196) = 2.5
|(3,4)||(.9928,-.1196)cos60 = 2.5
reverse the direction of the above vector to get
(-.9928, .1196) for a second vector, or
let a = some other value and repeat the above steps
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