Question

Find two non-parallel vectors v1 and v2 in R2 such that ||v1||2=||v2||2=1 and whose angle with [3,4] is 42 degrees. *Note: [3,4] is a 2 by 1 matrix with 3 on the top and 4 on the bottom.

Answer #1

let v = (x,y) be one such vector.

given ||v||^{2} = 1 =>
v*v = 1 => (x,y)*(x,y)=1
=>**x ^{2}+y^{2}= 1 ------ (1)**

Also Angle between v and (3,4) is 42 degrees.

||(3,4)|| =
(3^{2}+4^{2})^{1/2} = 5

Threfore v*(3,4) = ||v||*||(3,4)||*cos(42 degree)

=> (x,y)*(3,4) = 1*5*0.74

=> 3x+4y = 3.7

=> y = (3.7 - 3x)/4 =>
**y = 0.925 - 0.75x ------(2)**

put 2 in 1 we get

x^{2}+(0.925-0.75x)^{2}=1

=> x = -0.094 , 0.982

Using these values of x in (2) we get y = 0.995 , 0.188

hence our **vectors are
(-0.0925,0.995) and (0.982,0.188)**

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