Question

find the two unit vectors in 2-space that made an angle of 45 with 4i+3j

Answer 1

cos45 = ( (x,y)dot(4,3) ) / ( ((lengthof(x,y)) * (lengthof(4,3)) )

lengthof(x,y) = 1 they asked for unit vectors
lengthof(4,3) = (4^2 + 3^2)^0.5 = 5
so
cos45 = (4x + 3y)/5
4x + 3y = 2.5 * (2^0.5)
y = (2.5 * (2^0.5) - 4x) / 3
and also you have the equation :
x^2 + y^2 = 1
so plug the y into this equation :
x^2 + ((2.5*(2^0.5) - 4x)/3)^2 = 1
x^2 + (12.5 - 20*(2^0.5)x + 16x^2/9) = 1
9x^2 + 12.5 - 20*(2^0.5)x + 16x^2 = 9
25x^2 - 20*(2^0.5)x + 3.5 = 0
notice when you solve this quadratic equation you get under the square root 450
i made it 15*(2^0.5)
so the x's are : (2^0.5)/10 , ((7*(2^0.5))/10)
plug it into the y equation : y = (2.5 * (2^0.5) - 4x) / 3
you get y's : 0.7*(2^0.5) , -0.1*(2^0.5)

so the vectors are : ((2^0.5)/10,0.7*(2^0.5)) , (((7*(2^0.5))/10), -0.1*(2^0.5))

Answer 2

hope you are still here

Answer 3

thank you