Consider the vectors a = <1,1,0> and b = <0,1,-1>. Find a vector c which is perpendicular to a and makes an angle of 60 degree with b. How do i solve it? THANK YOU
I solve it myself, haha a = <1, 1, 0 > b = <0, 1, -1> Let c = <d, e, f> The dot product of orthogonal vectors is zero. a • c = <1, 1, 0 > • <d, e, f> = d + e = 0 e = -d So we have c = <d, -d, f>. ____________ b • c = <0, 1, -1> • <d, -d, f> = -d - f Take the magnitude of vectors b and c. | b | = √[0² + 1² + (-1)²] = √(0 + 1 + 1) = √2 | c | = √[d² + (-d)² + f²] = √(2d² + f²) The dot product can also be represented another way. b • c = | b | | c | cos60° because 60° is the angle between vectors b and c. b • c = | b | | c | cos60° -d - f = (√2) [√(2d² + f²)] (1/2) -d - f = [√(2d² + f²)] (1/√2) Square both sides. (-d - f)² = (2d² + f²) (1/2) 2(d² + 2df + f²) = 2d² + f² 2d² + 4df + 2f² = 2d² + f² 4df + f² = 0 f(f + 4d) = 0 f = 0, -4d The solution f = 0 is extraneous and therefore rejected. f = -4d So c = <d, -d, -4d> Let d = 1. Then c = <1, -1, -4> ________ Test. a • c = <1, 1, 0> • <1, -1, -4> = 1 - 1 + 0 = 0 So vectors a and c are perpendicular. Now vectors b and c. b • c = <0, 1, -1> • <1, -1, -4> = 0 - 1 + 4 = 3 | b | = √2 | c | = √(2d² + f²) = √(2*1 + 4²) = √(2 + 16) = √18 b • c =? | b | | c | cos60° 3 = (√2)(√18) cos60° 3 = (√36) (1/2) = 6 * (1/2) 3 = 3 So vector c = <1, -1, -4> is a solution.
They incorporated vectors in so people could get a more broader understanding of math. But like I said it is not calculus, part of it no doubt.