Using the formula for vector bisector:
#veca= ||vecc||vecb+||vecb||vecc#
Substitute #veca= -hati+hatj-hatk, vecb = 3hati+4hatj, and vecc = c_xhati+c_yhatj+c_zhatk#
#-hati+hatj-hatk = ||vecc||(3hati+4hatj)+ ||vecb||(c_xhati+c_yhatj+c_zhatk)#
Compute the magnitude of #vecb#:
#||vecb|| = sqrt(3^2+4^2) = sqrt(9+16) = sqrt25 = 5#
Substitute into the equation:
#-hati+hatj-hatk = ||vecc||(3hati+4hatj)+ 5(c_xhati+c_yhatj+c_zhatk)#
Divide both sides of the equation by 5:
#-1/5hati+1/5hatj-1/5hatk = ||vecc||(3/5hati+4/5hatj)+ (c_xhati+c_yhatj+c_zhatk)#
Subtract #||vecc||(3/5hati+4/5hatj)# from both sides:
#c_xhati+c_yhatj+c_zhatk = -1/5hati+1/5hatj-1/5hatk - ||vecc||(3/5hati+4/5hatj)#
It is obvious that #c_z = -1/5#, therefore, we remove it from both sides:
#c_xhati+c_yhatj = -1/5hati+1/5hatj - ||vecc||(3/5hati+4/5hatj)#
Substitute #||vecc|| = sqrt(c_x^2+c_y^2+(-1/5)^2)#
#c_xhati+c_yhatj = -1/5hati+1/5hatj - 3/5sqrt(c_x^2+c_y^2+(-1/5)^2)hati-4/5sqrt(c_x^2+c_y^2+(-1/5)^2)hatj#
We can write two equations.
Equation [1] is written with the #hati# terms:
#c_x = -1/5-3/5sqrt(c_x^2+c_y^2+(-1/5)^2)" [1]"#
Equation [2] is written with the #hatj# terms:
#c_y = 1/5-4/5sqrt(c_x^2+c_y^2+(-1/5)^2)" [2]"#
One can solve these two equations but why bother when you can substitute #c_x = x and c_y= y# and let WolframAlpha do it.
The vector components are:
#c_x = -11/10, c_y=-1, and c_z = -1/5#
This gives us the vector:
#vecc = -11/10hati-hatj-1/5hatk#
Check:
#||vecc|| = sqrt( (-11/10)^2+ (-1)^2+ (-1/5)^2)#
#||vecc|| = 3/2#
#||vecc||vecb+||vecb||c = 3/2(3hati+4hatj)+ 5(-11/10hati-hatj-1/5hatk)#
#||vecc||vecb+||vecb||c = (9/2-11/2)hati+(6-5)hatj-hatk#
#||vecc||vecb+||vecb||c = -hati+hatj-hatk#
This checks.
