Vector #-hati+hatj-hatk# bisects the angle between vectors #3hati+4hatj# and #vecc# find #vecc#?

1 Answer
May 7, 2018

#vecc = -11/10hati-hatj-1/5hatk#

Explanation:

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.