Equation of bisector between two straight lines given in symmetrical form

Question

Please help in solving the attached question.

I know in 2D, it is solved as $\dfrac{ax+by+c}{\sqrt{a^2+b^2}}=\pm \dfrac{px+qy+s}{\sqrt{p^2+q^2}}$

Not sure if we use the same formula for 3D? And if yes, what’s the value of the constant c and s here?

Answer 1

In $3D$ the equation of the two lines is given as follows

$ \dfrac{ x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c} $

for the first line, and

$ \dfrac{ x - x_1}{d} = \dfrac{y - y_1}{e} = \dfrac{z - z_1}{f} $

and these two symmetrical form equations correspond to the parametric equations

$P_1(t) = (x_1, y_1, z_1 ) + t (a,b,c) $

$P_2(t) = (x_1, y_1, z_1) + t (d, e, f) $

From here find the unit vectors

$u_1 = \dfrac{ (a,b,c) }{\sqrt{a^2 + b^2 + c^2} } = (a_1, b_1, c_1)$

$u_2 = \dfrac{(d,e,f) }{\sqrt{d^2 + e^2 + f^2}} = (d_1, e_1, f_1) $

The direction vectors along the two bisectors are

$ v_1 = u_1 + u_2 $ and $ v_2 = u_1 - u_2 $

And therefore, the symmetrical form equations of the two bisectors are

$ \dfrac{ x - x_1}{a_1 + d_1} = \dfrac{y - y_1}{b_1 + e_1} = \dfrac{z - z_1}{c_1 + f_1} $

$ \dfrac{ x - x_1}{a_1 - d_1} = \dfrac{y - y_1}{b_1 - e_1} = \dfrac{z - z_1}{c_1 - f_1} $

For the question quoted, $(x_1,y_1, z_1) = (3,-4, 5) $, $d_1 = (2,-1,-2) $, $d_2 = (4, -12, 3)$

From here $u_1 = \dfrac{ (2, -1, -2) }{\sqrt{2^2 + (-1)^2 + (-2)^2} } = (\dfrac{2}{3}, -\dfrac{1}{3}, - \dfrac{2}{3} ) $

and

$u_2 = \dfrac{ (4, -12, 3) }{\sqrt{4^2 + (-12)^2 + 3^2}} = (\dfrac{4}{13}, -\dfrac{12}{13}, \dfrac{3}{13} ) $

Hence,

$v_1 = u_1 + u_2 = ( \dfrac{38}{39} , -\dfrac{49}{39}, - \dfrac{17}{39} )$

$v_2 = u_1 - u_2 = ( \dfrac{14}{39} , \dfrac{23}{39} , -\dfrac{35}{39} )$

Normalizing $v_1, v_2$ by multiplying through by $39$, the equations of the bisectors are

$ \dfrac{ x - 3}{38} = \dfrac{y +4}{-49} = \dfrac{z - 5}{-17} $

and

$ \dfrac{ x - 3}{14} = \dfrac{y +4}{23} = \dfrac{z - 5}{-35} $