If $L_1$ and $L_2$ are two 3D lines represented by the equation ${L_1}:\frac{{x – 1}}{1} = \frac{y}{{ – 1}} = \frac{{z – 1}}{3}$ & ${L_2}:\frac{{x – 1}}{{ – 3}} = \frac{y}{{ – 1}} = \frac{{z – 1}}{1}$. If the line L bisects the acute angle between the lines $L_1$ and $L_2$. Then find the equation of the Line "L".
My approach is as follow the point of intersection is $(1,0,1)$.
${L_1}:\frac{{x – 1}}{1} = \frac{y}{{ – 1}} = \frac{{z – 1}}{3} \Rightarrow \overrightarrow r = \hat i + \hat k + \mu \left( {\hat i – \hat j + 3\hat k} \right) = \overrightarrow a + \mu \overrightarrow c $
${L_2}:\frac{{x – 1}}{{ – 3}} = \frac{y}{{ – 1}} = \frac{{z – 1}}{1} \Rightarrow \overrightarrow r = \hat i + \hat k + \mu \left( { – 3\hat i – \hat j + \hat k} \right) = \overrightarrow b + \lambda \overrightarrow d $
The direction vector of bisector between $L_1$ and $L_2$ is
$\overrightarrow T = \frac{{\overrightarrow c }}{{\left| {\overrightarrow c } \right|}} + \frac{{\overrightarrow d }}{{\left| {\overrightarrow d } \right|}} = \frac{{\hat i – \hat j + 3\hat k}}{{\left| {\hat i – \hat j + 3\hat k} \right|}} + \frac{{ – 3\hat i – \hat j + \hat k}}{{\left| { – 3\hat i – \hat j + \hat k} \right|}} = \frac{{ – 2\hat i – 2\hat j + 4\hat k}}{{\sqrt {11} }} = \left\langle {1,1, – 2} \right\rangle = \left\langle {\ell ,m, – 2} \right\rangle $
$\overrightarrow U = \frac{{\overrightarrow c }}{{\left| {\overrightarrow c } \right|}} – \frac{{\overrightarrow d }}{{\left| {\overrightarrow d } \right|}} = \frac{{\hat i – \hat j + 3\hat k}}{{\left| {\hat i – \hat j + 3\hat k} \right|}} – \frac{{ – 3\hat i – \hat j + \hat k}}{{\left| { – 3\hat i – \hat j + \hat k} \right|}} = \frac{{4\hat i + 2\hat k}}{{\sqrt {11} }} = \left\langle { – 4,0, – 2} \right\rangle = \left\langle {\ell ,m, – 2} \right\rangle $.
From here I am confused, my assumption is that if $\overrightarrow c .\overrightarrow d > 0$, then $\overrightarrow T $ is the direction vector that is acute angle and if $\overrightarrow c .\overrightarrow d < 0$, then $\overrightarrow U $ is the direction vector that is acute angle. Just want to clarify it.
Best Answer
We see that the direction vectors have the same length, and their dot product is positive. Thus, merely adding them will give a vector pointing in the direction of the bisector of the acute angle: $(1,-1,3)+(-3,-1,1)=(-2,-2,4)$. So the equation of $L$ is .$$(1,0,1)+t(1,1,-2)$$