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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.

Samar Imam Zaidi
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  • For 2-D line equation we have certain rules of finding acute angle bisector or obtuse angle bisector by keep the constant sign same and applying certain condition that is well documented. Any documentation regarding the same for 3D lines – Samar Imam Zaidi Oct 10 '20 at 09:10
  • If the dot product of directional vectors is positive, the acute angle bisector will be given by adding the unit vectors. – Math Lover Oct 10 '20 at 09:26

2 Answers2

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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 .$$(1,0,1)+t(1,1,-2)$$

Mikasa
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Parcly Taxel
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For two given lines, $\vec{a} \cdot \vec{b} = |a|.|b|\cos \theta$ where $\theta$ is the angle between the lines.

As, $\cos \theta \ge 0 \, $ for $(0 \le \theta \le {\pi/2}); \cos \theta \lt 0 \, $ for $({\pi/2} \lt \theta \le {\pi}), $
it is easy to remember that the if the dot product is positive, then you have an acute angle between the vectors. So just add the unit vectors for the angle bisector. If the dot product is negative, you have obtuse angle between them and you need one of the vectors in the opposite direction to find the acute angle so subtract the unit vectors to find the angle bisector of the acute angle between lines.

Math Lover
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