Given the lines L1:1x−1=−1y=3z−1and{L}_{2}: \dfrac{x - 1}{- 3} = \dfrac{y}{- 1} = \dfrac{z - 1}{1}.</p><p>Fromthegivenlines,wecansaythatboththelinespassthroughthepoint\left(1, 0, 1\right).</p><p>Now,let\overset{\to}{a}bethedirectionvectorofline{L}_{1}i.e.,\overset{\to}{a} = i - j + 3kandlet\overset{\to}{b}bethedirectionvectorofline{L}_{2}i.e.,\overset{\to}{b} = - 3i - j + k.</p><p>Wealsoknowthattheanglebisectorslieinthedirectionof(a→+b→)or(a→−b→).
Now, a→.b→=(1×−3)+(−1×−1)+(3×1)=−3+1+3=1>0.
So, direction ratios of the acute angle bisector between two lines will be in the direction of a→+b→ i.e., \left(1 - 3, - 1 - 1, 3 + 1\right) = \left(- 2, - 2, 4\right) = \left(1, 1, - 2\right).</p><p>Since,L: \dfrac{x - \alpha}{l} = \dfrac{y - 1}{m} = \dfrac{z - \gamma}{- 2}istheacuteanglebisectorbetweenthelines{L}_{1} \text{ and } {L}_{2}.</p><p>So,bycomparingthedirectionratios,weget</p><p>l = 1 \text{ and } m = 1</p><p>⇒ l + m = 2</p><p>So,equationoflineLwillbe\dfrac{x - \alpha}{1} = \dfrac{y - 1}{1} = \dfrac{z - \gamma}{- 2}.</p><p>GiventhatthelineLpassesthroughthepoint(1,0,1).
Putting the point in line L, we get
11−α=10−1=−21−γ
⇒1−α=−1=−21−γ
⇒1−α=−1 and −21−γ=−1
⇒α=2 and γ=−1
⇒α−γ=3
So, options \left(A\right) \text{ and } (B) are correct.
