Question

Let $L_{1}$ and $L_{2}$ be the following straight lines. $L_{1}: \frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{3}$ and $L_{2}: \frac{x-1}{-3}-\frac{y}{-1}=\frac{z-1}{1} .$ Suppose thestraight line $L: \frac{x-\alpha}{1}=\frac{y-1}{m}=\frac{z-\gamma}{-2}$ lies in the plane containing $L_{1}$ and $L_{2}$, and passes through thepoint of intersection of $L_{1}$ and $L_{2}$. If the line $L$ bisects the acute angle between the lines $L_{1}$ and $L_{2}$, then which of the following statements is/are TRUE?

• A. $\alpha-\gamma=3$
• B. $l+m=2$
• C. $\alpha-\gamma=1$
• D. $l+m=0$

Answer 1

Answer: (A), (B)

Clearly we can observe from the equations that $L_{1}$ and $L_{2}$ pass through $(1,0,1)$
Vectors in the directions of $L_{1}$ and $L_{2}$ are $(\hat{i}-\hat{j}+3 \hat{k})$ and $(-3 \hat{i}-\hat{j}+\hat{k})$ respectively which areequimodular and inclined at an acute angle, so their angle bisector will be along their resultant, which is $(-2 \hat{i}-2 \hat{j}+4 \hat{k})$. So equation of angle acute angle bisector will be
$\frac{x-1}{1}=\frac{y}{1}=\frac{z-1}{-2}($ Clearly $1=m=1)$
$\because$ Point $(\alpha, 1, \gamma)$ lies on this line, then $\alpha=2$ and $\gamma=-1$