Let $L_1: \frac{x-1}{1} = \frac{y}{-1} = \frac{z-1}{3}$ and $L_2: \frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1}$ be two lines. Let $L: \frac{x-\alpha}{l} = \frac{y-1}{m} = \frac{z-\gamma}{-2}$ be a line that lies in the plane containing $L_1$ and $L_2$, passes through their point of intersection, and bisects the acute angle between them. Then which of the following statements is/are TRUE?
Step-by-Step Solution
Key Concept: Lines intersect if their parametric equations yield consistent solutions for parameters. Non-parallel intersecting lines in 3D are coplanar.
1. **Find the point of intersection**:
The lines $L_1$ and $L_2$ intersect at $P = (1, 0, 1)$. Since $L$ passes through this intersection, the point $P(1, 0, 1)$ must lie on $L$:
$$\frac{1-\alpha}{l} = \frac{0-1}{m} = \frac{1-\gamma}{-2}$$
This relates $\alpha$, $\gamma$, $l$, and $m$.
2. **Find the acute angle bisector**:
The direction vectors of $L_1$ and $L_2$ are $\vec{v}_1 = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{v}_2 = -3\hat{i} - \hat{j} + \hat{k}$.
The unit vectors are:
$$\hat{v}_1 = \frac{\hat{i} - \hat{j} + 3\hat{k}}{\sqrt{11}}, \quad \hat{v}_2 = \frac{-3\hat{i} - \hat{j} + \hat{k}}{\sqrt{11}}$$
The angle is acute since $\vec{v}_1 \cdot \vec{v}_2 = 1 > 0$. The acute angle bisector vector $\vec{v}_b$ is parallel to $\hat{v}_1 + \hat{v}_2$:
$$\vec{v}_1 + \vec{v}_2 = \langle -2, -2, 4 \rangle \propto \langle 1, 1, -2 \rangle$$
Comparing this to the direction of $L$, $\langle l, m, -2 \rangle$, we get $l = 1$ and $m = 1$.
Therefore, $l + m = 2$ (Option B is TRUE).
3. **Find $\alpha$ and $\gamma$**:
Since $P(1, 0, 1)$ lies on $L$, we have:
$$\frac{1-\alpha}{1} = \frac{-1}{1} = \frac{1-\gamma}{-2} \implies 1-\alpha = -1 \implies \alpha = 2$$
and:
$$\frac{1-\gamma}{-2} = -1 \implies 1-\gamma = 2 \implies \gamma = -1$$
Compute $\alpha - \gamma$:
$$\alpha - \gamma = 2 - (-1) = 3$$
This matches Option A (Option A is TRUE).
4. **Conclusion**:
The correct options are A and B.
Correct Answer: A, B