If the line $\frac{x-2}{-1} = \frac{y+2}{1} = \frac{z+k^2-1}{4}$ is one of the angle bisector of the lines $\frac{x}{1} = \frac{y}{-2} = \frac{z}{3}$ and $\frac{x}{-2} = \frac{y}{3} = \frac{z}{1}$, then the value of $k$ is/are:
Step-by-Step Solution
Key Concept: The angle bisector passes through the point of intersection and has direction ratios derived from the normalized sum/difference of the line direction vectors.
The point of intersection of the given lines is $(0, 0, 0)$. For a point to lie on the angle bisector, its coordinates must satisfy $\frac{0 - 2}{-1} = \frac{0 + 2}{1} = \frac{0 + k^2 - 1}{4}$, which simplifies to $2 = 2 = \frac{k^2 - 1}{4}$, giving $k \pm 3$.
Correct Answer: The step-by-step solution contains an error in the approach. Let me solve this correctly.
For angle bisectors of two lines with direction ratios **dā** = (1, -2, 3) and **dā** = (-2, 3, 1), the direction ratios of the angle bisectors are given by:
**d**