Question

If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is

• A. $\frac{13}{6}$
• B. $\frac{5}{17}$
• C. $\frac{17}{5}$
• D. $\frac{6}{13}$

Answer 1

The Correct Option is A

We are given the equation of the line and the points \( (-1, k, 0), (2, k, -1), (1, 1, 2) \). The equation of the line is: \[ \frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1} \] From the given, we have: \[ \frac{x-1}{1} = \frac{y+1}{1} = \frac{z+1}{-1} \] Thus, the points \( A(-1, k, 0), B(2, k, -1), C(1, 1, 2) \) represent the positions on the plane. We calculate the vectors \( \overrightarrow{CA} \) and \( \overrightarrow{CB} \): \[ \overrightarrow{CA} = -2i + (k-1)j - 2k \hat{k} \] \[ \overrightarrow{CB} = i + (k-1)j - 3k \hat{k} \] Next, the cross product \( \overrightarrow{CA} \times \overrightarrow{CB} \) is calculated as: \[ \overrightarrow{CA} \times \overrightarrow{CB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-2 & k-1 & -2
1 & k-1 & -3 \end{vmatrix} \] This results in: \[ \overrightarrow{CA} \times \overrightarrow{CB} = \hat{i}(-3k+3+2k-2) - \hat{j}(6+2k-2k+2) + \hat{k}(-2k+2-k+1) \] \[ = (1-k) \hat{i} - 8 \hat{j} + (3-3k) \hat{k} \] The line \( \frac{x-1}{1} = \frac{y+1}{1} = \frac{z+1}{-1} \) is perpendicular to the plane. Using this information, we solve for the value of \( \frac{k^2+1}{(k-1)(k-2)} \), which simplifies to \( \frac{13}{6} \).