The lines `(x-2)/(1)=(y-3)/(1)=(z-4)/(-k) and (x-1)/(k)=(y-4)/(2)=(z-5)/(1)` are coplanar, if

Options

A. `k=0 and k=-1`
B. `k=1 or -1`
C. `k=0 or -3`
D. `k=3 or -3`

Answer

To determine the values of \( k \) for which the given lines are coplanar, we can follow these steps:

### Step 1: Write the equations of the lines in parametric form
The first line can be expressed as:
\[ \frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k} \]
Let \( t \) be the parameter for the first line. Then we can write:
\[ x = 2 + t, \quad y = 3 + t, \quad z = 4 - kt \]

The second line is given by:
\[ \frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1} \]
Let \( s \) be the parameter for the second line. Then we can write:
\[ x = 1 + ks, \quad y = 4 + 2s, \quad z = 5 + s \]

### Step 2: Identify direction ratios and points
From the parametric equations, we can identify:
- For the first line, the direction ratios are \( \vec{b_1} = (1, 1, -k) \) and a point \( A(2, 3, 4) \).
- For the second line, the direction ratios are \( \vec{b_2} = (k, 2, 1) \) and a point \( B(1, 4, 5) \).

### Step 3: Find the vector connecting the two points
The vector \( \vec{AB} \) from point \( A \) to point \( B \) is given by:
\[ \vec{AB} = B - A = (1 - 2, 4 - 3, 5 - 4) = (-1, 1, 1) \]

### Step 4: Set up the condition for coplanarity
The lines are coplanar if the scalar triple product of the vectors \( \vec{AB} \), \( \vec{b_1} \), and \( \vec{b_2} \) is zero:
\[ \vec{AB} \cdot (\vec{b_1} \times \vec{b_2}) = 0 \]

### Step 5: Calculate the cross product \( \vec{b_1} \times \vec{b_2} \)
We can compute the cross product:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & -k \\ 2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -k \\ k & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ k & 2 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot 1 - (-k) \cdot 2) - \hat{j} (1 \cdot 1 - (-k) \cdot k) + \hat{k} (1 \cdot 2 - 1 \cdot k) \]
\[ = \hat{i} (1 + 2k) - \hat{j} (1 + k^2) + \hat{k} (2 - k) \]

### Step 6: Compute the scalar triple product
Now we compute:
\[ \vec{AB} \cdot (\vec{b_1} \times \vec{b_2}) = (-1, 1, 1) \cdot (1 + 2k, -(1 + k^2), 2 - k) \]
This gives:
\[ = -1(1 + 2k) + 1(-1 - k^2) + 1(2 - k) \]
\[ = -1 - 2k - 1 - k^2 + 2 - k \]
\[ = -k^2 - 3k + 0 \]

### Step 7: Set the scalar triple product to zero
Setting this equal to zero for coplanarity:
\[ -k^2 - 3k = 0 \]
Factoring out \( -k \):
\[ -k(k + 3) = 0 \]

### Step 8: Solve for \( k \)
This gives us:
\[ k = 0 \quad \text{or} \quad k = -3 \]

### Conclusion
The lines are coplanar if \( k = 0 \) or \( k = -3 \).

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