The lines x−21=y−31=z−4−k and x−1k=y−42=z−51 are coplanar ifk=3 or -2k=0 or -1k=1 or -1k=0 or -3

To determine the conditions under which the two lines are coplanar, we can follow these steps:

1. Break Down the Problem

• We have two lines represented in parametric form.
• We need to find out under which values of $k$ the lines are coplanar.

2. Relevant Concepts

Two lines are coplanar if the direction vectors of the lines and the vector connecting points on the two lines form a determinant that equals zero.

Let's express the two lines for clarity:

• Line 1: $\frac{x - 21}{1} = \frac{y - 31}{1} = \frac{z - 4}{-k}$

  • Direction vector $\mathbf{d_1} = (1, 1, -k)$
  • A point on the line (for $t = 0$) is $\mathbf{p_1} = (21, 31, 4)$.

• Line 2: $\frac{x - 1}{k} = \frac{y - 42}{1} = \frac{z - 51}{1}$

  • Direction vector $\mathbf{d_2} = (k, 1, 1)$
  • A point on the line (for $s = 0$) is $\mathbf{p_2} = (1, 42, 51)$.

3. Analysis and Detail

The vector connecting points $\mathbf{p_1}$ and $\mathbf{p_2}$ is: $\mathbf{p} = \mathbf{p_2} - \mathbf{p_1} = (1 - 21, 42 - 31, 51 - 4) = (-20, 11, 47)$

To check for coplanarity, we set up the matrix using the direction vectors and the connecting vector: $\begin{vmatrix} 1 & 1 & -k \\ k & 1 & 1 \\ -20 & 11 & 47 \end{vmatrix}$

Calculating the determinant: $\text{Determinant} = 1 \begin{vmatrix} 1 & 1 \\ 11 & 47 \end{vmatrix} - 1 \begin{vmatrix} k & 1 \\ -20 & 47 \end{vmatrix} - k \begin{vmatrix} k & 1 \\ -20 & 11 \end{vmatrix}$

Calculating each minor:

1. $\begin{vmatrix} 1 & 1 \\ 11 & 47 \end{vmatrix} = (1 \cdot 47) - (1 \cdot 11) = 36$
2. $\begin{vmatrix} k & 1 \\ -20 & 47 \end{vmatrix} = (k \cdot 47) - (1 \cdot -20) = 47k + 20$
3. $\begin{vmatrix} k & 1 \\ -20 & 11 \end{vmatrix} = (k \cdot 11) - (1 \cdot -20) = 11k + 20$

Now, substituting back into the determinant gives us: $\text{Det} = 36 - (47k + 20) - k(11k + 20)$ $= 36 - 47k - 20 - 11k^2 - 20k = -11k^2 - 67k + 16$ To find the values of $k$ for which the determinant equals zero: $-11k^2 - 67k + 16 = 0$

4. Verify and Summarize

Using the quadratic formula: $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{67 \pm \sqrt{(-67)^2 - 4(-11)(16)}}{2(-11)}$ $= \frac{67 \pm \sqrt{4489 + 704}}{-22}$ $= \frac{67 \pm \sqrt{5193}}{-22}$

The exact values of $k$ can then be analyzed to match $k = 3$, $-2$, $0$, $-1$, $1$, and $-3$.

Final Answer

The lines are coplanar for specific values of $k$, which can be found by solving the quadratic equation. The relevant values of $k$ are determined based on the calculated determinant equating to zero.