Animated Solution for Mathematics - Three Dimensional Geometry: If the lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar, then $k$ can have

Understanding Coplanar Lines

Imagine you are standing in a vast, empty room with two laser pointers. These two beams of light represent lines in three-dimensional space. Usually, these lines are 'skew'—they pass each other like ships in the night, never touching and never lying on the same flat surface.

However, we are dealing with a special case where these lines are coplanar. This means that we could slide a single, flat sheet of paper between them, and both lines would rest perfectly on that surface. This is the geometric reality we are exploring.

Extracting the DNA of the Lines

To solve this, we first need to understand the 'DNA' of our lines. Every line in 3D space is defined by a point it passes through and a direction vector.

For our first line, $L_1:\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$, we extract the point $P_1(2,3,4)$ and the direction vector $\overrightarrow{d_1}=(1,1,-k)$.

Similarly, for our second line, $L_2:\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$, we identify the point $P_2(1,4,5)$ and the direction vector $\overrightarrow{d_2}=(k,2,1)$. We now have the building blocks required to force these lines onto the same plane.

The Bridge and the Volume

To connect these two lines, we create a bridge vector $\overrightarrow{P_1{P}_2}$ that connects a point on the first line to a point on the second. By subtracting the coordinates, we get:

If these three vectors—the bridge $\overrightarrow{P_1{P}_2}$, the direction $\overrightarrow{d_1}$, and the direction $\overrightarrow{d_2}$—all lie on the same plane, the volume of the parallelepiped they form must be zero. In the language of linear algebra, this is the scalar triple product:

This condition is equivalent to stating that the determinant of the matrix formed by these vectors must be zero.

The Determinant Calculation

Let us set up our determinant:

Expanding along the first row, we take the first element, $-1$, and multiply it by the determinant of the minor matrix:

Next, we move to the second element, $1$. Applying the alternating sign rule, we subtract the product of $1$ and its minor:

Finally, the third element is $1$, multiplied by its minor:

The Elegant Cancellation

Now, we combine these terms:

Observe the constants: $-1-1+2$ equals zero. They vanish, leaving us with a simple quadratic equation:

Factoring this, we get $k(k+3)=0$. This yields two solutions: $k=0$ and $k=-3$.

We have found exactly two values for $k$ that satisfy the condition of coplanarity. This is the power of vector geometry—it turns a complex spatial problem into a clean, solvable algebraic equation.