Animated Solution for Mathematics - Three Dimensional Geometry: If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is (are)

A 3D Odyssey

My dear student, welcome to the fascinating world of 3D geometry. Today, we are not just solving a problem; we are exploring the architecture of space itself.

We have two lines, $L_1$ and $L_2$, floating in three-dimensional space. They seem independent, yet they are bound by the condition of being coplanar. This means they share a common home—a plane. Let us embark on this journey to find that home.

Extracting the DNA of the Lines

Every line in 3D space carries its own DNA: a point it passes through and a direction in which it travels. Look at the symmetric equations provided:

By inspecting the numerators, we can instantly extract a point from each line. For $L_1$, we see the point $A(1,-1,0)$. For $L_2$, we find the point $B(-1,-1,0)$. These are our anchors.

Next, we look at the denominators. These are the components of our direction vectors. For $L_1$, the direction vector is $\overrightarrow{d_1}=2\hat{i}+k\hat{j}+2\hat{k}$. For $L_2$, it is $\overrightarrow{d_2}=5\hat{i}+2\hat{j}+k\hat{k}$.

Notice how the mystery parameter $k$ is woven into the very fabric of these directions. This is the variable we must unmask.

The Condition of Coplanarity

Imagine these two lines. To ensure they lie on the same plane, we need to connect them. Let us draw a vector $\overrightarrow{AB}$ from point $A$ to point $B$.

We calculate this by subtracting the coordinates of $A$ from $B$:

Now, here is the core geometric reality: if $\overrightarrow{AB}$, $\overrightarrow{d_1}$, and $\overrightarrow{d_2}$ all lie on the same plane, the volume of the parallelepiped they form must be zero. This is the Scalar Triple Product.

We set the determinant of these three vectors to zero:

Expanding this along the first row is our most efficient strategy. We get $-2(k^2-4)=0$. This simplifies beautifully to $k^2=4$, giving us two distinct possibilities: $k=2$ or $k=-2$.

Constructing the Planes

With $k$ values in hand, we now have two distinct scenarios. Let us tackle Case 1: $k=2$.

Our direction vectors become $\overrightarrow{d_1}=2\hat{i}+2\hat{j}+2\hat{k}$ and $\overrightarrow{d_2}=5\hat{i}+2\hat{j}+2\hat{k}$. To find the plane, we need a normal vector $\overrightarrow{n_1}$, which is perpendicular to both lines.

We find this via the cross product $\overrightarrow{n_1}=\overrightarrow{d_1}\times \overrightarrow{d_2}$. Calculating this, we get $\overrightarrow{n_1}=0\hat{i}+6\hat{j}-6\hat{k}$. We can simplify this to $\vec{n}=(0,1,-1)$.

Using point $A(1,-1,0)$, the plane equation becomes $0(x-1)+1(y+1)-1(z-0)=0$, which simplifies to $y-z=-1$.

Now, for Case 2: $k=-2$. Our direction vectors are $\overrightarrow{d_1}=2\hat{i}-2\hat{j}+2\hat{k}$ and $\overrightarrow{d_2}=5\hat{i}+2\hat{j}-2\hat{k}$.

Again, we find the normal vector $\overrightarrow{n_2}=\overrightarrow{d_1}\times \overrightarrow{d_2}$, resulting in $\overrightarrow{n_2}=0\hat{i}+14\hat{j}+14\hat{k}$. Simplifying, we get $\vec{n}=(0,1,1)$.

Using point $A(1,-1,0)$, the plane equation becomes $0(x-1)+1(y+1)+1(z-0)=0$, leading to $y+z=-1$.

We have arrived! We found that the lines are coplanar for two specific values of $k$, resulting in two beautiful planes: $y-z=-1$ and $y+z=-1$.

This problem taught us that geometry is not just about shapes; it is about the relationships between vectors and the conditions that force them to align. Keep practicing, keep visualizing, and never lose your curiosity.