Animated Solution for Mathematics - Vector Algebra: If the four points, whose position vectors are $3\hat{i}-4\hat{j}+2\hat{k},\hat{i}+2\hat{j}-\hat{k},-2\hat{i}-\hat{j}+3\hat{k}$ and $5\hat{i}-2\alpha \hat{j}+4\hat{k}$ are coplanar, then $\alpha$ is equal to

A Journey into 3D Space

Imagine you are standing in a vast, empty three-dimensional room. You have four points floating in this space, and you are told that these four points—let us call them $A$, $B$, $C$, and $D$—are special. They are coplanar.

This means that if you were to take a perfectly flat sheet of paper, you could place it in such a way that all four points touch the paper simultaneously. This is the essence of coplanarity.

In our problem, we are given the position vectors of these four points, but there is a catch: the point $D$ has an unknown coordinate, $\alpha$. Our mission is to find the value of $\alpha$ that forces point $D$ to lie on the same flat sheet as $A$, $B$, and $C$.

To solve this, we need to translate our geometric intuition into the language of vectors. If we anchor ourselves at point $A$, we can draw three vectors: $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$.

If these three vectors lie on the same plane, they cannot span any volume. In the world of vector algebra, the volume of the parallelepiped formed by three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ is given by their scalar triple product, denoted as $[\vec{u}\vec{v}\vec{w}]$.

If the points are coplanar, this volume must be zero. This is our golden key.

First, let us calculate these vectors by subtracting the position vector of $A$ from the others.

For $\overrightarrow{AB}$, we take $B-A$: $(1-3)\hat{i}+(2-(-4))\hat{j}+(-1-2)\hat{k}$, which simplifies to $\overrightarrow{AB}=-2\hat{i}+6\hat{j}-3\hat{k}$.

For $\overrightarrow{AC}$, we take $C-A$: $(-2-3)\hat{i}+(-1-(-4))\hat{j}+(3-2)\hat{k}$, resulting in $\overrightarrow{AC}=-5\hat{i}+3\hat{j}+1\hat{k}$.

For $\overrightarrow{AD}$, we take $D-A$: $(5-3)\hat{i}+(-2\alpha -(-4))\hat{j}+(4-2)\hat{k}$, which gives us $\overrightarrow{AD}=2\hat{i}+(4-2\alpha )\hat{j}+2\hat{k}$.

The Heart of the Problem

Now that we have our vectors, we construct the determinant. The scalar triple product is equivalent to the determinant of a $3\times 3$ matrix where each row represents one of our vectors:

This determinant represents the volume of the parallelepiped. By setting it to zero, we are mathematically enforcing the condition that the points must be coplanar.

Expanding along the first row, we get:

Let us simplify this step-by-step. Inside the first bracket, we have $6-4+2\alpha$, which simplifies to $2+2\alpha$. In the second bracket, we have $-10-2$, which is $-12$.

In the third bracket, we have $-20+10\alpha -6$, which simplifies to $10\alpha -26$. Substituting these back into our equation, we get:

Now, distribute the constants:

Grouping the $\alpha$ terms and the constants together, we have:

Finally, solving for $\alpha$:

And there we have it! By simply enforcing the geometric condition of coplanarity through the scalar triple product, we have found the exact value of $\alpha$ that keeps our points on that flat sheet.

The final result is $\alpha =\frac{73}{17}$.

It is a beautiful demonstration of how abstract vector algebra can describe the physical reality of space. Remember, whenever you face a problem involving coplanarity, think of the volume of the parallelepiped—it is your most powerful tool.