Animated Solution for Mathematics - Vector Algebra: Let a, b and c be distinct positive numbers. If the vectors $a\hat{i}+a\hat{j}+c\hat{k},\hat{i}+\hat{k}$ and $c\hat{i}+c\hat{j}+b\hat{k}$ are co-planar, then c is equal to:

Understanding Coplanarity

Imagine you are standing in a vast, three-dimensional space. You have three vectors, $\vec{u}$, $\vec{v}$, and $\vec{w}$, all originating from the same point.

Usually, these three vectors would reach out into space and define a volume—a parallelepiped. But today, the problem gives us a special condition: these vectors are coplanar.

They are trapped, forced to lie flat on a single two-dimensional plane. This is a powerful geometric constraint, meaning the volume of the parallelepiped they would otherwise form is exactly zero. This is the key to unlocking the problem.

The Scalar Triple Product

How do we translate this 'flatness' into algebra? We use the Scalar Triple Product, denoted as $[\vec{u}\vec{v}\vec{w}]$.

This product is the mathematical equivalent of calculating the volume of the parallelepiped formed by the vectors. If the volume is zero, the vectors are coplanar.

We define our vectors as $\vec{u}=a\hat{i}+a\hat{j}+c\hat{k}$, $\vec{v}=1\hat{i}+0\hat{j}+1\hat{k}$, and $\vec{w}=c\hat{i}+c\hat{j}+b\hat{k}$. To find the scalar triple product, we arrange their components into a determinant:

This determinant is our master equation. It encapsulates the entire geometric reality of the problem.

Unveiling the Algebra

Now, let's expand this determinant along the first row. We must be meticulous with our signs.

We take the first element, $a$, and multiply it by the determinant of the remaining $2\times 2$ matrix: $\left|\begin{array}{cc}0& 1\\ c& b\end{array}\right|$. This gives us $a(0\cdot b-1\cdot c)=a(0-c)=-ac$.

Next, we move to the second element, $a$, remembering the alternating sign convention (plus, minus, plus). So, we subtract $a$ multiplied by the minor $\left|\begin{array}{cc}1& 1\\ c& b\end{array}\right|$, which is $-a(1\cdot b-1\cdot c)=-a(b-c)$.

Finally, we add the third element, $c$, multiplied by its minor $\left|\begin{array}{cc}1& 0\\ c& c\end{array}\right|$, resulting in $+c(1\cdot c-0\cdot c)=+c(c)=c^2$. Combining these, we get the equation:

Let's distribute the terms and watch the magic happen. Expanding $-a(b-c)$ gives us $-ab+ac$.

Now, our full equation looks like this: $-ac-ab+ac+c^2=0$. Look closely at the terms $-ac$ and $+ac$; they are exact opposites and cancel each other out perfectly.

This leaves us with a beautifully simple expression:

This is the moment where the complexity of the 3D geometry collapses into a simple algebraic relationship. We rearrange this to find $c^2=ab$, which leads us directly to $c=\sqrt{ab}$.

The Geometric Mean

We have arrived at our destination. The value of $c$ is the square root of the product of $a$ and $b$.

In the language of sequences and series, we recognize this as the Geometric Mean of $a$ and $b$. This result is not just a number; it is the condition required for these three vectors to exist in perfect, flat harmony on a single plane.