Animated Solution for Mathematics - Vector Algebra: Let the vectors $\vec{a},\vec{b},\vec{c}$ represent three coterminous edges of a parallelopiped of volume $V$. Then the volume of the parallelopiped, whose coterminous edges are represented by $\vec{a},\vec{b}+\vec{c}$ and $\vec{a}+2\vec{b}+3\vec{c}$ is equal to

Imagine you are standing in a vast, empty 3D space. You have a parallelepiped—a simple, slanted box—defined by three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, meeting at a single corner.

These vectors are the DNA of your box. In the world of vector algebra, the volume $V$ of this box is not just length times width times height; it is the scalar triple product, denoted as $V=[\vec{a}\vec{b}\vec{c}]$.

This is our foundational truth.

Now, the problem introduces a challenge. We are asked to construct a new parallelepiped where the edges are transformed into:

At first glance, you might be tempted to start calculating cross products and dot products manually. Resist that urge!

In JEE Advanced, time is your most precious resource. When you see new vectors defined as linear combinations of the old ones, you are looking at a linear transformation. The volume of the new parallelepiped, $V^{\prime }$, is simply the original volume $V$ scaled by the determinant of the transformation matrix.

This is where the magic happens. We can represent our new vectors as rows in a matrix, where each column corresponds to the coefficients of $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Let us extract the coefficients carefully: - For $\vec{u}=1\vec{a}+0\vec{b}+0\vec{c}$, the row is $(1,0,0)$. - For $\vec{v}=0\vec{a}+1\vec{b}+1\vec{c}$, the row is $(0,1,1)$. - For $\vec{w}=1\vec{a}+2\vec{b}+3\vec{c}$, the row is $(1,2,3)$.

Our determinant $D$ is defined as:

Expanding this determinant is a breeze if you choose the right row. Let us expand along the first row, which is filled with zeros.

This simplifies our work significantly:

The determinant is exactly $1$. This means the volume of our new, distorted parallelepiped is $V^{\prime }=1\cdot V=V$.

Despite the stretching and the complex linear combinations, the volume remains unchanged.

This problem teaches us that geometry is often simpler than it appears. By using the properties of the scalar triple product and the determinant, we bypassed the tedious algebra and arrived at the truth with elegance.

Remember, in JEE, the most complex-looking problems often have the most beautiful, simple solutions hidden just beneath the surface. Keep looking for those shortcuts!