Animated Solution for Mathematics - Vector Algebra: Let $\vec{a},\vec{b}$ and $\vec{c}$ be three unit vectors such that $|\vec{a}-\vec{b}{|}^2+|\vec{a}-\vec{c}{|}^2=8$. Then $|\vec{a}+2\vec{b}{|}^2+|\vec{a}+2\vec{c}{|}^2$ is equal to

Welcome, future engineer! Today, we are going to dive into the elegant world of vectors. When we talk about unit vectors $\vec{a},\vec{b},$ and $\vec{c}$, we are talking about objects with a magnitude of exactly $1$.

This is our anchor. It means that whenever you see $|\vec{a}{|}^2$, $|\vec{b}{|}^2$, or $|\vec{c}{|}^2$, you can immediately replace them with $1$. This is the first step in simplifying the complex into the manageable.

Now, let us look at the given equation:

This looks intimidating, but it is just a test of your algebraic toolkit. We use the fundamental identity for the squared magnitude of a difference:

By applying this to both terms, we get:

Notice how the terms start to collapse. Since $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$, the equation becomes:

This simplifies beautifully to:

Here is where the magic happens. We have isolated the sum of the dot products. By subtracting $4$ from both sides, we get:

Dividing by $-2$, we find that:

This is our secret ingredient! We do not need to know the individual values of $\vec{a}\cdot \vec{b}$ or $\vec{a}\cdot \vec{c}$; we only need their sum. This is a powerful realization in JEE problems—often, the path to the answer lies in finding a specific combination of variables rather than solving for each one individually.

Now, we turn our attention to the target expression:

We apply the expansion identity again, this time for the sum:

Expanding the terms gives us:

This simplifies to:

Substituting our unit magnitudes, we get:

Finally, we plug in our secret ingredient:

And there you have it! The complexity vanishes, leaving behind a simple, elegant result of 2. Keep practicing this method of breaking down expressions, and you will find that even the hardest problems start to yield.