Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}|=1$ and $|\vec{b}\times \vec{a}|=2$. Then $|(\vec{b}\times \vec{a})-\vec{b}{|}^2$ is equal to

Welcome, future engineers! Today, we are going to peel back the layers of a vector problem that, at first glance, might look like a messy algebraic trap. We are tasked with finding the value of $|(\vec{b}\times \vec{a})-\vec{b}{|}^2$, given that $|\vec{b}|=1$ and $|\vec{b}\times \vec{a}|=2$.

Visualizing the Cross Product

Imagine you are standing in a 3D space with two vectors, $\vec{a}$ and $\vec{b}$. When you perform the cross product $\vec{b}\times \vec{a}$, you are creating a new vector that is perpendicular to the plane containing both $\vec{a}$ and $\vec{b}$.

This is the "Spark" of the problem. Because the cross product is perpendicular to the plane, it is, by definition, perpendicular to every vector lying within that plane.

Therefore, we conclude that:

The Algebraic Toolkit

Now, let us look at the expression we need to evaluate: $|(\vec{b}\times \vec{a})-\vec{b}{|}^2$. In the world of vectors, we use the following identity for the square of the magnitude of a difference:

Let $\vec{u}=(\vec{b}\times \vec{a})$ and $\vec{v}=\vec{b}$. Substituting these into our identity, we obtain:

The 'Aha!' Moment

Look closely at the last term: $2((\vec{b}\times \vec{a})\cdot \vec{b})$. We established in Phase 1 that $(\vec{b}\times \vec{a})$ is perpendicular to $\vec{b}$.

When two vectors are perpendicular, the angle $\theta$ between them is $90^{\circ }$. Since the dot product is defined as $|\vec{u}||\vec{v}|\cos (\theta )$ and $\cos (90^{\circ })=0$, the entire dot product term vanishes:

The Final Calculation

We are left with a beautifully simple equation:

Substituting the given values $|\vec{b}\times \vec{a}|=2$ and $|\vec{b}|=1$:

The final answer is 5. By trusting the geometric properties of the cross product and the algebraic identity of the dot product, we navigated through the problem with ease.