Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}$ and $\vec{b}$ be two non-collinear unit vectors. If $\vec{u}=\vec{a}-(\vec{a}\cdot \vec{b})\vec{b}$ and $\vec{v}=\vec{a}\times \vec{b}$, then $|\vec{v}|$ is

Imagine you are standing in a vast, open space, and before you lie two unit vectors, $\vec{a}$ and $\vec{b}$. They are the fundamental building blocks of our vector space.

Because they are unit vectors, their magnitudes are exactly $1$. Since they are non-collinear, they define a plane, separated by an angle $\theta$ that lies strictly between $0$ and $\pi$.

The Area of Possibility

Our first protagonist is $\vec{v}=\vec{a}\times \vec{b}$. The magnitude of a cross product is the product of the magnitudes of the vectors times the sine of the angle between them.

Since $|\vec{a}|=1$ and $|\vec{b}|=1$, the magnitude $|\vec{v}|$ simplifies elegantly to:

This is our target benchmark. We must now determine if our second vector, $\vec{u}$, shares this same magnitude.

Consider the vector $\vec{u}=\vec{a}-(\vec{a}\cdot \vec{b})\vec{b}$. The term $(\vec{a}\cdot \vec{b})\vec{b}$ represents the vector projection of $\vec{a}$ onto $\vec{b}$, effectively the shadow of $\vec{a}$ cast upon $\vec{b}$.

When you subtract this shadow from the original vector $\vec{a}$, you are left with the component of $\vec{a}$ that is perfectly perpendicular to $\vec{b}$. This is the "rejection" of $\vec{a}$ from $\vec{b}$, forming a right-angled triangle.

To find the magnitude of $\vec{u}$, we calculate its square: $|\vec{u}{|}^2=\vec{u}\cdot \vec{u}$. Expanding this expression, we have:

Using the distributive property, we obtain:

Given that $\vec{a}\cdot \vec{b}=\cos \theta$ and $|\vec{a}|=|\vec{b}|=1$, this simplifies to:

Applying the fundamental trigonometric identity, we find:

We have arrived at the climax. Since $|\vec{v}|=\sin \theta$ and $|\vec{u}|=\sin \theta$, we conclude that $|\vec{v}|=|\vec{u}|$.

To verify further properties, we calculate the dot product of $\vec{u}$ and $\vec{b}$:

Because $\vec{u}\cdot \vec{b}=0$, the expression $|\vec{u}|+|\vec{u}\cdot \vec{b}|$ simplifies to $|\vec{u}|$. You have successfully navigated the geometry and algebra, proving that these expressions are two sides of the same coin.