Animated Solution for Mathematics - Vector Algebra: Let $\vec{a},\vec{b},\vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{31},4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a}\times \vec{b})=3(\vec{c}\times \vec{a})$. If the angle between $\vec{b}$ and $\vec{c}$ is $2\pi /3$, then $(\frac{\vec{a}\times \vec{c}}{\vec{a}\cdot \vec{b}}{)}^2$ is equal to

A Journey into Space

Imagine you are standing in a vast, three-dimensional coordinate system. You have three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, floating in space. They aren't just arrows; they are the building blocks of the geometry we are about to unravel.

The problem provides a set of constraints—magnitudes and a cross-product relation—that might seem daunting at first. As we peel back the layers, we will find that this is a story of symmetry and elegant cancellation.

The Cross Product Mystery

We are given the relation $2(\vec{a}\times \vec{b})=3(\vec{c}\times \vec{a})$. At first glance, this looks like a messy equation.

Remember the fundamental nature of the cross product: it is anti-commutative. If you swap the order of the vectors, you pick up a negative sign. Thus, $\vec{c}\times \vec{a}$ is simply $-(\vec{a}\times \vec{c})$.

Substituting this into our equation, we get $2(\vec{a}\times \vec{b})=-3(\vec{a}\times \vec{c})$. If we bring everything to one side, we have $2(\vec{a}\times \vec{b})+3(\vec{a}\times \vec{c})=0$.

By the distributive property of the cross product, this simplifies beautifully to:

This is our "Aha!" moment. The cross product of $\vec{a}$ and the vector $(2\vec{b}+3\vec{c})$ is zero. Geometrically, this means $\vec{a}$ is perfectly parallel to the vector $(2\vec{b}+3\vec{c})$.

The Geometric Bridge

Since $\vec{a}$ is parallel to $(2\vec{b}+3\vec{c})$, we can write $\vec{a}=\lambda (2\vec{b}+3\vec{c})$ for some scalar $\lambda$. This is the bridge that connects our unknown vector $\vec{a}$ to the known vectors $\vec{b}$ and $\vec{c}$.

Now, we need to find $\lambda$. We know $|\vec{a}|=\sqrt{31}$, so let's square both sides:

Expanding the right side, we get:

The Magnitude Hunt

We have the magnitudes: $|\vec{b}|=1/2$ and $|\vec{c}|=2$. The dot product $\vec{b}\cdot \vec{c}$ is $|\vec{b}||\vec{c}|\cos (2\pi /3)$.

Since $\cos (2\pi /3)=-1/2$, the dot product is $(1/2)(2)(-1/2)=-1/2$. Plugging these values into our expanded equation:

This simplifies to $31={\lambda }^2(1+36-6)$, which is $31=31{\lambda }^2$. Thus, ${\lambda }^2=1$.

The Final Assembly

Now, we need to calculate the ratio ${\left(\frac{\vec{a}\times \vec{c}}{\vec{a}\cdot \vec{b}}\right)}^2$. Let's look at the numerator:

The magnitude squared is $|\vec{a}\times \vec{c}{|}^2=4{\lambda }^2|\vec{b}\times \vec{c}{|}^2$. Using the formula $|\vec{b}\times \vec{c}{|}^2=|\vec{b}{|}^2|\vec{c}{|}^2{\sin }^2(2\pi /3)$, we get:

So, $|\vec{a}\times \vec{c}{|}^2=4(1)(3/4)=3$.

For the denominator, we calculate:

Squaring this gives $(\vec{a}\cdot \vec{b}{)}^2={\lambda }^2=1$. The final ratio is $3/1=3$.

We have arrived at the destination, and the final answer is 3.