Animated Solution for Mathematics - Vector Algebra: If $\vec{a}=2\hat{i}+\hat{j}+2\hat{k}$, then the value of $|\hat{i}\times (\vec{a}\times \hat{i}){|}^2+|\hat{j}\times (\vec{a}\times \hat{j}){|}^2+|\hat{k}\times (\vec{a}\times \hat{k}){|}^2$ is equal to

Welcome, future engineers! Today, we are going to unravel a problem that, at first glance, looks like a messy algebraic nightmare.

We are given a vector $\vec{a}=2\hat{i}+\hat{j}+2\hat{k}$ and asked to evaluate the sum of the squared magnitudes of three triple cross products: $|\hat{i}\times (\vec{a}\times \hat{i}){|}^2+|\hat{j}\times (\vec{a}\times \hat{j}){|}^2+|\hat{k}\times (\vec{a}\times \hat{k}){|}^2$.

If you try to calculate each cross product manually, you will find yourself drowning in a sea of components. But fear not! We are going to use the elegance of the Vector Triple Product (VTP) to turn this into a simple, beautiful calculation.

The 'BAC-CAB' Rule

Whenever you see a cross product nested inside another, like $\vec{A}\times (\vec{B}\times \vec{C})$, your brain should immediately jump to the Vector Triple Product identity.

We often call this the 'BAC-CAB' rule because it helps us remember the expansion:

This identity is our master key. It transforms a complex vector operation into a combination of dot products and scalar multiplications, which are much easier to handle.

Let's focus on the first term: $\hat{i}\times (\vec{a}\times \hat{i})$. Here, $\vec{A}=\hat{i}$, $\vec{B}=\vec{a}$, and $\vec{C}=\hat{i}$. Applying the BAC-CAB rule, we get:

We know that $\hat{i}\cdot \hat{i}=1$. Furthermore, the dot product $\hat{i}\cdot \vec{a}$ is simply the $x$-component of $\vec{a}$, which is $x$.

So, the expression simplifies to $\vec{a}-x\hat{i}$. If we substitute $\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}$, we get:

Geometrically, this is fascinating! We have stripped away the $x$-component of $\vec{a}$, leaving only the projection of the vector onto the $yz$-plane. The magnitude squared of this resulting vector is simply $y^2+z^2$.

Now, do we need to repeat this for the other two terms? Absolutely not! This is where the beauty of symmetry comes into play.

By following the exact same logic for the second term, $\hat{j}\times (\vec{a}\times \hat{j})$, we will find that it leaves us with the $x$ and $z$ components: $x\hat{i}+z\hat{k}$. Its magnitude squared is $x^2+z^2$.

Similarly, the third term, $\hat{k}\times (\vec{a}\times \hat{k})$, leaves us with $x\hat{i}+y\hat{j}$, and its magnitude squared is $x^2+y^2$. See how the structure repeats?

Now, let's sum these squared magnitudes together:

Grouping the terms, we get $2x^2+2y^2+2z^2$, which is $2(x^2+y^2+z^2)$. Notice that $x^2+y^2+z^2$ is exactly the square of the magnitude of our original vector $\vec{a}$, denoted as $|\vec{a}{|}^2$.

So, the entire expression simplifies to $2|\vec{a}{|}^2$. Given $\vec{a}=2\hat{i}+\hat{j}+2\hat{k}$, we have $x=2,y=1,z=2$.

Thus, $|\vec{a}{|}^2=2^2+1^2+2^2=4+1+4=9$. The final answer is $2\times 9=18$.

This wasn't just a random calculation. You have just proven a standard identity in vector algebra: the sum of the squares of these cross products with the basis vectors is always equal to twice the square of the vector's magnitude.

Keep this in your arsenal, and you will tackle similar problems with confidence and speed. You have done excellent work today!