Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=2\hat{i}-\hat{j}+5\hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2\hat{k}$. If $((\vec{a}\times \vec{b})\times \hat{i})\cdot \hat{k}=\frac{23}{2}$, then $|\vec{b}\times 2\hat{j}|$ is equal to

Imagine you are standing in a 3D coordinate system, looking at two vectors, $\vec{a}=2\hat{i}-\hat{j}+5\hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2\hat{k}$.

We are asked to evaluate a complex expression involving a cross product nested within another cross product:

Many students would immediately dive into the matrix determinant method to calculate $\vec{a}\times \vec{b}$. While that works, it is a path filled with potential for algebraic slips.

Instead, let us embrace the elegance of the Vector Triple Product identity:

By applying this identity to our expression, we treat $(\vec{a}\times \vec{b})$ as our first vector and $\hat{i}$ as our third. The expansion becomes:

Suddenly, the terrifying cross products have vanished, replaced by simple dot products. We know that $\vec{a}\cdot \hat{i}$ is just the x-component of $\vec{a}$, which is $2$.

Similarly, $\vec{b}\cdot \hat{i}$ is the x-component of $\vec{b}$, which is $\alpha$. Our expression simplifies to $2\vec{b}-\alpha \vec{a}$.

This is the power of mathematical identity—it turns a complex geometric operation into a simple linear combination.

Now, we are told that the dot product of this result with $\hat{k}$ is $\frac{23}{2}$. Dotting with $\hat{k}$ is just a filter that extracts the z-component of our vectors.

For $\vec{b}$, the z-component is $2$. For $\vec{a}$, the z-component is $5$. Our equation becomes:

Solving this linear equation is straightforward: $4-5\alpha =\frac{23}{2}$, which leads to $-5\alpha =\frac{15}{2}$, and finally $\alpha =-\frac{3}{2}$.

With $\alpha$ in hand, we turn to the final task: finding the magnitude of $\vec{b}\times 2\hat{j}$. We substitute $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2\hat{k}$ and cross it with $2\hat{j}$.

As we distribute the cross product, we see that $\beta (\hat{j}\times \hat{j})$ becomes zero. The $\beta$ component, which seemed so important, simply evaporates.

We are left with $2\alpha (\hat{i}\times \hat{j})+4(\hat{k}\times \hat{j})$, which simplifies to $2\alpha \hat{k}-4\hat{i}$. The magnitude of this vector is:

Substituting $\alpha =-\frac{3}{2}$, we get:

It is a perfect, clean integer. This problem reminds us that in physics and mathematics, complexity is often just a mask for underlying simplicity. When you see a nested cross product, do not panic; look for the identity that will set you free. The final answer is 5.