Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}-2\hat{k}$ and $\vec{c}=-\hat{i}+4\hat{j}+3\hat{k}$. If $\vec{d}$ is a vector perpendicular to both $\vec{b}$ and $\vec{c}$, and $\vec{a}\cdot \vec{d}=18$, then $|\vec{a}\times \vec{d}{|}^2$ is equal to

Imagine you are standing in a 3D coordinate system with three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. We seek a vector $\vec{d}$ that is perpendicular to both $\vec{b}$ and $\vec{c}$.

When a vector is perpendicular to two others, it must lie along the normal to the plane containing those two vectors. The most elegant way to find this normal vector is via the cross product.

By calculating $\vec{b}\times \vec{c}$, we determine the direction of $\vec{d}$.

We know $\vec{d}$ is parallel to $\vec{b}\times \vec{c}$, but its magnitude is unknown. We introduce a scalar $\lambda$ and define:

Given the vectors $\vec{b}=\hat{i}-2\hat{j}-2\hat{k}$ and $\vec{c}=-\hat{i}+4\hat{j}+3\hat{k}$, we set up the determinant to compute the cross product:

Next, we apply the constraint $\vec{a}\cdot \vec{d}=18$. Substituting $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$ and $\vec{d}=\lambda (2\hat{i}-\hat{j}+2\hat{k})$, the dot product becomes:

This simplifies to $9\lambda =18$, which yields $\lambda =2$. Consequently, our vector $\vec{d}$ is fully revealed as:

Lagrange's Identity

We now need to find $|\vec{a}\times \vec{d}{|}^2$. While one could compute the cross product directly, we can utilize Lagrange's Identity for a more efficient calculation:

First, we calculate the squared magnitudes of the vectors:

Plugging these values into the identity, we obtain:

The final result is 720. This journey demonstrates that with the right tools—the cross product for direction, the dot product for scaling, and Lagrange's Identity for efficiency—complex 3D problems become straightforward.