Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}$ and $\vec{b}$ be two vector such that $|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$ and $|\vec{a}\times \vec{b}|=\sqrt{48}$. Then $(\vec{a}\cdot \vec{b}{)}^2$ is equal to

We are given two vectors, $\vec{a}$ and $\vec{b}$, with magnitudes $|\vec{a}|=\sqrt{14}$ and $|\vec{b}|=\sqrt{6}$. We are also provided with the magnitude of their cross product, $|\vec{a}\times \vec{b}|=\sqrt{48}$.

Our objective is to determine the value of $(\vec{a}\cdot \vec{b}{)}^2$. While one could solve for the angle $\theta$ between the vectors, there is a more direct path using the fundamental relationship between these vector operations.

Lagrange's Identity is an essential tool in vector calculus that relates the dot product and the cross product of two vectors. It is expressed as:

This identity acts as a geometric version of the Pythagorean theorem. It demonstrates that the square of the area of the parallelogram formed by the vectors (the cross product) and the square of their projection (the dot product) sum to the product of the squares of their magnitudes.

A Symphony of Simplification

First, we calculate the squares of the given magnitudes: $|\vec{a}{|}^2=(\sqrt{14}{)}^2=14$ $|\vec{b}{|}^2=(\sqrt{6}{)}^2=6$ $|\vec{a}\times \vec{b}{|}^2=(\sqrt{48}{)}^2=48$

Substituting these values into the identity, we obtain:

Simplifying the right side of the equation:

To isolate the desired term, we subtract $48$ from both sides of the equation:

The final result is 36. By utilizing Lagrange's Identity, we bypassed the need for trigonometric calculations, demonstrating the efficiency of recognizing underlying vector symmetries.