Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}$ and $\vec{b}$ be two vectors. Let $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a}\cdot \vec{b}=2$. If $\vec{c}=(2\vec{a}\times \vec{b})-3\vec{b}$, then the value of $\vec{b}\cdot \vec{c}$ is

We are given two vectors, $\vec{a}$ and $\vec{b}$, with the following properties: $|\vec{a}|=1$, $|\vec{b}|=4$, and $\vec{a}\cdot \vec{b}=2$.

Our objective is to evaluate the expression $\vec{b}\cdot \vec{c}$, where the vector $\vec{c}$ is defined as:

While it may be tempting to resolve these vectors into components, the most efficient approach in JEE mathematics is to leverage the fundamental properties of vector algebra.

We begin by substituting the definition of $\vec{c}$ into our target expression:

Applying the distributive property of the dot product over vector subtraction, we obtain:

By pulling the scalar constants to the front, the expression simplifies to:

Consider the first term: $\vec{b}\cdot (\vec{a}\times \vec{b})$.

By definition, the cross product $\vec{a}\times \vec{b}$ results in a vector that is perpendicular to the plane containing both $\vec{a}$ and $\vec{b}$. Consequently, this resulting vector is orthogonal to $\vec{b}$.

Since the dot product of any two orthogonal vectors is zero, the term vanishes:

We are now left with the second term of our expression:

Recalling that the dot product of a vector with itself is the square of its magnitude, we have $\vec{b}\cdot \vec{b}=|\vec{b}{|}^2$. Given $|\vec{b}|=4$, we calculate:

Substituting this value back into our equation:

The final result is -48.