Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{c}=\hat{j}-\hat{k}$ and a vector $\vec{b}$ be such that $\vec{a}\times \vec{b}=\vec{c}$ and $\vec{a}\cdot \vec{b}=3$. Then $|\vec{b}|$ equals :

Welcome, future engineers! Today, we are going to dive into a problem that might look like a simple algebraic exercise, but it is actually a beautiful dance of geometry and vector identities.

We are given two vectors, $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{j}-\hat{k}$. We are told that there exists a vector $\vec{b}$ such that $\vec{a}\times \vec{b}=\vec{c}$ and $\vec{a}\cdot \vec{b}=3$.

Our goal is to find the magnitude $|\vec{b}|$. Let's break this down step by step.

The Setup

First, let's visualize what we have. The cross product $\vec{a}\times \vec{b}=\vec{c}$ tells us that the vector $\vec{c}$ is perpendicular to the plane formed by $\vec{a}$ and $\vec{b}$.

We also have the dot product $\vec{a}\cdot \vec{b}=3$, which gives us information about the projection of $\vec{b}$ onto $\vec{a}$. To solve this, we need to connect these two pieces of information.

We start by calculating the squared magnitudes of the vectors we know: For $\vec{a}=1\hat{i}+1\hat{j}+1\hat{k}$, the magnitude squared is $|\vec{a}{|}^2=(1{)}^2+(1{)}^2+(1{)}^2=3$. Similarly, for $\vec{c}=0\hat{i}+1\hat{j}-1\hat{k}$, the magnitude squared is $|\vec{c}{|}^2=(0{)}^2+(1{)}^2+(-1{)}^2=2$.

The Bridge

Now, here is the secret weapon of vector algebra: Lagrange's Identity. This identity is a favorite of the JEE examiners because it elegantly bridges the gap between the cross product and the dot product.

It states that:

Since the cross product $\vec{a}\times \vec{b}$ is equal to $\vec{c}$, we know that $|\vec{a}\times \vec{b}{|}^2$ must be equal to $|\vec{c}{|}^2$. We already calculated $|\vec{c}{|}^2=2$, so we have our first piece of the puzzle: $|\vec{a}\times \vec{b}{|}^2=2$.

The Calculation

Now, let's substitute everything we know into Lagrange's Identity. We have $|\vec{a}\times \vec{b}{|}^2=2$, $\vec{a}\cdot \vec{b}=3$, and $|\vec{a}{|}^2=3$.

Plugging these into the equation, we get:

Three squared is nine, and two plus nine is eleven. So, our equation simplifies to:

To isolate $|\vec{b}{|}^2$, we divide both sides by three:

Finally, we take the square root of both sides to find the magnitude of $\vec{b}$:

And there you have it! By using Lagrange's Identity, we turned a potentially messy system of equations into a clean, elegant calculation.

This is the power of knowing your vector identities. Keep practicing, keep visualizing, and most importantly, keep falling in love with the mathematics behind these problems. You have got this! The final answer is $|\vec{b}|=\sqrt{\frac{11}{3}}$.