Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=3\hat{i}+2\hat{j}+\hat{k}$, $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b})\times \vec{c}=2(\vec{a}\times \vec{b})+24\hat{j}-6\hat{k}$ and $(\vec{a}-\vec{b}+\hat{i})\cdot \vec{c}=-3$. Then $|\vec{c}{|}^2$ is equal to _______.

Unlocking the Mystery of $\vec{c}$

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of 3D vector space.

When you look at a problem like this, it is easy to feel overwhelmed by the symbols—the cross products, the dot products, the hats on the unit vectors. But I want you to take a deep breath. Behind every symbol is a geometric reality.

We are looking for a vector $\vec{c}$ that exists in a space defined by $\vec{a}$ and $\vec{b}$. Let us break this down, step by step, and turn this complexity into clarity.

The Foundation

We begin with our building blocks: $\vec{a}=3\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$. These are our anchors.

The problem asks us to work with $\vec{a}+\vec{b}$ and $\vec{a}\times \vec{b}$. Adding vectors is the most intuitive part of our journey. We simply combine the components:

This is our new vector, the sum of our anchors.

Next, we calculate the cross product $\vec{a}\times \vec{b}$. Think of the cross product as a machine that generates a vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$.

Using the determinant method, we set up our matrix with $\hat{i},\hat{j},\hat{k}$ in the first row, followed by the components of $\vec{a}$ and $\vec{b}$. Expanding this, we get:

This vector is the heart of our right-hand side.

The Cross Product Challenge

Now, let us look at the equation $(\vec{a}+\vec{b})\times \vec{c}=2(\vec{a}\times \vec{b})+24\hat{j}-6\hat{k}$. We already know $\vec{a}\times \vec{b}$, so let us calculate the full right-hand side (RHS).

Multiplying our cross product by 2 gives $14\hat{i}-14\hat{j}-14\hat{k}$. Adding $24\hat{j}-6\hat{k}$ to this, we get:

Now, we introduce our unknown vector $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$. The left-hand side (LHS) is the cross product of $(5\hat{i}+\hat{j}+4\hat{k})$ and $(x\hat{i}+y\hat{j}+z\hat{k})$.

Expanding this determinant, we get:

Do not be intimidated by these variables! They are just placeholders for the coordinates we need to find.

The System of Equations

Since the LHS must equal the RHS, we equate the components. This gives us three beautiful, simple linear equations:

1) $z-4y=14$ 2) $4x-5z=10$ 3) $5y-x=-20$

From the third equation, we can easily express $x$ in terms of $y$: $x=5y+20$. This is our first breakthrough. We are slowly peeling back the layers of this problem.

The Dot Product Key

We have one more condition: $(\vec{a}-\vec{b}+\hat{i})\cdot \vec{c}=-3$. First, let us simplify $\vec{a}-\vec{b}+\hat{i}$.

This is $(3-2+1)\hat{i}+(2-(-1))\hat{j}+(1-3)\hat{k}=2\hat{i}+3\hat{j}-2\hat{k}$. Now, we take the dot product with $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$, which gives:

This is the missing piece of our puzzle!

The Final Calculation

We have $x=5y+20$. From the first equation, $z=4y+14$. Substituting these into our dot product equation:

Expanding this, we get $10y+40+3y-8y-28=-3$, which simplifies to $5y+12=-3$. Thus, $5y=-15$, and $y=-3$.

With $y=-3$, we find $x=5(-3)+20=5$ and $z=4(-3)+14=2$. Our vector $\vec{c}$ is $5\hat{i}-3\hat{j}+2\hat{k}$.

Finally, the magnitude squared is:

We have arrived at the destination. You did it! The elegance of the math is in how everything cancels out perfectly. Keep this confidence, and you will conquer any problem the JEE throws at you.