Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k},\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$ and $\vec{c}$ be vector such that $\vec{a}\times \vec{c}=\vec{b}$. If $\vec{c}\cdot (\hat{i}+\hat{j}+\hat{k})=4$, then $|\vec{a}+\vec{c}{|}^2$ is equal to :

Unlocking the Mystery of $\vec{c}$

Imagine standing in a three-dimensional coordinate system. You have two fixed vectors, $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$.

They are locked in place, rigid and unmoving. Then, there is a ghost—a vector $\vec{c}$—that we cannot see, but whose behavior is dictated by the laws of cross products and dot products.

Our mission is to capture this ghost. This is not just a calculation; it is a detective story where we use the language of linear algebra to reveal the hidden identity of $\vec{c}$.

The Assumption of Form

When we face an unknown vector, we must give it a name and a structure. We assume $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$.

Because we do this, we transform an abstract vector equation into a concrete system of linear equations. We are essentially saying, "I don't know where you are, $\vec{c}$, but I know you have components $x,y,$ and $z$."

This is the first step toward mastery: turning the unknown into the manageable.

The Cross Product—The Determinant's Dance

The problem gives us a powerful clue: $\vec{a}\times \vec{c}=\vec{b}$. This is the heart of the problem.

To solve this, we invoke the determinant method. We set up our $3\times 3$ matrix:

As we expand this, we are not just doing arithmetic; we are calculating the area of the parallelogram formed by $\vec{a}$ and $\vec{c}$. Expanding along the first row, we get:

Simplifying this, we arrive at:

The System Emerges

Now, we equate this to $\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$. For two vectors to be equal, their components must match perfectly.

This gives us three beautiful equations:

1. $2z-2y=8⟹z-y=4$ 2. $z+2x=7$ 3. $-y-2x=-3⟹y+2x=3$

We have three equations, but are they enough? These equations are linearly dependent, meaning we need one more piece of information to pin down the values of $x,y,$ and $z$.

The Dot Product—The Final Anchor

The problem provides the final key: $\vec{c}\cdot (\hat{i}+\hat{j}+\hat{k})=4$. This is the dot product, the scalar projection that anchors our vector $\vec{c}$ in space.

Substituting our components, we get:

Now, the system is complete. We have four equations and three variables. The fog is lifting.

The Resolution

Solving this system is a test of patience and precision. From equation (1), we have $z=y+4$.

Substituting this into our dot product equation ($x+y+z=4$), we get $x+y+(y+4)=4$, which simplifies to $x+2y=0$, or $x=-2y$.

Now, we substitute $x=-2y$ into equation (3): $y+2(-2y)=3$. This yields $-3y=3$, so $y=-1$.

With $y$ in hand, the rest falls like dominoes: $x=-2(-1)=2$, and $z=-1+4=3$. Our vector $\vec{c}$ is revealed: $\vec{c}=2\hat{i}-\hat{j}+3\hat{k}$.

The question asks for $|\vec{a}+\vec{c}{|}^2$. First, we find the sum:

Finally, the magnitude squared is the sum of the squares of the components:

We have arrived at the answer: 27. This journey shows that even the most complex vector problems are just puzzles waiting to be solved by breaking them down into their fundamental components.