Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=2\hat{i}-\hat{j}+3\hat{k}$, $\vec{b}=3\hat{i}-5\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a}\times \vec{c}=\vec{c}\times \vec{b}$ and $(\vec{a}+\vec{c})\cdot (\vec{b}+\vec{c})=168$ Then the maximum value of $|\vec{c}{|}^2$ is:

Welcome, fellow traveler on the JEE Advanced journey. Today, we are going to dissect a problem that, at first glance, looks like a messy algebraic nightmare.

You see a vector $\vec{c}$ defined by a cross product condition and a dot product constraint. Your instinct might be to dive straight into component-wise calculation, but stop and take a breath.

In vector algebra, the most powerful tool isn't your ability to calculate; it is your ability to see.

Let us look at the first condition: $\vec{a}\times \vec{c}=\vec{c}\times \vec{b}$.

Most students see this and immediately start writing out the determinant form of the cross product. Please, do not do that yet, as you will be drowning in a sea of components before you even start.

Instead, look at the symmetry. We know that the cross product is anti-commutative, which means $\vec{c}\times \vec{b}=-(\vec{b}\times \vec{c})$.

If we substitute this back into our equation, we get:

Moving the term to the left, we arrive at:

By the distributive property of the cross product, this simplifies beautifully to:

This is the "Aha!" moment. What does it mean for the cross product of two vectors to be the zero vector?

It means they are collinear. It means $\vec{c}$ is parallel to the resultant vector $(\vec{a}+\vec{b})$. We have just reduced a complex vector relationship into a simple geometric statement: $\vec{c}$ lies along the same line as $(\vec{a}+\vec{b})$.

Now that we know $\vec{c}$ is parallel to $(\vec{a}+\vec{b})$, we can define $\vec{c}$ using a scalar parameter, $\lambda$. This is a classic JEE technique—turning a vector problem into a scalar one.

Let us calculate the vector $(\vec{a}+\vec{b})$ using the components provided: $\vec{a}=2\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}-5\hat{j}+\hat{k}$.

And its squared magnitude is:

Keep this number, $77$, in your pocket. It is going to be the anchor for our final calculation.

Now, let us tackle the second condition: $(\vec{a}+\vec{c})\cdot (\vec{b}+\vec{c})=168$. Do not be intimidated by the dot product; treat it exactly like the expansion of $(x+y)(z+y)$.

Rearranging the middle terms, we get:

We already know $\vec{a}\cdot \vec{b}=(2)(3)+(-1)(-5)+(3)(1)=6+5+3=14$. Now, substitute $\vec{c}=\lambda (\vec{a}+\vec{b})$ into the equation:

Since $(\vec{a}+\vec{b})\cdot (\vec{a}+\vec{b})=|\vec{a}+\vec{b}{|}^2=77$, the equation becomes:

We are in the home stretch. Subtract $14$ from both sides:

Divide the entire equation by $77$:

Factoring this quadratic gives us $(\lambda +2)(\lambda -1)=0$, so $\lambda =1$ or $\lambda =-2$.

Finally, we need the maximum value of $|\vec{c}{|}^2$. Since $|\vec{c}{|}^2=|\lambda (\vec{a}+\vec{b}){|}^2={\lambda }^2|\vec{a}+\vec{b}{|}^2=77{\lambda }^2$, we test our values:

If $\lambda =1$, $|\vec{c}{|}^2=77(1{)}^2=77$.

If $\lambda =-2$, $|\vec{c}{|}^2=77(-2{)}^2=77(4)=308$.

The maximum value is clearly $308$.