Animated Solution for Mathematics - Vector Algebra: Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times (2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times \vec{d}$. If ${\lambda }_1,{\lambda }_2({\lambda }_1>{\lambda }_2)$ are the possible values of $(\vec{c}+\vec{d})\cdot (-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation $K^2{x}^2+(K^2-5K+{\lambda }_1)xy+(3K+\frac{{\lambda }_2}{2})y^2-8x+12y+{\lambda }_2=0$ represents a circle, for $K$ equal to :

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a problem; we are orchestrating a meeting between two distinct worlds: the abstract, directional elegance of Vector Algebra and the rigid, structured beauty of Coordinate Geometry.

I know that when you first look at this problem, the sight of cross products mixed with a quadratic equation in $x$ and $y$ might feel overwhelming. But take a deep breath. Let us break this down, step by step, and find the hidden harmony.

Our journey begins with the equation $\vec{c}\times (2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times \vec{d}$. Let us simplify our lives immediately. That vector $(2\hat{i}+3\hat{j}+4\hat{k})$ is a recurring character, so let us call it $\vec{a}$.

Now, the equation reads $\vec{c}\times \vec{a}=\vec{a}\times \vec{d}$. Here is where the trap lies for the unwary. The cross product is not commutative; it is anti-commutative. This means $\vec{a}\times \vec{d}=-\vec{d}\times \vec{a}$.

By substituting this, our equation transforms into $\vec{c}\times \vec{a}+\vec{d}\times \vec{a}=0$. By the distributive property, we arrive at $(\vec{c}+\vec{d})\times \vec{a}=0$.

What does this tell us? It tells us that the vector sum $(\vec{c}+\vec{d})$ is parallel to $\vec{a}$. Mathematically, this means:

for some scalar $t$. We have successfully tamed the vector beast!

Now, we need to find $t$. We are given that $|\vec{c}+\vec{d}|=\sqrt{29}$. Since $\vec{c}+\vec{d}=t\vec{a}$, we have $|t||\vec{a}|=\sqrt{29}$.

Calculating the magnitude of $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$, we find:

Thus, $|t|\sqrt{29}=\sqrt{29}$, which implies $|t|=1$. So, $t$ can be $1$ or $-1$.

This gives us two possible values for the dot product $\lambda =(\vec{c}+\vec{d})\cdot (-7\hat{i}+2\hat{j}+3\hat{k})$. Substituting $\vec{c}+\vec{d}=\pm \vec{a}$, we get:

Since ${\lambda }_1>{\lambda }_2$, we identify ${\lambda }_1=4$ and ${\lambda }_2=-4$.

We now enter the realm of Coordinate Geometry. We are given the equation:

For this to represent a circle, it must satisfy two fundamental laws: first, the coefficient of $xy$ must be zero, because a circle has no rotation or skew. Second, the coefficients of $x^2$ and $y^2$ must be equal.

Setting the $xy$ coefficient to zero:

Factoring this, we get $(K-1)(K-4)=0$, so $K=1$ or $K=4$.

Now, equating the $x^2$ and $y^2$ coefficients:

This gives $K^2-3K+2=0$, which factors to $(K-1)(K-2)=0$, yielding $K=1$ or $K=2$.

We have two sets of potential values for $K$: $\{1,4\}$ and $\{1,2\}$. For the equation to represent a circle, $K$ must satisfy both conditions simultaneously.

The only value that appears in both sets is $K=1$.

And there you have it! We navigated the vector cross product, solved for the scalar magnitude, determined our constants, and applied the geometric constraints of a circle. It is not just about the answer; it is about the elegance of the path. Keep practicing, keep questioning, and keep falling in love with the logic.