Animated Solution for Mathematics - Vector Algebra: Let $\vec{A}=2\hat{i}+\hat{k}$, $\vec{B}=\hat{i}+\hat{j}+\hat{k}$, and $\vec{C}=4\hat{i}-3\hat{j}+7\hat{k}$. Determine a vector $\vec{R}$ satisfying $\vec{R}\times \vec{B}=\vec{C}\times \vec{B}$ and $\vec{R}\cdot \vec{A}=0$.

Welcome, fellow explorer of the physical world! Today, we are not just solving a vector equation; we are uncovering a geometric truth hidden within the language of $\hat{i}$, $\hat{j}$, and $\hat{k}$.

We are given three vectors, $\vec{A}$, $\vec{B}$, and $\vec{C}$, and we are tasked with finding a mysterious vector $\vec{R}$ that obeys two strict rules: it must satisfy a cross-product condition and a dot-product condition. Let us embark on this journey together.

Our first condition is $\vec{R}\times \vec{B}=\vec{C}\times \vec{B}$. The immediate temptation for many is to 'cancel' the $\vec{B}$ from both sides.

But remember, in the world of vectors, the cross product is not a simple scalar multiplication. We cannot just divide by $\vec{B}$. Instead, let us bring everything to one side:

By the distributive property, this becomes $(\vec{R}-\vec{C})\times \vec{B}=0$. Now, pause and think: what does it mean when the cross product of two vectors is zero? It means they are parallel! Thus, the vector $(\vec{R}-\vec{C})$ must be parallel to $\vec{B}$.

Since $(\vec{R}-\vec{C})$ is parallel to $\vec{B}$, we can express this relationship using a scalar parameter, let us call it $\lambda$. We write:

Geometrically, this is a beautiful revelation. This equation describes a straight line in 3D space that passes through the tip of vector $\vec{C}$ and extends infinitely in the direction of vector $\vec{B}$.

Our unknown vector $\vec{R}$ is not just any vector; it is a point living on this specific line.

Now, we introduce the second condition: $\vec{R}\cdot \vec{A}=0$. This is the 'pinpoint' condition. A dot product of zero tells us that $\vec{R}$ must be perpendicular to $\vec{A}$.

We now have our strategy: we will substitute our parametric expression for $\vec{R}$ into this dot product equation to find the exact value of $\lambda$ that places $\vec{R}$ in the correct spot on the line.

Substituting $\vec{R}=\vec{C}+\lambda \vec{B}$ into $\vec{R}\cdot \vec{A}=0$, we get:

Expanding this using the distributive property, we arrive at:

Let us calculate these dot products. Given $\vec{A}=2\hat{i}+\hat{k}$, $\vec{B}=\hat{i}+\hat{j}+\hat{k}$, and $\vec{C}=4\hat{i}-3\hat{j}+7\hat{k}$:

1. $\vec{C}\cdot \vec{A}=(4)(2)+(-3)(0)+(7)(1)=8+0+7=15$. 2. $\vec{B}\cdot \vec{A}=(1)(2)+(1)(0)+(1)(1)=2+0+1=3$.

Substituting these back into our equation: $15+\lambda (3)=0$. Solving for $\lambda$, we find $3\lambda =-15$, so $\lambda =-5$.

We have found the key! Now, we plug $\lambda =-5$ back into our line equation: $\vec{R}=\vec{C}-5\vec{B}$.

Performing the final arithmetic:

And there it is! Our vector $\vec{R}=-\hat{i}-8\hat{j}+2\hat{k}$. You have successfully navigated the geometry of 3D space.

Remember, every equation tells a story—the cross product defined the path, and the dot product defined the destination. Keep visualizing, keep calculating, and keep falling in love with the elegance of physics!