Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$, $\vec{b}=3\hat{i}-\hat{j}+7\hat{k}$ and $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$. If a vector $\vec{d}$ satisfies $\vec{d}\times \vec{b}=\vec{c}\times \vec{b}$ and $\vec{d}\cdot \vec{a}=24$, then $|\vec{d}{|}^2$ is equal to

Welcome, fellow traveler of the mathematical cosmos. Today, we are not just solving a problem; we are embarking on a journey through the elegant, rigid, and beautiful world of three-dimensional vectors.

We are given three vectors: $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$ $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$ $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$

These are our anchors in space. We are searching for a mysterious fourth vector, $\vec{d}$, that obeys two strict laws.

The Cross Product Constraint

The first law is $\vec{d}\times \vec{b}=\vec{c}\times \vec{b}$. If we bring everything to one side, we get $\vec{d}\times \vec{b}-\vec{c}\times \vec{b}=\vec{0}$.

By the distributive property of the cross product, this becomes:

This is the 'Aha!' moment. The cross product of two vectors is zero if and only if they are collinear, meaning the vector $(\vec{d}-\vec{c})$ must be parallel to $\vec{b}$.

Mathematically, this means $(\vec{d}-\vec{c})=\lambda \vec{b}$ for some scalar $\lambda$. We have just unlocked the general form of $\vec{d}$:

Geometrically, this means $\vec{d}$ lies on a line passing through the tip of $\vec{c}$ and running parallel to $\vec{b}$. We have reduced an infinite set of possibilities down to a single line defined by the parameter $\lambda$.

The Dot Product Anchor

Now, we bring in the second law: $\vec{d}\cdot \vec{a}=24$. This is our anchor, which fixes the position of $\vec{d}$ on that line.

We substitute our parametric form of $\vec{d}$ into this equation:

Expanding this, we get:

Now, we compute the necessary dot products: $\vec{c}\cdot \vec{a}=(2)(1)+(-1)(4)+(4)(2)=2-4+8=6$ $\vec{b}\cdot \vec{a}=(3)(1)+(-2)(4)+(7)(2)=3-8+14=9$

Substituting these back, we get $6+9\lambda =24$. Subtracting 6 from both sides gives $9\lambda =18$, and dividing by 9 yields $\lambda =2$.

The Final Calculation

With $\lambda =2$ in hand, we can find the exact components of $\vec{d}$:

Distributing the 2, we get:

Finally, we calculate the square of the magnitude:

And there it is—413. A beautiful, clean result born from the interplay of geometric constraints and algebraic precision.