Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=\alpha \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\alpha \hat{k},\alpha >0$. If the projection of $\vec{a}\times \vec{b}$ on the vector $-\hat{i}+2\hat{j}-2\hat{k}$ is 30, then $\alpha$ is equal to

A Journey into 3D Space

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are navigating the elegant landscape of 3D vectors.

Imagine you are standing in a coordinate system where $\vec{a}$ and $\vec{b}$ are two arrows originating from the origin. They define a plane, and our first mission is to find the vector that stands perfectly perpendicular to this plane. This is the essence of the cross product, $\vec{a}\times \vec{b}$.

The Determinant as a Tool of Discovery

To find this perpendicular vector, we employ the determinant method. It is a powerful, systematic way to handle the components. We set up our matrix with the unit vectors $\hat{i},\hat{j},\hat{k}$ in the top row, followed by the components of $\vec{a}$ and $\vec{b}$.

As we expand this, we are essentially calculating the 'area' contributions in each dimension. For the $\hat{i}$ component, we hide the first row and column, giving us $(1)(-\alpha )-(-1)(1)=1-\alpha$.

For the $\hat{j}$ component, we must be careful with the alternating sign, leading us to $-[(\alpha )(-\alpha )-(-1)(2)]={\alpha }^2-2$. Finally, the $\hat{k}$ component gives us $(\alpha )(1)-(1)(2)=\alpha -2$.

We have now constructed our new vector: $\vec{v}=(1-\alpha )\hat{i}+({\alpha }^2-2)\hat{j}+(\alpha -2)\hat{k}$.

The Shadow of the Vector

Now, we introduce a third vector, $\vec{c}=-\hat{i}+2\hat{j}-2\hat{k}$. The problem asks for the 'projection' of our new vector $\vec{v}$ onto $\vec{c}$. Think of projection as the 'shadow' that $\vec{v}$ casts onto the line defined by $\vec{c}$.

The formula is the dot product of the two vectors, normalized by the magnitude of the vector we are projecting onto:

First, let us find the magnitude of $\vec{c}$. It is the square root of the sum of the squares of its components:

This is a beautifully clean number, which is often a sign that we are on the right track.

The Algebraic Climax

We are given that this projection equals $30$. So, we set up our equation:

Multiplying both sides by $3$, we get $90$ on the right. Now, let us expand the numerator with precision.

We have $(\alpha -1)+(2{\alpha }^2-4)+(-2\alpha +4)=90$. Notice how the constants $-4$ and $+4$ cancel out, leaving us with a much simpler expression: $2{\alpha }^2-\alpha -1=90$.

Bringing the $90$ over, we arrive at our quadratic equation: $2{\alpha }^2-\alpha -91=0$.

To solve $2{\alpha }^2-\alpha -91=0$, we look for factors. We need two numbers that multiply to $2\times (-91)=-182$ and add to $-1$. Those numbers are $-14$ and $13$.

Thus, we factor it as $(2\alpha +13)(\alpha -7)=0$. This gives us two potential candidates for $\alpha$: $7$ and $-\frac{13}{2}$.

But remember, the problem gave us the constraint $\alpha >0$. This is the final gatekeeper of our solution.

We must reject the negative value and embrace $\alpha =7$. You have successfully navigated the cross product, the projection, and the quadratic algebra.