Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $\vec{a}\cdot \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2\sqrt{2}$ and the angle between $(\vec{a}\times \vec{b})$ and $\vec{c}$ is $30^{\circ }$, then $|(\vec{a}\times \vec{b})\times \vec{c}|=$

Welcome, future engineer! Today, we are going to unravel a beautiful vector algebra problem. It is not just about crunching numbers; it is about understanding the spatial dance of vectors in three-dimensional space.

We are given two vectors, $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. We are also introduced to a mysterious third vector, $\vec{c}$, which is bound by specific geometric constraints.

Our goal is to find the magnitude of the double cross product $|(\vec{a}\times \vec{b})\times \vec{c}|$. Let us embark on this journey.

The Magnitude of $\vec{a}$

Before we dive into the complex relationships, let us ground ourselves by calculating the magnitude of $\vec{a}$. The magnitude of a vector is the length of the arrow representing it in space.

For $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$, the magnitude is given by the square root of the sum of the squares of its components:

Calculating this, we get $|\vec{a}|=\sqrt{4+1+4}=\sqrt{9}=3$. Keep this value, $3$, in your toolkit; we will need it soon.

The Algebraic Trap

Now, let us tackle the condition $|\vec{c}-\vec{a}|=2\sqrt{2}$. This represents the distance between the tips of vector $\vec{a}$ and vector $\vec{c}$.

To work with this, we square both sides: $|\vec{c}-\vec{a}{|}^2=(2\sqrt{2}{)}^2=8$. Expanding this using the vector identity, we get:

We know $|\vec{a}{|}^2=3^2=9$. The problem also tells us that $\vec{a}\cdot \vec{c}=|\vec{c}|$. Substituting these into our equation, we get $|\vec{c}{|}^2+9-2|\vec{c}|=8$.

Rearranging this, we arrive at $|\vec{c}{|}^2-2|\vec{c}|+1=0$. This is a perfect square: $(|\vec{c}|-1{)}^2=0$. Thus, the magnitude of $\vec{c}$ is exactly $1$.

The Cross Product

Next, we need the vector $\vec{a}\times \vec{b}$. We set up a determinant with $\hat{i},\hat{j},\hat{k}$ in the first row, the components of $\vec{a}$ in the second, and the components of $\vec{b}$ in the third:

Expanding this, we get $\hat{i}(0-(-2))-\hat{j}(0-(-2))+\hat{k}(2-1)=2\hat{i}-2\hat{j}+\hat{k}$. The magnitude of this vector is:

The Final Act

We are finally ready to find $|(\vec{a}\times \vec{b})\times \vec{c}|$. Using the definition of the cross product magnitude, $|\vec{u}\times \vec{v}|=|\vec{u}||\vec{v}|\sin \theta$, where $\vec{u}=\vec{a}\times \vec{b}$, $\vec{v}=\vec{c}$, and $\theta =30^{\circ }$.

We have $|\vec{a}\times \vec{b}|=3$, $|\vec{c}|=1$, and $\sin 30^{\circ }=\frac{1}{2}$. Thus, the magnitude is:

We have successfully navigated the complexities of this problem to find the answer: $1.5$, or $\frac{3}{2}$. Keep practicing, and remember that every vector problem is just a story waiting to be told!