To verify if the vectors $\vec{a} = \hat{i} + 2\hat{j}$, $\vec{b} = 2\hat{i} + 4\hat{j}$, and $\vec{c} = -\hat{i} - 2\hat{j}$ are linearly dependent, we need to check if there exists a non-trivial linear combination of these vectors that results in the zero vector.

Step 1: Express the linear combination

The vectors are linearly dependent if there exist scalars $x_1$, $x_2$, and $x_3$, not all zero, such that:

$x_1 \vec{a} + x_2 \vec{b} + x_3 \vec{c} = \vec{0}$

In component form:

$x_1(\hat{i} + 2\hat{j}) + x_2(2\hat{i} + 4\hat{j}) + x_3(-\hat{i} - 2\hat{j}) = \vec{0}$

Step 2: Set up the system of equations

By breaking this into components for $\hat{i}$ and $\hat{j}$, we get two equations:

For $\hat{i}$: $x_1(1) + x_2(2) + x_3(-1) = 0$ $x_1 + 2x_2 - x_3 = 0 \quad \text{(Equation 1)}$

For $\hat{j}$: $x_1(2) + x_2(4) + x_3(-2) = 0$ $2x_1 + 4x_2 - 2x_3 = 0 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations

We now solve this system:

From Equation 1:

$x_1 + 2x_2 - x_3 = 0 \quad \Rightarrow \quad x_1 = x_3 - 2x_2$

Substitute $x_1 = x_3 - 2x_2$ into Equation 2:

$2(x_3 - 2x_2) + 4x_2 - 2x_3 = 0$ Expand: $2x_3 - 4x_2 + 4x_2 - 2x_3 = 0$ Simplify: $0 = 0$

This equation is always true, so there are infinitely many solutions. This implies that the system is linearly dependent.

Step 4: Conclusion

Since there are infinitely many solutions (not just the trivial solution where all scalars are zero), the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are linearly dependent.

Final Answer:

Yes, the vectors $\vec{a} = \hat{i} + 2\hat{j}$, $\vec{b} = 2\hat{i} + 4\hat{j}$, and $\vec{c} = -\hat{i} - 2\hat{j}$ are linearly dependent.

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Do you want further details or have any questions?

Here are five related questions to deepen your understanding:

1. How can we geometrically interpret linearly dependent vectors in two dimensions?
2. Can three linearly dependent vectors form a basis in $\mathbb{R}^2$?
3. What conditions must hold for three 2D vectors to be linearly independent?
4. How would this process change if we were working in $\mathbb{R}^3$ instead?
5. What is the role of the determinant in identifying linear dependence for vectors?

Tip: Linearly dependent vectors in 2D are collinear, meaning they lie along the same line.