To determine the angle between $\mathbf{w}$ and $\mathbf{v}$, we use the fact that the projection of one vector onto another is proportional to the cosine of the angle between them. Here's the step-by-step solution:

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Step 1: Projections

1. The projection of $\mathbf{w}$ onto $\mathbf{v}$ is given as: $\text{Proj}_{\mathbf{v}} \mathbf{w} = \frac{\mathbf{w} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v}$ From the problem, this is equal to $-3 \mathbf{v}$. Hence: $\frac{\mathbf{w} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} = -3 \mathbf{v}.$ Cancel $\mathbf{v}$ (assuming $\mathbf{v} \neq 0$): $\frac{\mathbf{w} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} = -3.$ Therefore: $\mathbf{w} \cdot \mathbf{v} = -3 \|\mathbf{v}\|^2.$

2. Similarly, the projection of $\mathbf{v}$ onto $\mathbf{w}$ is given as: $\text{Proj}_{\mathbf{w}} \mathbf{v} = \frac{\mathbf{w} \cdot \mathbf{v}}{\|\mathbf{w}\|^2} \mathbf{w}.$ From the problem, this is equal to $-\frac{1}{4} \mathbf{w}$. Hence: $\frac{\mathbf{w} \cdot \mathbf{v}}{\|\mathbf{w}\|^2} \mathbf{w} = -\frac{1}{4} \mathbf{w}.$ Cancel $\mathbf{w}$ (assuming $\mathbf{w} \neq 0$): $\frac{\mathbf{w} \cdot \mathbf{v}}{\|\mathbf{w}\|^2} = -\frac{1}{4}.$ Therefore: $\mathbf{w} \cdot \mathbf{v} = -\frac{1}{4} \|\mathbf{w}\|^2.$

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Step 2: Equating the Dot Product Expressions

From the two equations for $\mathbf{w} \cdot \mathbf{v}$, we have: $-3 \|\mathbf{v}\|^2 = -\frac{1}{4} \|\mathbf{w}\|^2.$ Eliminate the negative signs: $3 \|\mathbf{v}\|^2 = \frac{1}{4} \|\mathbf{w}\|^2.$ Multiply through by 4: $12 \|\mathbf{v}\|^2 = \|\mathbf{w}\|^2.$

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Step 3: Using the Dot Product Formula

The dot product is related to the cosine of the angle $\theta$ between the vectors: $\mathbf{w} \cdot \mathbf{v} = \|\mathbf{w}\| \|\mathbf{v}\| \cos \theta.$ From either projection equation, we know: $\mathbf{w} \cdot \mathbf{v} = -3 \|\mathbf{v}\|^2.$ Substitute $\|\mathbf{w}\| = \sqrt{12} \|\mathbf{v}\|$ (from $\|\mathbf{w}\|^2 = 12 \|\mathbf{v}\|^2$): $-3 \|\mathbf{v}\|^2 = (\sqrt{12} \|\mathbf{v}\|)(\|\mathbf{v}\|) \cos \theta.$ Simplify: $-3 \|\mathbf{v}\|^2 = \sqrt{12} \|\mathbf{v}\|^2 \cos \theta.$ Cancel $\|\mathbf{v}\|^2$ (nonzero): $-3 = \sqrt{12} \cos \theta.$ Solve for $\cos \theta$: $\cos \theta = \frac{-3}{\sqrt{12}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}.$

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Step 4: Determine the Angle

The cosine value $\cos \theta = -\frac{\sqrt{3}}{2}$ corresponds to an angle: $\theta = 150^\circ.$

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Final Answer:

The angle between $\mathbf{w}$ and $\mathbf{v}$ is: $\boxed{150^\circ}.$

Let me know if you would like further details or clarification!

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Related Questions:

1. What is the significance of the projection scalar coefficients in this problem?
2. How does the dot product relate to projections geometrically?
3. Can the angle between two vectors ever be greater than 180°?
4. How would the solution change if the projections were given as positive scalars?
5. How can this method be applied in three-dimensional vector spaces?

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Tip:

Always check the consistency of your derived relationships when working with projections and dot products!