To calculate the projection of a vector $\mathbf{u}$ onto a vector $\mathbf{v}$, you use the formula:

$\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}$

Where:

• $\mathbf{u} \cdot \mathbf{v}$ is the dot product of the vectors $\mathbf{u}$ and $\mathbf{v}$,
• $\mathbf{v} \cdot \mathbf{v}$ is the dot product of $\mathbf{v}$ with itself (which gives the square of the magnitude of $\mathbf{v}$).

Step-by-step calculation:

1. Calculate the dot product $\mathbf{u} \cdot \mathbf{v}$:

[ \mathbf{u} = [2, -3, 1], \quad \mathbf{v} = [1, 3, -2] ]

$\mathbf{u} \cdot \mathbf{v} = (2)(1) + (-3)(3) + (1)(-2) = 2 - 9 - 2 = -9$

2. Calculate the dot product $\mathbf{v} \cdot \mathbf{v}$:

$\mathbf{v} \cdot \mathbf{v} = (1)(1) + (3)(3) + (-2)(-2) = 1 + 9 + 4 = 14$

3. Now, use the formula to compute the projection:

$\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{-9}{14} \mathbf{v}$

4. Multiply the vector $\mathbf{v}$ by the scalar $\frac{-9}{14}$:

[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{-9}{14} [1, 3, -2] = \left[\frac{-9}{14}, \frac{-27}{14}, \frac{18}{14}\right] ]

So, the projection of $\mathbf{u}$ onto $\mathbf{v}$ is:

$\text{proj}_{\mathbf{v}} \mathbf{u} = \left[ \frac{-9}{14}, \frac{-27}{14}, \frac{9}{7} \right]$

Summary:

The projection of $\mathbf{u} = [2, -3, 1]$ onto $\mathbf{v} = [1, 3, -2]$ is $\left[ \frac{-9}{14}, \frac{-27}{14}, \frac{9}{7} \right]$.

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Would you like more details on any part of the process? Or would you like to try a related problem?

Related Questions:

1. How do you calculate the magnitude of a vector?
2. What is the cross product of two vectors, and how does it relate to projections?
3. How would the projection change if $\mathbf{v}$ were a unit vector?
4. Can you apply projections in higher dimensions (e.g., 4D vectors)?
5. How can projections be used in physics or engineering?

Tip: A vector projection always gives you a vector that is parallel to the vector onto which you're projecting.