To find the decomposition of the vector $\mathbf{a} = \langle -1, 2, 0 \rangle$ with respect to $\mathbf{b} = \langle 2, 0, 1 \rangle$, we need to express $\mathbf{a}$ as the sum of two components: one parallel to $\mathbf{b}$ and one perpendicular to $\mathbf{b}$.

That is, we want to find: $\mathbf{a} = \mathbf{a}_{\parallel \mathbf{b}} + \mathbf{a}_{\perp \mathbf{b}}$ where:

• $\mathbf{a}_{\parallel \mathbf{b}}$ is the projection of $\mathbf{a}$ onto $\mathbf{b}$,
• $\mathbf{a}_{\perp \mathbf{b}}$ is the component of $\mathbf{a}$ perpendicular to $\mathbf{b}$.

Step 1: Find the projection of $\mathbf{a}$ onto $\mathbf{b}$.

The projection of vector $\mathbf{a}$ onto vector $\mathbf{b}$ is given by the formula: $\mathbf{a}_{\parallel \mathbf{b}} = \text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b}$ where $\mathbf{a} \cdot \mathbf{b}$ is the dot product of $\mathbf{a}$ and $\mathbf{b}$, and $\mathbf{b} \cdot \mathbf{b}$ is the dot product of $\mathbf{b}$ with itself.

1.1 Calculate $\mathbf{a} \cdot \mathbf{b}$:

$\mathbf{a} \cdot \mathbf{b} = (-1)(2) + (2)(0) + (0)(1) = -2$

1.2 Calculate $\mathbf{b} \cdot \mathbf{b}$:

$\mathbf{b} \cdot \mathbf{b} = (2)(2) + (0)(0) + (1)(1) = 4 + 1 = 5$

1.3 Compute the projection:

$\mathbf{a}_{\parallel \mathbf{b}} = \frac{-2}{5} \mathbf{b} = \frac{-2}{5} \langle 2, 0, 1 \rangle = \langle \frac{-4}{5}, 0, \frac{-2}{5} \rangle$

Step 2: Find the component of $\mathbf{a}$ perpendicular to $\mathbf{b}$.

The component of $\mathbf{a}$ perpendicular to $\mathbf{b}$ is given by: $\mathbf{a}_{\perp \mathbf{b}} = \mathbf{a} - \mathbf{a}_{\parallel \mathbf{b}}$ Substitute the values of $\mathbf{a}$ and $\mathbf{a}_{\parallel \mathbf{b}}$: $\mathbf{a}_{\perp \mathbf{b}} = \langle -1, 2, 0 \rangle - \langle \frac{-4}{5}, 0, \frac{-2}{5} \rangle = \langle -1 + \frac{4}{5}, 2 - 0, 0 + \frac{2}{5} \rangle$ Simplifying: $\mathbf{a}_{\perp \mathbf{b}} = \langle \frac{-1}{5}, 2, \frac{2}{5} \rangle$

Final Answer:

Thus, the decomposition of $\mathbf{a}$ with respect to $\mathbf{b}$ is: $\mathbf{a} = \mathbf{a}_{\parallel \mathbf{b}} + \mathbf{a}_{\perp \mathbf{b}} = \langle \frac{-4}{5}, 0, \frac{-2}{5} \rangle + \langle \frac{-1}{5}, 2, \frac{2}{5} \rangle$ where: $\mathbf{a}_{\parallel \mathbf{b}} = \langle \frac{-4}{5}, 0, \frac{-2}{5} \rangle \quad \text{and} \quad \mathbf{a}_{\perp \mathbf{b}} = \langle \frac{-1}{5}, 2, \frac{2}{5} \rangle$

Would you like further clarification on any of the steps, or do you have any questions? Here are some related questions for further exploration:

1. How would you calculate the angle between two vectors using their dot product?
2. What is the geometric interpretation of the projection of one vector onto another?
3. How do you find the magnitude of a vector?
4. Can the concept of vector projection be applied to other spaces, such as in physics?
5. What is the relationship between the dot product and orthogonality of vectors?

Tip: The projection of a vector onto another is a key operation in many areas of mathematics and physics, especially in decomposing forces, velocities, and other vector quantities.