https://file.aiquickdraw.com/mathbot/file/17259312295298hy1pia2.jpg

Let's go through the problem step by step to find both the projections requested.

### Given:
\[
\mathbf{u} = \langle 3, -2, -1 \rangle \quad \text{and} \quad \mathbf{v} = \langle 2, -3, -1 \rangle
\]

### 1. **Projection of \(\mathbf{u}\) along \(\mathbf{v}\):**

The formula for the projection of vector \(\mathbf{u}\) onto vector \(\mathbf{v}\) is:
\[
\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}
\]

#### Step 1: Dot product \(\mathbf{u} \cdot \mathbf{v}\)
\[
\mathbf{u} \cdot \mathbf{v} = (3)(2) + (-2)(-3) + (-1)(-1) = 6 + 6 + 1 = 13
\]

#### Step 2: Dot product \(\mathbf{v} \cdot \mathbf{v}\)
\[
\mathbf{v} \cdot \mathbf{v} = (2)^2 + (-3)^2 + (-1)^2 = 4 + 9 + 1 = 14
\]

#### Step 3: Projection of \(\mathbf{u}\) onto \(\mathbf{v}\)
\[
\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{13}{14} \mathbf{v} = \frac{13}{14} \langle 2, -3, -1 \rangle
\]
Multiplying each component by \(\frac{13}{14}\):
\[
\text{proj}_{\mathbf{v}} \mathbf{u} = \langle \frac{26}{14}, -\frac{39}{14}, -\frac{13}{14} \rangle = \langle 1.857, -2.786, -0.929 \rangle
\]

### 2. **Projection of \(\mathbf{u}\) orthogonal to \(\mathbf{v}\):**

This projection is found using:
\[
\mathbf{u}_{\perp \mathbf{v}} = \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u}
\]

Subtracting the projection from \(\mathbf{u}\):
\[
\mathbf{u}_{\perp \mathbf{v}} = \langle 3, -2, -1 \rangle - \langle 1.857, -2.786, -0.929 \rangle
\]
\[
\mathbf{u}_{\perp \mathbf{v}} = \langle 3 - 1.857, -2 + 2.786, -1 + 0.929 \rangle = \langle 1.143, 0.786, -0.071 \rangle
\]

---

### Final Answers:
1. The projection of \(\mathbf{u}\) along \(\mathbf{v}\) is approximately:
\[
\langle 1.857, -2.786, -0.929 \rangle
\]
2. The projection of \(\mathbf{u}\) orthogonal to \(\mathbf{v}\) is approximately:
\[
\langle 1.143, 0.786, -0.071 \rangle
\]

Would you like further details or clarification on any steps?

---

Here are 5 related questions to explore further:
1. How can the projection formula be derived geometrically?
2. What is the significance of the dot product in vector projections?
3. How would the projections change if the vectors \(\mathbf{u}\) and \(\mathbf{v}\) were in 2D instead of 3D?
4. How do vector projections relate to the concept of orthogonality?
5. Can projections be used to solve systems of linear equations?

**Tip**: Vector projections are useful in breaking down vectors into components in the direction of another vector and perpendicular to it, often applied in physics and engineering problems.

Suppose u = <3, -2, -1> and v = <2, -3, -1>. Then: 1. The projection of u along v is _____. 2. The projection of u orthogonal to v is _____.

Learn how to compute the projection of vector u along vector v and find the orthogonal component. Given u = <3, -2, -1> and v = <2, -3, -1>, the step-by-step solution includes the dot product calculations and vector subtraction.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation