Let's walk through each part of the problem in the image:

Part a.

Given the coordinate vectors:
$\vec{a} = \langle 3, 3 \rangle, \quad \vec{b} = \langle -2, 1 \rangle, \quad \vec{c} = \langle 2, -1 \rangle$

These vectors are already identified with their coordinate values.

Part b.

You are asked to draw the vector sum $\vec{a} + \vec{b} + \vec{c}$ and verify your answer through vector addition.

Adding the vectors: $\vec{a} + \vec{b} + \vec{c} = \langle 3, 3 \rangle + \langle -2, 1 \rangle + \langle 2, -1 \rangle$

Perform the component-wise addition: $(3 - 2 + 2, 3 + 1 - 1) = \langle 3, 3 \rangle$

Thus, the result is: $\boxed{\langle 3, 3 \rangle}$

Part c.

You are asked to find the sum $2\vec{a} + 3\vec{b} - \vec{c}$ and verify your answer.

Calculating the scalar products: $2\vec{a} = 2 \cdot \langle 3, 3 \rangle = \langle 6, 6 \rangle$ $3\vec{b} = 3 \cdot \langle -2, 1 \rangle = \langle -6, 3 \rangle$ $-\vec{c} = -\langle 2, -1 \rangle = \langle -2, 1 \rangle$

Adding these vectors: $2\vec{a} + 3\vec{b} - \vec{c} = \langle 6, 6 \rangle + \langle -6, 3 \rangle + \langle -2, 1 \rangle$

Component-wise: $(6 - 6 - 2, 6 + 3 + 1) = \langle -2, 10 \rangle$

Thus, the result is: $\boxed{\langle -2, 10 \rangle}$

Part d.

Find the sum $x_1 \vec{a} + x_2 \vec{b} + x_3 \vec{c}$ using the given scalars: $x_1 = -2, \quad x_2 = 4, \quad x_3 = -1$

Calculating the scalar products: $-2\vec{a} = -2 \cdot \langle 3, 3 \rangle = \langle -6, -6 \rangle$ $4\vec{b} = 4 \cdot \langle -2, 1 \rangle = \langle -8, 4 \rangle$ $-1\vec{c} = -1 \cdot \langle 2, -1 \rangle = \langle -2, 1 \rangle$

Adding these vectors: $x_1 \vec{a} + x_2 \vec{b} + x_3 \vec{c} = \langle -6, -6 \rangle + \langle -8, 4 \rangle + \langle -2, 1 \rangle$

Component-wise: $(-6 - 8 - 2, -6 + 4 + 1) = \langle -16, -1 \rangle$

Thus, the result is: $\boxed{\langle -16, -1 \rangle}$

Part e.

Find a vector $\vec{d}$ such that: $\vec{a} + \vec{c} + \vec{d} = \vec{0}$

Rearrange the equation: $\vec{d} = -(\vec{a} + \vec{c})$

First, add $\vec{a} + \vec{c}$: $\vec{a} + \vec{c} = \langle 3, 3 \rangle + \langle 2, -1 \rangle = \langle 5, 2 \rangle$

Now negate the result: $\vec{d} = -\langle 5, 2 \rangle = \langle -5, -2 \rangle$

Thus, the vector $\vec{d}$ is: $\boxed{\langle -5, -2 \rangle}$

Summary of Answers

• b. $\langle 3, 3 \rangle$
• c. $\langle -2, 10 \rangle$
• d. $\langle -16, -1 \rangle$
• e. $\langle -5, -2 \rangle$

Would you like further explanations on any part of these calculations?

Related Questions:

1. How do you find the magnitude of a vector?
2. What is the geometric interpretation of adding vectors?
3. How do scalar multiples affect a vector’s direction?
4. What are some applications of vector addition in physics?
5. How does the dot product differ from vector addition?

Tip: When adding vectors graphically, make sure to align them tip-to-tail to visualize the resulting vector.