Question

use the figure to evaluate ( a + b ), ( a - b ), and ( - a ).( a + b = langle square, square
angle )

Explanation:

Step1: Determine vectors \(\mathbf{a}\) and \(\mathbf{b}\)

From the graph, let's find the components of vectors \(\mathbf{a}\) and \(\mathbf{b}\).

• For vector \(\mathbf{a}\): Starting from the origin (or the common vertex), it moves left 2 units (x - component: \(-2\)) and up 3 units (y - component: \(3\)). So \(\mathbf{a}=\langle - 2,3

angle\).

• For vector \(\mathbf{b}\): Starting from the common vertex, it moves right 3 units (x - component: \(3\)) and up 1 unit (y - component: \(1\)). So \(\mathbf{b}=\langle3,1

angle\).

Step2: Calculate \(\mathbf{a}+\mathbf{b}\)

To add two vectors \(\mathbf{a}=\langle a_1,a_2
angle\) and \(\mathbf{b}=\langle b_1,b_2
angle\), we use the formula \(\mathbf{a}+\mathbf{b}=\langle a_1 + b_1,a_2 + b_2
angle\).
Substitute \(a_1=-2\), \(a_2 = 3\), \(b_1=3\), \(b_2 = 1\) into the formula:
\(\mathbf{a}+\mathbf{b}=\langle-2 + 3,3+1
angle=\langle1,4
angle\)

(Note: We can also calculate \(\mathbf{a}-\mathbf{b}\) and \(-\mathbf{a}\) using similar vector operations. For \(\mathbf{a}-\mathbf{b}=\mathbf{a}+(-\mathbf{b})\), where \(-\mathbf{b}=\langle - 3,-1
angle\), so \(\mathbf{a}-\mathbf{b}=\langle-2-3,3 - 1
angle=\langle-5,2
angle\). For \(-\mathbf{a}\), we multiply each component of \(\mathbf{a}\) by \(- 1\), so \(-\mathbf{a}=\langle2,-3
angle\). But since the question first asks for \(\mathbf{a}+\mathbf{b}\), we focus on that for the first part.)

Answer:

\(\langle1,4
angle\)