Question

tiles
a + {b}, where a = <-3, 5> and b = <2, 1>
c + {d}, where c = <2, 1> and d = <3, 5>
e + {f}, where e = <6, 1> and f = <7, 4>
g + {h}, where g = <3, 3> and h = <5, -1>
m + {n}, where m = <2, -5> and n = <-2, 4>
o + {p}, where o = <-1, -5> and p = <3, 4>
sequence
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Explanation:

Step1: Recall Vector Addition Rule

To add two vectors \(\mathbf{u}=\langle u_1, u_2
angle\) and \(\mathbf{v}=\langle v_1, v_2
angle\), we use the rule \(\mathbf{u}+\mathbf{v}=\langle u_1 + v_1, u_2 + v_2
angle\).

Step2: Solve \( \mathbf{a}+\mathbf{b} \) (First Problem)

Given \(\mathbf{a}=\langle - 3,5
angle\) and \(\mathbf{b}=\langle2,1
angle\).
Using the vector addition rule:
\(u_1=-3, v_1 = 2\), so \(u_1 + v_1=-3 + 2=-1\);
\(u_2 = 5, v_2=1\), so \(u_2 + v_2=5 + 1 = 6\).
Thus, \(\mathbf{a}+\mathbf{b}=\langle-1,6
angle\).

Step3: Solve \( \mathbf{c}+\mathbf{d} \) (Second Problem)

Given \(\mathbf{c}=\langle2,7
angle\) and \(\mathbf{d}=\langle3,5
angle\).
\(u_1 = 2, v_1=3\), so \(u_1 + v_1=2 + 3=5\);
\(u_2=7, v_2 = 5\), so \(u_2 + v_2=7+5 = 12\).
Thus, \(\mathbf{c}+\mathbf{d}=\langle5,12
angle\).

Step4: Solve \( \mathbf{e}+\mathbf{f} \) (Third Problem)

Given \(\mathbf{e}=\langle6,1
angle\) and \(\mathbf{f}=\langle7,4
angle\).
\(u_1=6, v_1 = 7\), so \(u_1 + v_1=6 + 7=13\);
\(u_2=1, v_2 = 4\), so \(u_2 + v_2=1+4 = 5\).
Thus, \(\mathbf{e}+\mathbf{f}=\langle13,5
angle\).

Step5: Solve \( \mathbf{g}+\mathbf{h} \) (Fourth Problem)

Given \(\mathbf{g}=\langle3,3
angle\) and \(\mathbf{h}=\langle5,-1
angle\).
\(u_1=3, v_1 = 5\), so \(u_1 + v_1=3 + 5=8\);
\(u_2=3, v_2=-1\), so \(u_2 + v_2=3+( - 1)=2\).
Thus, \(\mathbf{g}+\mathbf{h}=\langle8,2
angle\).

Step6: Solve \( \mathbf{m}+\mathbf{n} \) (Fifth Problem)

Given \(\mathbf{m}=\langle2,-5
angle\) and \(\mathbf{n}=\langle-2,4
angle\).
\(u_1=2, v_1=-2\), so \(u_1 + v_1=2+( - 2)=0\);
\(u_2=-5, v_2 = 4\), so \(u_2 + v_2=-5 + 4=-1\).
Thus, \(\mathbf{m}+\mathbf{n}=\langle0,-1
angle\).

Step7: Solve \( \mathbf{q}+\mathbf{p} \) (Sixth Problem)

Given \(\mathbf{q}=\langle-1,-5
angle\) and \(\mathbf{p}=\langle3,4
angle\).
\(u_1=-1, v_1 = 3\), so \(u_1 + v_1=-1 + 3=2\);
\(u_2=-5, v_2 = 4\), so \(u_2 + v_2=-5 + 4=-1\).
Thus, \(\mathbf{q}+\mathbf{p}=\langle2,-1
angle\).

Answer:

• \(\mathbf{a}+\mathbf{b}=\boldsymbol{\langle - 1,6

angle}\)

• \(\mathbf{c}+\mathbf{d}=\boldsymbol{\langle5,12

angle}\)

• \(\mathbf{e}+\mathbf{f}=\boldsymbol{\langle13,5

angle}\)

• \(\mathbf{g}+\mathbf{h}=\boldsymbol{\langle8,2

angle}\)

• \(\mathbf{m}+\mathbf{n}=\boldsymbol{\langle0,-1

angle}\)

• \(\mathbf{q}+\mathbf{p}=\boldsymbol{\langle2,-1

angle}\)