Problem 1.44

a) จาก \(|\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^{2} + |\vec{B}|^{2} + 2|\vec{A}||\vec{B}|\cos \theta}\)

\(|\vec{A} + \vec{B}| =\sqrt{(1)^{2} + (3)^{2} + 2(1)(3)\cos (30º)}=7\;\blacksquare\)

b) จาก \(|\vec{B} + \vec{C}| = \sqrt{|\vec{B}|^{2} + |\vec{C}|^{2} + 2|\vec{B}||\vec{C}|\cos \theta}\)

\(|\vec{B} + \vec{C}| =\sqrt{(3)^{2} + (3)^{2} + 2(3)(3)\cos (60º)}= 3 \;\blacksquare\)

c) จาก \(|\vec{C} + \vec{A}| = \sqrt{|\vec{C}|^{2} + |\vec{A}|^{2} + 2|\vec{C}||\vec{A}|\cos \theta}\)

\(|\vec{C} + \vec{A}| =\sqrt{(3)^{2} + (1)^{2} + 2(3)(1)\cos (90º)}= 2 \;\blacksquare\)

d) จาก \(|\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^{2} + |\vec{B}|^{2} - 2|\vec{A}||\vec{B}|\cos \theta}\)

\(|\vec{A} - \vec{B}| =\sqrt{(1)^{2} + (3)^{2} - 2(1)(3)\cos (30º)}= 1 \;\blacksquare\)

e) จาก \(|\vec{B} - \vec{C}| = \sqrt{|\vec{B}|^{2} + |\vec{C}|^{2} - 2|\vec{B}||\vec{C}|\cos \theta}\)

\(|\vec{B} - \vec{C}| =\sqrt{(3)^{2} + (3)^{2} - 2(3)(3)\cos (60º)}=3\;\blacksquare\)

f) จาก \(|\vec{C} - \vec{A}| = \sqrt{|\vec{C}|^{2} + |\vec{A}|^{2} - 2|\vec{C}||\vec{A}|\cos \theta}\)

\(|\vec{C} - \vec{A}| =\sqrt{(3)^{2} + (1)^{2} - 2(3)(1)\cos (90º)}= 2 \;\blacksquare\)

î

g) จาก \(\vec{A} + \vec{C} = 1 + 3ĵ\)

\(3\)

\(\therefore \theta _{x}= \tan ^{-1}(\frac{3}{1}) = 60° \;\blacksquare\)

\(\frac{\theta}{1}\)

\(|\vec{A}|\)

î

h) จาก \(\vec{A} + \vec{C} = 1 + 3ĵ\theta\)

\(3\)

\(\therefore \theta _{y}= \tan ^{-1}(\frac{1}{3}) = 30° \;\blacksquare\)

\(\frac{\theta}{1}\)

\(|\vec{A}|\)

î

i) จากรูปจะได้ \(\vec{A} = 1\)

îĵîĵ

\(\vec{B} =3\cos (30º) +3\sin (30º) =\frac{3}{2}+\frac{3}{2}\)

\(\vec{C} =3ĵ\)

îîĵ

\(\therefore \vec{A} + \vec{B} + \vec{C} = (1 ) + (\frac{3}{2}+\frac{3}{2}) + (3ĵ)\)

î

\(\vec{A}+\vec{B}+\vec{C}=(1+\frac{3}{2})+(\frac{3}{2}+3)ĵ\)

îĵ

\(\vec{A} + \vec{B} + \vec{C} =\frac{5}{2}+\frac{3 3}{2}\)

\(\therefore |\vec{A} + \vec{B} + \vec{C}| =\sqrt{(\frac{5}{2})^{2}+(332)^{2}}\)

\(|\vec{A} + \vec{B} + \vec{C}| =\sqrt{13}\;\blacksquare\)