Problem 1.43

î

จากรูปจะได้ \(\vec{A} = |\vec{A}|\)

îĵ

และ \(\vec{B} = |\vec{B}|cosθ + |\vec{B}|sinθ\)

îîĵ

\(\therefore \vec{A} - \vec{B} = |\vec{A}| - (|\vec{B}|\cos \theta + |\vec{B}|\sin \theta )\)

îĵ

\(\vec{A} - \vec{B} = (|\vec{A}| - |\vec{B}|\cos \theta ) - |\vec{B}|\sin \theta\)

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

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

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

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