1. Diketahui Jal. (b) dan la-bl berturut- turut adalah 4, 6 dan 2√19. Nilai |a+b=
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Jawab:
Penjelasan dengan langkah-langkah:
[tex]|a| = 4\\|b| = 6\\|a-b|=2\sqrt{19}\\\\|a-b| = \sqrt{|a|^2+|b|^2-2|a||b| \cos \theta}\\(2\sqrt{19})^2 = 4^2+6^2-2(4)(6) \cos \theta\\76 = 16+36-48 \cos\theta\\24 = -48 \cos \theta\\\cos \theta = -\frac{1}{2}\\\\|a+b| = \sqrt{|a|^2+|b|^2+2|a||b| \cos \theta}\\|a+b| = \sqrt{16+36+2(4)(6)(-\frac{1}{2})}\\|a+b| = \sqrt{52-24}\\|a+b| = \sqrt{28}\\|a+b| = 2\sqrt{7}[/tex]
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