mohon bantuannya kak no 4,5,6

Berikut ini adalah pertanyaan dari daaaaa19 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Mohon bantuannya kak

no 4,5,6
mohon bantuannya kak no 4,5,6

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\purple{\huge{4.}}$

$\sf \overrightarrow{AB}=((-4-3)~,~(-2-(-1))~,~(0-4))$

$\sf \overrightarrow{AB}=(-7~,~-1~,~-4)$

$\sf \left|\overrightarrow{AB}\right|=\sqrt{(-7)^2+(-1)^2+(-4)^2}$ $\sf =\sqrt{49+1+16}$ $\sf =\sqrt{66}$

$\sf \overrightarrow{AC}=((3-3)~,~(-2-(-1))~,~(1-4))$

$\sf \overrightarrow{AC}=(0~,~-1~,~-3)$

$\sf \left|\overrightarrow{AC}\right|=\sqrt{0^2+(-1)^2+(-3)^2}$ $\sf =\sqrt{0+1+9}$ $\sf =\sqrt{10}$

$[tex]\sf \overrightarrow{AB}.\overrightarrow{AC}=(-7\times 0)+(-1\times -1)+(-4\times -3)$ $\sf =0+1+12$ $\sf =13$

$\sf cos~\angle~BAC=\frac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|\times \left|\overrightarrow{AC}\right|}$

$\sf cos~\angle~BAC=\frac{13}{\sqrt{66}\times \sqrt{10}}=\frac{13}{\sqrt{660}}$ $\sf \approx 0,506$

$\sf \angle~BAC\approx$ $arc.\sf cos~0,506$

$\red{\huge{\sf \angle~BAC\approx 60\degree}}$

$\\$

$\purple{\huge{5.}}$

$\sf cos~\frac{\pi}{6}=$ $\frac{\vec{a}.\vec{b}}{\sqrt{3}\times 1}$

$\vec{a}.\vec{b}\sf ~=\sqrt{3}\times \frac{1}{2}\sqrt{3}$

$\vec{a}.\vec{b}\sf ~=\frac{3}{2}$

$\left|\vec{a}+\vec{b}\right|=\sqrt{\left|\vec{a}\right|^2+\left|\vec{b}\right|^2+2\times \left|\vec{a}\right|\times \left|\vec{b}\right|\times \sf cos~\frac{\pi}{6}}$

$\left|\vec{a}+\vec{b}\right|=\sqrt{\left(\sqrt{3}\right)^2+1^2+2\times \sqrt{3}\times 1\times \frac{1}{2}\sqrt{3}}$ $=\sqrt{3+1+2\times \frac{3}{2}}$ $=\sqrt{3+1+3}$

$\left|\vec{a}+\vec{b}\right|=\sqrt{7}$

$\left|\vec{a}-\vec{b}\right|=\sqrt{\left|\vec{a}\right|^2+\left|\vec{b}\right|^2-2\times \left|\vec{a}\right|\times \left|\vec{b}\right|\times \sf cos~\frac{\pi}{6}}$ $=\sqrt{\left(\sqrt{3}\right)^2+1^2-2\times \sqrt{3}\times 1\times \frac{1}{2}\sqrt{3}}$ $=\sqrt{3+1-2\times \frac{3}{2}}$ $=\sqrt{3+1-3}$