SBMPTN Zone : Vektor

Berikut ini adalah kumpulan soal mengenai vektor tipe standar. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

No. 1

Diketahui vektor \vec{a} dan \vec{b} dengan \left|\vec{a}\right|=\sqrt5 dan \left(\vec{a}+\vec{b}\right)\left(\vec{a}-\vec{b}\right)=0. Jika \vec{a}\left(\vec{a}-\vec{b}\right)=4 dan sudut antara \vec{a} dan \vec{b} adalah \alpha, maka \tan\alpha=

1. \sqrt6
2. 2\sqrt6
   
3. 2
   

4. 2\sqrt3
5. \sqrt3
   

$\begin{array}{rl}(\overrightarrow{a}+\overrightarrow{b})(\overrightarrow{a}-\overrightarrow{b})& =0\\ {\overrightarrow{a}}^2-{\overrightarrow{b}}^2& =0\\ {\overrightarrow{a}}^2& ={\overrightarrow{b}}^2\\ {\left|\overrightarrow{a}\right|}^2& ={\left|\overrightarrow{b}\right|}^2\\ \left|\overrightarrow{a}\right|& =\left|\overrightarrow{b}\right|\\ \left|\overrightarrow{b}\right|& =\sqrt{5}\end{array}$

$\begin{array}{rl}\overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})& =4\\ {\overrightarrow{a}}^2-\overrightarrow{a}\cdot \overrightarrow{b}& =4\\ {\left|\overrightarrow{a}\right|}^2-\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \alpha & =4\\ {\left(\sqrt{5}\right)}^2-\left(\sqrt{5}\right)\left(\sqrt{5}\right)\cos \alpha & =4\\ 5-5\cos \alpha & =4\\ \cos \alpha & ={\displaystyle \frac{1}{5}}\end{array}$

$\begin{array}{rl}de& =\sqrt{5^2-1^2}\\ & =\sqrt{24}\\ & =2\sqrt{6}\end{array}$

$\begin{array}{rl}\tan \alpha & ={\displaystyle \frac{2\sqrt{6}}{1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2\sqrt{6}}}}}\end{array}$

Jadi, \tan\alpha=2\sqrt6.
JAWAB: B