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SMK Teknologi dan Rekayasa

Soal 50 Besar Sudut antara Dua Vektor


Vektor\ \bar{a}=3\bar{i}+\bar{j}\ dan\ \bar{b}=2\bar{j}-2\bar{k}.\\Jika\ sudut\ antara\ kedua\ vektor\ adalah\ \alpha,\ maka\ cos\ \alpha\ adalah\ ....\\a.\ \ \sqrt{5}\ \ \ \ \ \ \ \ \ \ b.\ \ \frac{1}{5}\sqrt{5}\ \ \ \ \ \ \ \ \ \ c.\ -\frac{1}{5}\sqrt{5}\ \ \ \ \ \ \ \ \ \ d.\ \ \frac{1}{10}\sqrt{5}\ \ \ \ \ \ \ \ \ \ e.\ -\frac{1}{10}\sqrt{5}
Jawaban: d

\bar{a}=3\bar{i}+\tilde{j}=3\bar{i}+1\tilde{j}+0\tilde{k}
\bar{b}=2\bar{j}-2\tilde{k}=0\bar{i}+2\tilde{j}-2\tilde{k}
\bar{a}.\bar{b}=3.0+1.2+0.(-2)=0+2+0=2
\begin{vmatrix} \bar{a} \end{vmatrix}=\sqrt{3^{2}+1^{2}+0^{2}}=\sqrt{9+1+0}=\sqrt{10}
\begin{vmatrix} \bar{b} \end{vmatrix}=\sqrt{0^{2}+2^{2}+(-2)^{2}}=\sqrt{0+4+4}=\sqrt{8}
\begin{array}{rcl}cos\ \alpha &=&\frac{\bar{a}.\bar{b}}{\begin{vmatrix} \bar{a} \end{vmatrix}.\begin{vmatrix} \bar{b} \end{vmatrix}}\\\\&=&\frac{2}{\sqrt{10}.\sqrt{8}}\\\\&=&\frac{2}{\sqrt{80}}\\\\&=&\frac{2}{\sqrt{16.5}}\\\\&=&\frac{2}{\sqrt{16}.\sqrt{5}}\\\\&=&\frac{2}{4.\sqrt{5}}\\\\&=&\frac{1}{2.\sqrt{5}}\ \ \ \ \ \times \ \ \frac{\sqrt{5}}{\sqrt{5}}\\\\&=&\frac{1}{10}\sqrt{5}\end{array}

About Kalakay

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2 thoughts on “Soal 50 Besar Sudut antara Dua Vektor

  1. Besar sudutnya berapa?

    Posted by Ayik | 14 Juni 2016, 7:18 AM

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