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SMK Teknologi dan Rekayasa

Soal 49 Besar Sudut antara Dua Vektor


Vektor\ \bar{a}=\bar{i}+\bar{j}\ dan\ \bar{b}=\bar{i}+\bar{k}.\\Besar\ sudut\ yang\ dibentuk\ oleh\ kedua\ vektor\ tersebut\ adalah\ ....\\a.\ \ 0^{^{\circ}}\ \ \ \ \ \ \ \ \ \ b.\ \ 30^{^{\circ}}\ \ \ \ \ \ \ \ \ \ c.\ \ 45^{^{\circ}}\ \ \ \ \ \ \ \ \ \ d.\ \ 60^{^{\circ}}\ \ \ \ \ \ \ \ \ \ e.\ \ 90^{^{\circ}}
Jawaban: d

\bar{a}=\bar{i}+\tilde{j}=1\bar{i}+1\tilde{j}+0\tilde{k}
\bar{b}=\bar{i}+\tilde{k}=1\bar{i}+0\tilde{j}+1\tilde{k}
\bar{a}.\bar{b}=1.1+1.0+0.1=1+0+0=1
\begin{vmatrix} \bar{a} \end{vmatrix}=\sqrt{1^{2}+1^{2}+0^{2}}=\sqrt{1+1+0}=\sqrt{2}
\begin{vmatrix} \bar{b} \end{vmatrix}=\sqrt{1^{2}+0^{2}+1^{2}}=\sqrt{1+0+1}=\sqrt{2}
Misalkan\ \alpha \ adalah\ besar\ sudut\ yang\ dibentuk\ vektor\ \bar{a}\ dan\ \bar{b},\ maka:
\begin{array}{rcl}cos\ \alpha &=&\frac{\bar{a}.\bar{b}}{\begin{vmatrix} \bar{a} \end{vmatrix}.\begin{vmatrix} \bar{b} \end{vmatrix}}\\\\&=&\frac{1}{\sqrt{2}.\sqrt{2}}\\\\&=&\frac{1}{2}\\\\\alpha &=&60^{^{\circ}}\end{array}

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