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SMK Teknologi dan Rekayasa

Solusi Paket 59 UN 2011 Matematika SMK Teknologi (17)


Diketahui\ vektor\ \vec{a}=\left ( \begin{matrix} 1\\ 1\\ 0 \end{matrix} \right )\ dan\ vektor\ \vec{b}=\left ( \begin{matrix} 1\\ 0\\ 1 \end{matrix} \right ).\\Besar\ sudut\ antara\ \vec{a}\ dan\ \vec{b}\ adalah\ ....\\\\A.\ 30^\circ \ \ \ \ \ \ \ \ \ \ B.\ 45^\circ \ \ \ \ \ \ \ \ \ \ C.\ \ 60^\circ \ \ \ \ \ \ \ \ \ \ D.\ \ 90^\circ \ \ \ \ \ \ \ \ \ \ E.\ \ 180^\circ
Jawaban: c
Misalkan\ \alpha \ adalah\ besar\ sudut\ yang\ dibentuk\ vektor\ \vec{a}\ dan\ \vec{b},\ maka:
cos\ \alpha =\frac{\vec{a}.\vec{b}}{\begin{vmatrix} \vec{a} \end{vmatrix}.\begin{vmatrix} \vec{b} \end{vmatrix}}
Diperoleh:
\vec{a}.\vec{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}
sehingga:
\begin{array}{rcl}cos\ \alpha &=&\frac{1}{\sqrt{2}.\sqrt{2}}\\\\cos\ \alpha &=&\frac{1}{2}\\\\\alpha &=&60^{^{\circ}}\end{array}

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