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SMK Teknologi dan Rekayasa

Indikator UN 2011 Matematika SMK Teknologi (15)


Indikator 15
Menentukan besar sudut antara dua vektor

Besar sudut antara vektor \bar{a}=2\bar{i}-\bar{j}+3\bar{k} dan \bar{b}=\bar{i}+3\bar{j}-2\bar{k} adalah ….
a. \frac{1}{8}\pi
b. \frac{1}{4}\pi
c. \frac{1}{3}\pi
d. \frac{1}{2}\pi
e. \frac{2}{3}\pi
Jawaban: e
cos\ \alpha =\frac{\bar{a}.\bar{b}}{|\bar{a}|.|\bar{b}|}
\begin{array}{rcl}\bar{a}.\bar{b}&=&2.1+(-1).3+3.(-2)\\&=&2-3-6\\&=&-7\end{array}
\begin{array}{rcl}|\bar{a}|&=&\sqrt{2^2+(-1)^2+3^2}\\&=&\sqrt{4+1+9}\\&=&\sqrt{14}\end{array}
\begin{array}{rcl}|\bar{b}|&=&\sqrt{1^2+3^2+(-2)^2}\\&=&\sqrt{1+9+4}\\&=&\sqrt{14}\end{array}
\begin{array}{rcl} cos\ \alpha &=&\frac{-7}{\sqrt{14}.\sqrt{14}}\\\\&=&\frac{-7}{14}\\\\&=&-\frac{1}{2}\end{array}
Karena nilai cos\ \alpha  bertanda negatif maka \alpha  adalah sudut tumpul (sudut di kuadran II).
Ingat: cos\ \frac{\pi}{3}=\frac{1}{2}
Diperoleh:
\begin{array}{rcl} \alpha &=&\pi-\frac{\pi}{3}\\&=&\frac{2}{3}\pi\end{array}

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