Diketahui segitiga PQR dengan P (1,5,1) Q (3,4,1) dan R(2,2,1) tentukan besar sudut PQR Vektor QR = Vektor QP = |QR| = |QP| = COS sudut Q =
Matematika
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Pertanyaan
Diketahui segitiga PQR dengan P (1,5,1) Q (3,4,1) dan R(2,2,1) tentukan besar sudut PQR
Vektor QR =
Vektor QP =
|QR| =
|QP| =
COS sudut Q =
Vektor QR =
Vektor QP =
|QR| =
|QP| =
COS sudut Q =
1 Jawaban
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1. Jawaban nallsst16
QR = [tex] \left[\begin{array}{ccc}2-4\\2-4\\1-1\end{array}\right]= \left[\begin{array}{ccc}-1\\-2\\0\end{array}\right] [/tex]
QP = [tex]\left[\begin{array}{ccc}2-1\\2-5\\1-1\end{array}\right]= \left[\begin{array}{ccc}1\\-3\\0\end{array}\right][/tex]
|QR| = [tex] \sqrt{(-1)^{2}+(-2)^{2}+0^{2}}= \sqrt{1+4+0}= \sqrt{5} [/tex]
|QP| = [tex]\sqrt{(1)^{2}+(-3)^{2}+0^{2}}= \sqrt{1+9+0}= \sqrt{10} [/tex]
Cos Q = [tex] \frac{QR.QP}{|QR||QP|} [/tex]
= [tex] \frac{(-i-2j)(i-3j)}{(\sqrt{5})( \sqrt{10})} [/tex]
= [tex] \frac{-1+6}{ \sqrt{50}} [/tex]
= [tex] \frac{5}{5\sqrt{2}} [/tex]
= [tex] \frac{1}{ \sqrt{2}}= \frac{1}{2} \sqrt{2} [/tex]