2. Tiga Dimensi (R3) Ø
Persamaan Garis Titik A (xA,yA,zA) dan titik B (xB,yB,zB) terletak pada satu garis. Jika titik P (xP,yP,zP) terletak di tengah titik A dan B, secara vektor dituliskan :
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Jadi persamaan garis yang melalui titik A dan titik B dituliskan dituliskan dalam bentuk persamaan parametrik :
x P = k( x B– B– A) + x A x A) B– y P = k( y B– y A) A) + y A B– z P = k( z B– z A) A) + z A Contoh : Tentukan persamaan garis yang melalui titik (2,-1,0) dan
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Jawab : Gunakan persamaan garis melalui kedua titik tersebut . x = k( x B– B– x A) A) + x A = k(2 – 1)+2= k + 2 y P = k( y B– B– y A) A) + y A = k(– 1–(–1)+(–1) = k – 1 z P = k( z B– B– z A) A) + z A = k(0 – 1)+ 0= – k
Persamaan garis di ruang 3 dimensi adalah persamaan parametrik. Variabel A dan B dapat ditukar, yang mem bedakan adalah arah garisnya
Perhatika n:
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Ø
Persamaan bidang
Bidang merupakan suatu permukaan datar. datar. Untuk membentuk suatu persamaan garis dibutuhkan 2 titik, sedangkan untuk membentuk persamaan bidang dibutuhkan 3 titik atau satu titik dan vektor normal dari bidang tersebut. Jika terdapat satu bidang yang melalui titik P (xP,yP,zP) dan memiliki vektor normal n = (a,b,c), maka bila ingin mencari persamaan dari bidang tersebut diperlukan suatu titik sembarang Q(x,y,z) yang terletak pada bidang tersebut.
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Dari definisi bahwa vektor normal tegak lurus terhadap bidang, maka Q P. n 0
=
x P − x y − y P z P − z
.
a x + by by
cz cz ( − ax
+
a b = 0 c b+y
c z+)
0
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Contoh : Tentukan persamaan bidang yang melalui titik (1,2,1) dan memiliki vektor normal (-1,2,3). Jawab : Langsung digunakan persamaan umum dengan mensubstitusi vektor normal :
Untuk mencari nilai d, dilakukan substitusi titik (1,2,1) ke persamaan, karena titik tersebut terletak di bidang. Maka :
− x+ 2 y + 3 z + d = 0
Jadi persamaan bidang yang dicari adalah :
ax + by+cz+ = d
0
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Bagaimana mencari persamaan bidang jika yang diketahui adalah 3 buah titik? Contoh : Tentukan persamaan bidang yang melalui titik A(-1,2,1), B(2,1,1) dan C(-2,-1,3). Jawab :
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Dengan menggunakan eliminasi Gauss-Jordan diperoleh :
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Mencari vektor normal n dengan Cara menggunakan perkalian silang vektor AB 2. dan vektor BC.
3 AB = -1 ;
-1 AC = -3
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Ø
Jarak titik terhadap bidang
Vektor normal n pada bidang ax + by + cz+ d = 0 dapat ditulis sebagai ( a,b,c). Titik A(xA, yA) berada di luar bidang, sedangkan sembarang titik P(x,y,z) pada
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Contoh :
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Ø
Sudut antara dua bidang
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Jika koefisien koefisien tidak mempunyai nilai n ilai yang sama, s ama, maka
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Contoh :
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