Diketahui titik vektor a(1,0, -2), b(2,1,-1) dan c(2,0,-3) sudut antara vektor AB dengan AC adalah

T h e v er t e x f or m : y = a ( x − k ) 2 + h w h ere ( k ; h ) a re coor d ina t es o f v er t e x − − − − − − − − − − − − − − − − − − − − − − − − y = a ( x − 4 ) 2 − 3 t h e p o in t o n p a r ab o l a ( 5 ; − 6 ) P u t x = 5 an d y = − 6 t o e q u a t i o n : a ( 5 − 4 ) 2 − 3 = − 6 a ( 1 ) 2 = − 6 + 3 a = − 3 A n s w er : y = − 3 ( x − 4 ) 2 − 3 = − 3 ( x 2 − 8 x + 16 ) − 3 = − 3 x 2 + 24 x − 48 − 3 = − 3 x 2 + 24 x − 51

Answered by Anonymous | 2024-06-10

The coefficient of the squared expression in the parabola's equation, given the vertex at (4, -3) and a point (5, -6), is -3. This indicates that the parabola opens downwards. The equation of the parabola is of the form y = − 3 ( x − 4 ) 2 − 3 .
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Answered by Anonymous | 2024-12-24

Jawaban:[tex]{90^\circ}[/tex]Vektor AB dan AC saling tegak lurus.Penjelasan dengan langkah-langkah:Diketahui tiga titik vektor:Titik A: (1,0,-2) Titik B: (2, 1,-1) Titik C: (2, 0,-3) Langkah-langkah:1. Hitung vektor AB dan ACAB = B-A = (2 - 1, 1 - 0, -1 - (-2)) = (1, 1, 1)AC = C-A= (2 - 1, 0 - 0, -3 - (-2)) = (1, 0, -1) 2. Gunakan rumus sudut antara dua vektor:[tex]\cos \theta = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| \cdot |\vec{AC}|}[/tex]Dot product:[tex]\vec{AB} \cdot \vec{AC} = (1)(1) + (1)(0) + (1)(-1) = 1 + 0 - 1 = 0[/tex]Panjang vektor:[tex]|\vec{AB}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}[/tex][tex]|\vec{AC}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2} [/tex]3. Masukkan ke rumus:[tex]\cos \theta = \frac{0}{\sqrt{3} \cdot \sqrt{2}} = 0\Rightarrow \theta = \cos^{-1}(0) = 90^\circ[/tex]

Answered by Sunariyadi | 2025-07-07