Jawab:
enjelasan dengan langkah-langkah:
[tex]2x-y-5z=-6 = 2\cdot 1 - 3 - 5\cdot 1 = 2\cdot 0 - 1-5\cdot 1\\\to \bold{n_1} = \left[\begin{array}{ccc}2\\-1\\-5\end{array}\right] ,A_1=(1,3,1), A_2=(0,1,1)\\4x+5y+4z = 9 \to \bold{n_2} = \left[\begin{array}{ccc}4\\5\\4\end{array}\right], B_1 = A_2 \\[/tex]
Karena B1 = A2, maka titik B1/A2 akan berada di garis potong kedua bidang
Mencari garis potong:
Metode 1 : dengan mencari nullspace (menggunakan eliminasi jordan
[tex]\bold{L} = \bold{n_1\times n_2} \to \bold{N} = \left[\begin{array}{ccc}\bold{n_1}^T\\\bold{n_2}^T\end{array}\right] \\\\ \bold{Nv} = \left[\begin{array}{c}0\\0\end{array}\right]\to \left[\begin{array}{ccc|c}2&-1&-5&0\\4&5&4&0\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc|c}2&-1&-5&0\\4&5&4&0\end{array}\right] \xrightarrow{\textstyle 1) B_2-2B_1 \to B_1, 2)7B_1+B_2\to B_1}\left[\begin{array}{ccc|c}14&0&-21&0\\0&7&14&0\end{array}\right] \\\\[/tex]
[tex]14 t_1 = 21t_3, 7t_2 = -14t_3\\t_1 = \dfrac{3}{2}\;t_3, t_2= -2\;t_3\to t_3 = 2\to t_1 = 3, t_2 = -4\\\\ \bold{L}(t) = \left[\begin{array}{ccc}3t\\-4t\\2t\end{array}\right][/tex]
Garis potong nya :
[tex]\bold{L'}(t) = \bold{A_2}+\bold{L}(t) = \left[\begin{array}{ccc}0\\1\\1\end{array}\right]+\left[\begin{array}{ccc}3t\\-4t\\2t\end{array}\right]\\\\\boxed{\boxed{\bold{L'}(t) = \left[\begin{array}{ccc}3t\\1-4t\\1+2t\end{array}\right] }}[/tex]
Metode 2 : cross product
[tex]\bold{L} = \bold{n_1\times n_2} = \left[\begin{array}{ccc}\ \left|\begin{array}{cc}-1&-5\\5&4\end{array}\right|\\- \left|\begin{array}{cc}2&-5\\4&4\end{array}\right|\\ \left|\begin{array}{cc}2&-1\\4&5\end{array}\right|\end{array}\right]\\\bold{L} = 7\cdot\left[\begin{array}{ccc} 3 \\-4\\ 2\end{array}\right], \bold{L'} = \left[\begin{array}{ccc} 3 \\-4\\ 2\end{array}\right][/tex]
Lalu tinggal jumlahkan L atau L' (terserah mau yang mana) dengan vektor A2 :
[tex]\boxed{\boxed{\bold{L'}(t) = \left[\begin{array}{ccc}21t\\1-28t\\1+14t\end{array}\right] }}[/tex]
atau (sama dengan metode 1) :
[tex]\boxed{\boxed{\bold{L'}(t) = \left[\begin{array}{ccc}3t\\1-4t\\1+2t\end{array}\right] }}[/tex]
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