第13章 13.16

(1)

(2,3,4)(2,3,4) (1,−2,−3)(1,-2,-3) (4,3,−2)(4,3,-2)

法線ベクトル n=(a,b,c)n=(a,b,c) とおき、平面の方程式を ax+by+cz+d=0ax+by+cz+d=0 とおくと

(2341−2−343−2)(abc)=(−d−d−d)\begin{pmatrix}2&3&4\\1&-2&-3\\4&3&-2\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}-d\\-d\\-d\end{pmatrix}

(234−d1−2−3−d43−2−d)→(0710d1−2−3−d011103d)→(0710d1−2−3−d0402d)→(0710d1−2−3−d010d2)\left(\begin{array}{ccc|c}2&3&4&-d\\1&-2&-3&-d\\4&3&-2&-d\end{array}\right)\to\left(\begin{array}{ccc|c}0&7&10&d\\1&-2&-3&-d\\0&11&10&3d\end{array}\right)\to\left(\begin{array}{ccc|c}0&7&10&d\\1&-2&-3&-d\\0&4&0&2d\end{array}\right)\to\left(\begin{array}{ccc|c}0&7&10&d\\1&-2&-3&-d\\0&1&0&\dfrac{d}{2}\end{array}\right)

→(0010−52d10−30010d2)→(001−d410−30010d/2)→(001−d4100−34d010d2)\to\left(\begin{array}{ccc|c}0&0&10&-\dfrac{5}{2}d\\1&0&-3&0\\0&1&0&\dfrac{d}{2}\end{array}\right)\to\left(\begin{array}{ccc|c}0&0&1&-\dfrac{d}{4}\\1&0&-3&0\\0&1&0&d/2\end{array}\right)\to\left(\begin{array}{ccc|c}0&0&1&-\dfrac{d}{4}\\1&0&0&-\dfrac{3}{4}d\\0&1&0&\dfrac{d}{2}\end{array}\right)

これから −34dx+d2y−d4z+d=0-\dfrac{3}{4}dx+\dfrac{d}{2}y-\dfrac{d}{4}z+d=0 −3x+2y−z+4=0-3x+2y-z+4=0

(2)

(3,2,1)(3,2,1) (1,9,2)(1,9,2) (−4,4,−3)(-4,4,-3)

(321−d192−d−44−3−d)→(0−25−52d192−d0405−5d)→(0−25−52d192−d0150−3d)→(0−25−52d192−d010−d5)\left(\begin{array}{ccc|c}3&2&1&-d\\1&9&2&-d\\-4&4&-3&-d\end{array}\right)\to\left(\begin{array}{ccc|c}0&-25&-5&2d\\1&9&2&-d\\0&40&5&-5d\end{array}\right)\to\left(\begin{array}{ccc|c}0&-25&-5&2d\\1&9&2&-d\\0&15&0&-3d\end{array}\right)\to\left(\begin{array}{ccc|c}0&-25&-5&2d\\1&9&2&-d\\0&1&0&-\dfrac{d}{5}\end{array}\right)

→(00−5−3d10245d010−d5)→(00135d10245d010−d5)→(00135d100−25d010−d5)\to\left(\begin{array}{ccc|c}0&0&-5&-3d\\1&0&2&\dfrac{4}{5}d\\0&1&0&-\dfrac{d}{5}\end{array}\right)\to\left(\begin{array}{ccc|c}0&0&1&\dfrac{3}{5}d\\1&0&2&\dfrac{4}{5}d\\0&1&0&-\dfrac{d}{5}\end{array}\right)\to\left(\begin{array}{ccc|c}0&0&1&\dfrac{3}{5}d\\1&0&0&-\dfrac{2}{5}d\\0&1&0&-\dfrac{d}{5}\end{array}\right)

これから −25dx−d5y+35dz+d=0-\dfrac{2}{5}dx-\dfrac{d}{5}y+\dfrac{3}{5}dz+d=0 2x+y−3z−5=02x+y-3z-5=0

(3)

(1,4,2)(1,4,2) (2,1,3)(2,1,3) (3,−2,0)(3,-2,0)

(142−d213−d3−20−d)→(142−d0−7−1d0−14−62d)→(142−d071−d073−d)→(1−100d071−d0−1402d)→(1−100d071−d070−d)\left(\begin{array}{ccc|c}1&4&2&-d\\2&1&3&-d\\3&-2&0&-d\end{array}\right)\to\left(\begin{array}{ccc|c}1&4&2&-d\\0&-7&-1&d\\0&-14&-6&2d\end{array}\right)\to\left(\begin{array}{ccc|c}1&4&2&-d\\0&7&1&-d\\0&7&3&-d\end{array}\right)\to\left(\begin{array}{ccc|c}1&-10&0&d\\0&7&1&-d\\0&-14&0&2d\end{array}\right)\to\left(\begin{array}{ccc|c}1&-10&0&d\\0&7&1&-d\\0&7&0&-d\end{array}\right)

(1−100d0010070−d)→(1−100d0010010−d7)→(100−37d0010010−d7)\left(\begin{array}{ccc|c} 1 & -10 & 0 & d \\ 0 & 0 & 1 & 0 \\ 0 & 7 & 0 & -d \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & -10 & 0 & d \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & -\dfrac{d}{7} \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & 0 & 0 & -\dfrac{3}{7}d \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & -\dfrac{d}{7} \end{array}\right)

これから −37dx−d7y+d=0-\dfrac{3}{7}dx - \dfrac{d}{7}y + d = 0 3x+y−7=03x + y - 7 = 0

解説: r31bn1z