{"id":3074,"date":"2024-11-10T15:06:21","date_gmt":"2024-11-10T06:06:21","guid":{"rendered":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=3074"},"modified":"2024-11-20T14:08:36","modified_gmt":"2024-11-20T05:08:36","slug":"%e6%95%b0%e5%ad%a6ii_3q%e6%a8%a1%e6%93%ac","status":"publish","type":"post","link":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=3074","title":{"rendered":"\u6570\u5b66II_3Q\u6a21\u64ec"},"content":{"rendered":"\n\n\n\n\n
\u518d\u751f\u30ea\u30b9\u30c8<\/span><\/a><\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
4H1_\u524d\u534a<\/a><\/td>\n4H1_\u5f8c\u534a<\/a><\/td>\n2H1<\/a><\/td>\n4H2_\u524d\u534a<\/a><\/td>\n4H2_\u5f8c\u534a<\/a><\/td>\n2H2<\/a><\/td>\n<\/tr>\n
4H11_pdf<\/a><\/td>\n4H12_pdf<\/a><\/td>\n2H1_pdf<\/a><\/td>\n4H21_pdf<\/a><\/td>\n4H22_pdf<\/a><\/td>\n2H2_pdf<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
4H1_\u524d\u534a<\/span>4H1_\u5f8c\u534a<\/span>2H1<\/span>4H2_\u524d\u534a<\/span>4H2_\u5f8c\u534a<\/span>2H2<\/span><\/div>
\n

\u7b2c\uff12\u6b21\u5c0e\u95a2\u6570<\/span><\/p>\n

1\uff0e\u6b21\u306e\u95a2\u6570\u306e\u7b2c\uff12\u6b21\u5c0e\u95a2\u6570\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $y=(3x-5)^4$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\{(ax+b)^n\\}^{\\prime}=an(ax+b)^{n-1}$\u3000\u3092\u7528\u3044\u308b
\n$y^{\\prime}=3\\cdot4(3x-5)^3=12(3x-5)^3$
\n$y^{\\prime\\prime}=12\\{3\\cdot3(3x-5)^2\\}=108(3x-5)^2$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(2) $y=e^{5x^2-2}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\left(e^{f}\\right)^{\\prime}=f^{\\prime}e^{f}$\u3001$(f\\cdot g)’=f’g+fg’$\u3000\u3092\u7528\u3044\u308b
\n$y^{\\prime}=10xe^{5x^2-2}$
\n$y^{\\prime\\prime}=10e^{5x^2-2}+10x\\left(10xe^{5x^2-2}\\right)=10(10x^2+1)e^{5x^2-2}$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(3) $y=\\mathrm{log}(x+1)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(\\mathrm{log}f)^{\\prime}=\\dfrac{f’}{f}$\u3001$\\left(\\dfrac{1}{g} \\right)^{\\prime}=\\dfrac{-g’}{g^2}$\u3000\u3092\u7528\u3044\u308b
\n$y^{\\prime}=\\dfrac{1}{x+1}$
\n$y^{\\prime\\prime}=\\dfrac{-1}{(x+1)^2}$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(4) $y=\\mathrm{sin}3x$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(\\mathrm{sin}(ax+b))^{\\prime}=a\\mathrm{cos}(ax+b)\\;\\left((\\mathrm{cos}(ax+b))^{\\prime}=-a\\mathrm{sin}(ax+b)\\right)$\u3000\u3092\u7528\u3044\u308b
\n$y^{\\prime}=3\\mathrm{cos}3x$
\n$y^{\\prime\\prime}=-9\\mathrm{sin}3x$ (\u7b54)<\/div><\/div><\/span><\/p>\n

2\u6b21\u8fd1\u4f3c<\/span><\/p>\n

2\uff0e\u6b21\u306e\u95a2\u6570\u306e( )\u5185\u306e\u5024\u306b\u304a\u3051\u308b2\u6b21\u8fd1\u4f3c\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $f(x)=x^5+1\\;(x=1)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$x=a$ \u3067\u306e2\u6b21\u8fd1\u4f3c\u306e\u516c\u5f0f\u3000$f(x)\\fallingdotseq f(a)+f'(a)(x-a)+\\dfrac{1}{2}f^{\\prime\\prime}(a)(x-a)^2$\u3000\u3092\u7528\u3044\u308b
\n$f(x)=x^5+1,f'(x)=5x^4,\\;f^{\\prime\\prime}(x)=20x^3\\;\\to\\;f(1)=2,f'(1)=5,f^{\\prime\\prime}(1)=20$
\n$f(x)\\fallingdotseq f(1)+f'(1)(x-1)+\\dfrac{1}{2}f^{\\prime\\prime}(1)(x-1)^2$
\n$f(x)\\fallingdotseq 2+5(x-1)+10(x-1)^2$ (\u7b54)<\/span><\/div><\/div>\n

(2) $f(x)=e^{2x}\\;(x=0)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$x=a$ \u3067\u306e2\u6b21\u8fd1\u4f3c\u306e\u516c\u5f0f\u3000$f(x)\\fallingdotseq f(a)+f'(a)(x-a)+\\dfrac{1}{2}f^{\\prime\\prime}(a)(x-a)^2$\u3000\u3092\u7528\u3044\u308b
\n$f(x)=e^{2x},f'(x)=2e^{2x},\\;f^{\\prime\\prime}(x)=4e^{2x}\\;\\to\\;f(0)=1,f'(0)=2,f^{\\prime\\prime}(0)=4$
\n$f(x)\\fallingdotseq f(0)+f'(0)(x-0)+\\dfrac{1}{2}f^{\\prime\\prime}(0)(x-0)^2$
\n$f(x)\\fallingdotseq 1+2x+2x^2$ (\u7b54)<\/span><\/div><\/div>\n

(3) $f(x)=\\mathrm{log}(x+1)\\;(x=0)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$x=a$ \u3067\u306e2\u6b21\u8fd1\u4f3c\u306e\u516c\u5f0f\u3000$f(x)\\fallingdotseq f(a)+f'(a)(x-a)+\\dfrac{1}{2}f^{\\prime\\prime}(a)(x-a)^2$\u3000\u3092\u7528\u3044\u308b
\n$f(x)=\\mathrm{log}(x+1),f'(x)=\\dfrac{1}{x+1},\\;f^{\\prime\\prime}(x)=\\dfrac{-1}{(x+1)^2}$
\n$\\to\\;f(0)=\\mathrm{log}1=0,f'(0)=1,f^{\\prime\\prime}(0)=-1$
\n$f(x)\\fallingdotseq f(0)+f'(0)(x-0)+\\dfrac{1}{2}f^{\\prime\\prime}(0)(x-0)^2$
\n$f(x)\\fallingdotseq x-\\dfrac{1}{2}x^2$ (\u7b54)<\/span><\/div><\/div>\n

\u95a2\u6570\u306e\u30b0\u30e9\u30d5(\u5897\u6e1b\u3068\u51f9\u51f8)<\/span><\/p>\n

3\uff0e\u6b21\u306e\u95a2\u6570\u306e\u5897\u6e1b\u30fb\u6975\u5024\u30fb\u51f9\u51f8\u30fb\u5909\u66f2\u70b9\u3092\u8abf\u3079\u3066\u30b0\u30e9\u30d5\u3092\u66f8\u3051\u3002<\/p>\n

(1) $y=x^3-3x^2$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u65b9\u91dd\uff1a\u65b9\u7a0b\u5f0f $y’=0$ \u304a\u3088\u3073 $y^{\\prime\\prime}=0$ \u3092\u89e3\u3044\u3066\u3001\u5897\u6e1b\u8868\u3092\u304b\u304f
\n$y’=3x^2-6x=3x(x-2)=0\\;\\to \\begin{cases}x=0,\\hspace{7pt} 2\\\\y=0,-4\\end{cases}$
\n$y^{\\prime\\prime}=6x-6=6(x-1)=0\\;\\to \\begin{cases}x=\\hspace{7pt}1\\\\y=-2\\end{cases}$
\n\u3053\u308c\u3088\u308a\u5897\u6e1b\u8868\u3092\u4f5c\u6210\u3057\u3066\u30b0\u30e9\u30d5\u3092\u66f8\u304f\u3068
\n\"\"
\n\u6975\u5927\u5024 $0\\;(x=0)$\u3001\u6975\u5c0f\u5024 $-4\\;(x=2)$\u3001\u5909\u66f2\u70b9 $(1,-2)$ (\u7b54)<\/span><\/div><\/div>\n

(2) $y=\\dfrac{1}{4}x^4-\\dfrac{3}{2}x^2$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u65b9\u91dd\uff1a\u65b9\u7a0b\u5f0f $y’=0$ \u304a\u3088\u3073 $y^{\\prime\\prime}=0$ \u3092\u89e3\u3044\u3066\u3001\u5897\u6e1b\u8868\u3092\u304b\u304f
\n$y’=x^3-3x=x(x^2-3)=0\\;\\to \\begin{cases}x=-\\sqrt{3},0, \\sqrt{3} \\\\y=\\dfrac{-9}{4}, \\hspace{6pt} 0, \\dfrac{-9}{4} \\end{cases}$
\n$y^{\\prime\\prime}=3x^2-3=3(x^2-1)=0\\;\\to \\begin{cases}x=-1,\\hspace{6pt} 1\\\\y=\\dfrac{-5}{4},\\dfrac{-5}{4}\\end{cases}$
\n\u3053\u308c\u3088\u308a\u5897\u6e1b\u8868\u3092\u4f5c\u6210\u3057\u3066\u30b0\u30e9\u30d5\u3092\u66f8\u304f\u3068
\n\"\"
\n\u6975\u5927\u5024 $0\\;(x=0)$\u3001\u6975\u5c0f\u5024 $\\dfrac{-9}{4}\\;(x=\\pm\\sqrt{3})$\u3001\u5909\u66f2\u70b9 $\\left(\\pm 1,\\dfrac{-5}{4}\\right)$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u3044\u308d\u3044\u308d\u306a\u95a2\u6570\u306e\u30b0\u30e9\u30d5<\/span><\/p>\n

1\uff0e\u6b21\u306e\u95a2\u6570\u306e\u5897\u6e1b\u30fb\u6975\u5024\u30fb\u51f9\u51f8\u30fb\u5909\u66f2\u70b9\u3092\u8abf\u3079\u3066\u30b0\u30e9\u30d5\u3092\u66f8\u3051\u3002<\/p>\n

(1) $y=x\\mathrm{log}x\\;(x>0)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
3Q_4H3\u53c2\u7167(3Q_4H3\u306e\u6f14\u7fd2\u3068\u540c\u3058\u554f\u984c\u3067\u3059)<\/span><\/div><\/div>\n

(2) $y=\\dfrac{1}{x^2-1}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
3Q_4H4\u53c2\u7167(3Q_4H4\u306e\u6f14\u7fd2\u3068\u540c\u3058\u554f\u984c\u3067\u3059)<\/span><\/div><\/div>\n

(3) $y=x\\sqrt{4-x^2}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
3Q_4H5\u53c2\u7167(3Q_4H5\u306e\u6f14\u7fd2\u3068\u540c\u3058\u554f\u984c\u3067\u3059)<\/span><\/div><\/div>\n

2\uff0e\u6b21\u306e\u95a2\u6570\u306e\u30b0\u30e9\u30d5\u306e\u5909\u66f2\u70b9\u3092\u6c42\u3081\u3088\u3002(\u5206\u6570\u95a2\u6570\u306b\u3064\u3044\u3066\u306f\u3001\u6f38\u8fd1\u7dda\u3082\u6c42\u3081\u3088\u3002)<\/p>\n

(1) $y=\\dfrac{1}{x^2+3}-\\dfrac{5}{4}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u5b9a\u7fa9\uff1a\u5909\u66f2\u70b9\u3068\u306f\u51f9\u51f8\u304c\u5909\u308f\u308b\u70b9\u3001\u3059\u306a\u308f\u3061 $y^{\\prime\\prime}$ \u306e\u7b26\u53f7\u304c\u5909\u308f\u308b\u70b9
\n\u516c\u5f0f $\\left(\\dfrac{1}{g}\\right)^{\\prime}=\\dfrac{-g’}{g^2},\\;\\left(\\dfrac{f}{g^n}\\right)^{\\prime}=\\dfrac{f’g-nfg’}{g^{n+1}}$ \u3092\u7528\u3044\u308b
\n$\\displaystyle \\lim_{n\\to \\pm\u3000\\infty}y=\\displaystyle \\lim_{n\\to \\pm\u3000\\infty}\\left(\\dfrac{1}{x^2+3}-\\dfrac{5}{4}\\right)=\\left(=\\dfrac{1}{\\infty}-\\dfrac{5}{4}\\right)-\\dfrac{5}{4}$
\n\u5f93\u3063\u3066\u3001\u6f38\u8fd1\u7dda\u306f $y=-\\dfrac{5}{4}$ (\u7b54)
\n$y’=\\dfrac{-2x}{(x^2+3)^2}\\;\\to\\;y^{\\prime\\prime}=\\dfrac{-2(x^2)+3-2(-2x)(2x)}{(x^2+3)^3}=\\dfrac{6(x^2-1)}{(x^2+3)^3}$
\n$(y^{\\prime\\prime}\\;\u306e\u5206\u5b50)=6(x^2-1)=0$ \u3088\u308a $x=\\pm 1$ \u3060\u304b\u3089\u3001\u51f9\u51f8\u8868\u3092\u66f8\u304f\u3068\u6b21\u306e\u3088\u3046\u306b\u306a\u308b\uff1a
\n$\\begin{array}{c|c c c c c}x&\\cdots&-1&\\cdots&1&\\cdots \\\\ \\hline y^{\\prime\\prime}&+&0&-&0&+ \\\\ y&\u4e0b\u306b\u51f8&-1&\u4e0a\u306b\u51f8&-1&\u4e0b\u306b\u51f8\\end{array}$
\n\u8868\u3088\u308a\u5909\u66f2\u70b9\u306f $(\\pm 1, -1)$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(2) $y=x+\\mathrm{sin}x\\;(0\\lt x \\lt 2\\pi)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u5b9a\u7fa9\uff1a\u5909\u66f2\u70b9\u3068\u306f\u51f9\u51f8\u304c\u5909\u308f\u308b\u70b9\u3001\u3059\u306a\u308f\u3061 $y^{\\prime\\prime}$ \u306e\u7b26\u53f7\u304c\u5909\u308f\u308b\u70b9
\n$y’=1+\\mathrm{cos}x\\;\\to\\;y^{\\prime\\prime}=-\\mathrm{sin}x=0\\;\\to\\;\\mathrm{sin}x=0\\;(0\\lt x \\lt 2\\pi)\\;\\to\\;x=\\pi$
\n\u5f93\u3063\u3066\u3001\u51f9\u51f8\u8868\u3092\u66f8\u304f\u3068\u6b21\u306e\u3088\u3046\u306b\u306a\u308b\uff1a
\n$\\begin{array}{c|c c c c c}x&0&\\cdots&\\pi&\\cdots&2\\pi&\\cdots \\\\ \\hline y^{\\prime\\prime}&0&-&0&+&0 \\\\ y&0&\u4e0a\u306b\u51f8&\\pi&\u4e0b\u306b\u51f8&2\\pi\\end{array}$
\n\u8868\u3088\u308a\u5909\u66f2\u70b9\u306f $(\\pi, \\pi)$ (\u7b54)<\/div><\/div><\/span><\/p>\n<\/div>\n
\n

\u6210\u5206\u8868\u793a\u3068\u548c\u30fb\u5b9a\u6570\u500d\u30fb\u5185\u7a4d<\/span><\/p>\n

1\uff0e$\\vec{a}=\\begin{pmatrix}2\\\\2\\\\-3\\end{pmatrix}$\u3001$\\vec{b}=\\begin{pmatrix}1\\\\-2\\\\2\\end{pmatrix}$ \u306e\u3068\u304d $\\vec{a}\\cdot(3\\vec{a}-2\\vec{b})$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\vec{a}\\cdot(3\\vec{a}-2\\vec{b})=\\begin{pmatrix}2\\\\2\\\\-3\\end{pmatrix}\\cdot\\left\\{3\\begin{pmatrix}2\\\\2\\\\-3\\end{pmatrix}-2\\begin{pmatrix}1\\\\-2\\\\2\\end{pmatrix}\\right\\}=\\begin{pmatrix}2\\\\2\\\\-3\\end{pmatrix}\\cdot\\left\\{\\begin{pmatrix}6\\\\6\\\\-9\\end{pmatrix}-\\begin{pmatrix}2\\\\-4\\\\4\\end{pmatrix}\\right\\}$
\n$=\\begin{pmatrix}2\\\\2\\\\-3\\end{pmatrix}\\cdot\\begin{pmatrix}4\\\\10\\\\-13\\end{pmatrix}=2\\cdot4+2\\cdot10+(-3)\\cdot(-13)=67$ (\u7b54)<\/span><\/div><\/div>\n

\u5185\u7a4d\u3068\u306a\u3059\u89d2<\/span><\/p>\n

2\uff0e$\\vec{a}=\\begin{pmatrix}-2\\\\4\\\\-4\\end{pmatrix}$\u3001$\\vec{b}=\\begin{pmatrix}2\\\\-2\\\\0\\end{pmatrix}$ \u306e\u3068\u304d $\\vec{a}$\u3001$\\vec{b}$ \u306e\u306a\u3059\u89d2 $\\theta\\;(0\\leqq \\theta \\leqq \\pi)$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>

\n$\\begin{matrix}\\theta & 0^{\\circ} & 30^{\\circ} & 45^{\\circ} & 60^{\\circ} & 90^{\\circ}\\\\ {\\mathrm{cos}}\\theta & 1&\u3000\\dfrac{\\sqrt{3}}{2} & \\dfrac{\\sqrt{2}}{2} & \\dfrac{1}{2} & 0 \\end{matrix}$ $\\hspace{20pt}$(\u516c\u5f0f) $\\mathrm{Cos}^{-1}(-x)=\\pi-\\mathrm{Cos}^{-1}x$
\n(\u516c\u5f0f) ${\\mathrm{cos}}\\theta=\\dfrac{\\vec{a}\\cdot\\vec{b}}{|\\vec{a}||\\vec{b}|}$\u3000\u3088\u308a $\\theta={\\mathrm{Cos}}^{-1}\\left(\\dfrac{\\vec{a}\\cdot\\vec{b}}{|\\vec{a}||\\vec{b}|}\\right)$ \u3092\u7528\u3044\u308b
\n$\\vec{a}\\cdot\\vec{b}=(-2)\\cdot2+4\\cdot(-2)+(-4)\\cdot 0=-12$\u3001
\n$|\\vec{a}|=\\sqrt{2^2+4^2+(-4)^2}=6$\u3001$|\\vec{b}|=\\sqrt{2^2+(-2)^2+0^2}=2\\sqrt{2}$ \u3088\u3063\u3066\u3001
\n$\\theta={\\mathrm{Cos}}^{-1}\\left( \\dfrac{-12}{6\\cdot\u30002\\sqrt{2}} \\right)={\\mathrm{Cos}}^{-1}\\left(\\dfrac{-\\sqrt{2}}{2}\\right)=\\pi-{\\mathrm{Cos}}^{-1}\\left(\\dfrac{\\sqrt{2}}{2}\\right)=\\pi-\\dfrac{\\pi}{4}=\\dfrac{3\\pi}{4}$ (\u7b54)<\/span><\/div><\/div>\n

\u5206\u70b9\u306e\u4f4d\u7f6e\u30d9\u30af\u30c8\u30eb<\/span><\/p>\n

3\uff0e2\u70b9 $\\mathrm{A}(5,5,1)$\u3001$\\mathrm{B}(1,-1,1)$ \u306b\u3064\u3044\u3066\u3001\u7dda\u5206$\\mathrm{AB}$ \u3092$3:1$\u306b\u5916\u5206\u3059\u308b\u70b9$\\mathrm{P}$ \u306e\u5ea7\u6a19\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u5916\u5206\u306a\u306e\u3067\u3001 $m:n=3:-1$ \u306b\u5206\u3051\u308b\u70b9\u3068\u3057\u3066\u3001\u5206\u70b9\u306e\u4f4d\u7f6e\u30d9\u30af\u30c8\u30eb\u306e\u516c\u5f0f\u3092\u7528\u3044\u308b\u3002
\n$\\vec{p}=\\dfrac{n\\vec{a}+m\\vec{b}}{m+n}=\\dfrac{(-1)\\vec{a}+3\\vec{b}}{3+(-1)}=\\dfrac{1}{2}\\left(-\\begin{pmatrix}5\\\\5\\\\1\\end{pmatrix}+3\\begin{pmatrix}1\\\\-1\\\\1\\end{pmatrix}\\right)$
\n$=\\dfrac{1}{2}\\begin{pmatrix}-2\\\\8\\\\2\\end{pmatrix}=\\begin{pmatrix}-1\\\\-4\\\\1\\end{pmatrix}$ \u3088\u308a $\\mathrm{P}(-1,-4,1)$ (\u7b54)<\/span><\/div><\/div>\n

\u5e73\u884c\u6761\u4ef6\u30fb\u5782\u76f4\u6761\u4ef6<\/span><\/p>\n

4. $\\vec{a}=\\begin{pmatrix}2\\\\x\\\\1\\end{pmatrix}$\u3001$\\vec{b}=\\begin{pmatrix}x\\\\y\\\\3\\end{pmatrix}$\u3000\u304c\u5e73\u884c\u3067\u3042\u308b\u3088\u3046\u306b $x$\u3001$y$ \u306e\u5024\u3092\u5b9a\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\vec{a}, \\vec{b}$ \u304c\u5e73\u884c\u3060\u304b\u3089\u3001$\\vec{b}=t\\vec{a}\\;(\u307e\u305f\u306f\u3001\\vec{a}=t\\vec{b})$ \u3088\u308a
\n$\\vec{b}=\\begin{pmatrix}x\\\\y\\\\3\\end{pmatrix}=t\\vec{a}=t\\begin{pmatrix}2t\\\\tx\\\\t\\end{pmatrix}\\;\\Rightarrow\\;\\begin{cases}x=2t \\\\y=tx\u3000\\\\3=t
\n\\end{cases}$ \u3053\u308c\u3092\u89e3\u3044\u3066
\n$t=3$ \u3088\u308a\u3000$x=2t=2\\cdot3=6, y=tx=3\\cdot6=18$(\u7b54)<\/span><\/div><\/div>\n

5\uff0e$\\vec{a}=\\begin{pmatrix}2\\\\x\\\\1\\end{pmatrix}$\u3001$\\vec{b}=\\begin{pmatrix}x\\\\1\\\\3\\end{pmatrix}$\u3000\u304c\u5782\u76f4\u3067\u3042\u308b\u3088\u3046\u306b $x$ \u306e\u5024\u3092\u5b9a\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\vec{a}, \\vec{b}$ \u304c\u5782\u76f4\u3060\u304b\u3089\u3001\u5185\u7a4d $\\vec{a}\\cdot\\vec{b}=0$ \u3088\u308a
\n$\\vec{a}\\cdot\\vec{b}=\\begin{pmatrix}2\\\\x\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}x\\\\1\\\\3\\end{pmatrix}=2\\cdot x+x\\cdot 1 +1\\cdot 3=3x+3=0$ \u3088\u308a $x=-1$ (\u7b54)<\/span><\/div><\/div>\n

\u7403\u9762\u306e\u65b9\u7a0b\u5f0f<\/span><\/p>\n

6. \u70b9 $(1,-2,1)$ \u3092\u4e2d\u5fc3\u3068\u3059\u308b\u3001\u534a\u5f84 $\\sqrt{6}$ \u306e\u7403\u9762\u306e\u65b9\u7a0b\u5f0f\u306e\u4e00\u822c\u5f62\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u4e2d\u5fc3 $(a, b, c)$ \u534a\u5f84 $r$ \u306e\u7403\u9762\u306e\u65b9\u7a0b\u5f0f: $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ \u3092\u7528\u3044\u308b
\n$(x-1)^2+(y+2)^2+(z-1)^2=(\\sqrt{6})^2$ \u3060\u304b\u3089\u3001\u4e00\u822c\u5f62\u306f\u3053\u308c\u3092\u5c55\u958b\u3057\u3066
\n$x^2+y^2+z^2-2x+4y-2z=0$ (\u7b54)<\/span><\/div><\/div>\n

7. \u65b9\u7a0b\u5f0f $x^2+y^2+z^2-4x+2y-2z+1=0$ \u3067\u8868\u3055\u308c\u308b\u7403\u9762\u306e\u4e2d\u5fc3\u3068\u534a\u5f84\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u5e73\u65b9\u5b8c\u6210\u3057\u3066\u6a19\u6e96\u5f62\u306b\u3059\u308b\u3068 $\\underbrace{x^2-4x}_{(x-2)^2-4}+\\underbrace{y^2+2y}_{(y+1)^2-1}+\\underbrace{z^2-2z}_{(z-1)^2-1}+1=0\\; \\to\\; (x-2)^2+(y+1)^2+(z-1)^2=5$ \u3088\u308a
\n\u4e2d\u5fc3 $(2, -1, 1)$ \u534a\u5f84 $\\sqrt{5}$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u4e0d\u5b9a\u7a4d\u5206(\u57fa\u672c)<\/span><\/p>\n

1\uff0e \u6b21\u306e\u4e0d\u5b9a\u7a4d\u5206\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $\\displaystyle \\int \\left(x^2+2x+\\dfrac{3}{x}+\\dfrac{2}{x^3}\\right) \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f$\\displaystyle \\int x^n \\,dx = \\dfrac{1}{n+1}x^{n+1}+C\\;(n\\neq -1),\\;\\displaystyle \\int \\dfrac{1}{x} \\,dx =\\mathrm{log}|x|+C$ \u3092\u7528\u3044\u308b
\n$\\displaystyle \\int \\left(x^2+2x+\\dfrac{3}{x}+2x^{-3}\\right) \\,dx=\\dfrac{1}{3}x^3+x^2+3\\mathrm{log}|x|-\\dfrac{1}{x^2}+C$ (\u7b54)<\/span><\/div><\/div>\n

(2) $\\displaystyle \\int (5x-3)^7 \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\displaystyle \\int (ax+b)^n \\,dx = \\dfrac{1}{a(n+1)}(ax+b)^{n+1}+C\\;(n\\neq-1)$ \u3092\u7528\u3044\u308b
\n$\\displaystyle \\int (5x-3)^7 \\,dx=\\dfrac{1}{5\\cdot8}(5x-3)^8+C=\\dfrac{1}{40}(5x-3)^8+C$ (\u7b54)<\/span><\/div><\/div>\n

(3) $\\displaystyle \\int \\dfrac{1}{\\sqrt[5]{4x+7}} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>

\n\u516c\u5f0f $\\displaystyle \\int (ax+b)^n \\,dx =\\dfrac{1}{a} \\cdot \\dfrac{1}{n+1} (ax+b)^{n+1}+C\\;(n\\neq -1)$ \u3092\u7528\u3044\u308b
\n$\\displaystyle \\int (4x+7)^{\\tiny{\\dfrac{-1}{5}}} \\,dx=\\dfrac{1}{4}\\dfrac{5}{4}(4x+7)^{\\tiny{\\dfrac{4}{5}}}+C=\\dfrac{5}{16}\\sqrt[5]{(4x+t)^4}$ (\u7b54)<\/span><\/div><\/div>\n

(4) $\\displaystyle \\int \\mathrm{sin}(7x-2) \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\displaystyle \\int \\mathrm{sin}(ax+b) \\,dx =-\\dfrac{1}{a} \\mathrm{cos}(ax+b)+C$
\n$\\hspace{20pt}\\displaystyle \\int \\mathrm{cos}(ax+b) \\,dx =\\dfrac{1}{a} \\mathrm{sin}(ax+b)+C$ \u7b49\u3092\u7528\u3044\u308b
\n$\\displaystyle \\int \\mathrm{sin}(7x-2) \\,dx=-\\dfrac{1}{7}\\mathrm{cos}(7x-2)+C$ (\u7b54)<\/span><\/div><\/div>\n

(5) $\\displaystyle \\int e^{-2x+3} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>

\n\u516c\u5f0f $\\displaystyle \\int e^{ax+b} \\,dx=\\dfrac{1}{a}e^{ax+b}+C$ \u3092\u7528\u3044\u308b
\n$\\displaystyle \\int e^{-2x+3} \\,dx=\\dfrac{-1}{2}e^{-2x+3}+C$ (\u7b54)<\/span><\/div><\/div>\n

(6) $\\displaystyle \\int \\dfrac{1}{\\mathrm{cos}^2(4x+1)} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>

\n\u516c\u5f0f $\\displaystyle \\int \\dfrac{1}{\\mathrm{cos}^2(ax+b)} \\,dx=\\dfrac{1}{a}\\mathrm{tan}(ax+b)+C$ \u3092\u7528\u3044\u308b
\n$\\displaystyle \\int \\dfrac{1}{\\mathrm{cos}^2(4x+1)} \\,dx=\\dfrac{1}{4}\\mathrm{tan}(4x+1)+C$ (\u7b54)<\/span><\/div><\/div>\n

(7) $\\displaystyle \\int \\dfrac{\\mathrm{cos}x}{5+\\mathrm{sin}x} \\, dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>

\n\u516c\u5f0f $\\displaystyle \\int \\dfrac{f’}{f} \\, dx=\\mathrm{log}|f|+C$ \u3092\u7528\u3044\u308b
\n$\\displaystyle \\int \\dfrac{\\mathrm{cos}x}{5+\\mathrm{sin}x} \\, dx=\\displaystyle \\int \\dfrac{(5+\\mathrm{sin}x)^{\\prime}}{5+\\mathrm{sin}x} \\, dx=\\mathrm{log}|5+\\mathrm{sin}x|+C$ (\u7b54)<\/span><\/div><\/div>\n

(8) $\\displaystyle \\int \\dfrac{x^4}{x^5-5} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\displaystyle \\int \\dfrac{f’}{f} \\, dx=\\mathrm{log}|f|+C$ \u3092\u7528\u3044\u308b
\n$\\displaystyle \\int \\dfrac{x^4}{x^5-5} \\,dx=\\dfrac{1}{5}\\displaystyle \\int \\dfrac{5x^4}{x^5-5} \\,dx=\\dfrac{1}{5}\\displaystyle \\int \\dfrac{(x^5-5)’}{x^5-5} \\,dx=\\dfrac{1}{5}\\mathrm{log}|x^5-5|+C$ (\u7b54)<\/span><\/div><\/div>\n

\u4e0d\u5b9a\u7a4d\u5206(\u4e09\u89d2\u95a2\u6570\u306e\u516c\u5f0f\u3092\u5229\u7528\u3059\u308b\u554f\u984c)<\/span><\/p>\n

2\uff0e\u6b21\u306e\u4e0d\u5b9a\u7a4d\u5206\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $\\displaystyle \\int \\mathrm{tan}x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\;\\mathrm{tan}x=\\dfrac{\\mathrm{sin}x}{\\mathrm{cos}x},\\;\\mathrm{cot}x=\\dfrac{\\mathrm{cos}x}{\\mathrm{sin}x}\\;$ \u7b49\u3092\u7528\u3044\u308b
\n$\\displaystyle \\int \\mathrm{tan}x \\,dx=-\\displaystyle \\int \\dfrac{-\\mathrm{sin}x}{\\mathrm{cos}x} \\,dx=-\\displaystyle \\int \\dfrac{(\\mathrm{cos}x)’}{\\mathrm{cos}x} \\,dx=-\\mathrm{log}|\\mathrm{cos}x|+C$ (\u7b54)<\/span><\/div><\/div>\n

(2) $\\displaystyle \\int \\mathrm{tan}^2x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\;\\mathrm{tan}^2x=\\dfrac{1}{\\mathrm{cos}^2x}-1,\\;\\mathrm{cot}^2x=\\dfrac{1}{\\mathrm{sin}^2x}-1\\;$ \u7b49\u3092\u7528\u3044\u308b
\n$\\displaystyle \\int \\mathrm{tan}^2x \\,dx=\\displaystyle \\int \\left(\\dfrac{1}{\\mathrm{cos}^2x}-1\\right) \\,dx=\\mathrm{tan}x-x+C$ (\u7b54)<\/span><\/div><\/div>\n

(3) $\\displaystyle \\int \\mathrm{cos}7x\\,\\mathrm{cos}3x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\;\\mathrm{cos}\\alpha\\,\\mathrm{cos}\\beta=\\dfrac{1}{2}\\left(\\mathrm{cos}(\\alpha+\\beta)+\\mathrm{cos}(\\alpha-\\beta)\\right)$
\n$\\hspace{22pt} \\mathrm{sin}\\alpha \\,\\mathrm{cos}\\beta=\\dfrac{1}{2}\\left(\\mathrm{sin}(\\alpha+\\beta)+\\mathrm{sin}(\\alpha-\\beta)\\right)$
\n$\\hspace{22pt}\\mathrm{sin}\\alpha \\,\\mathrm{sin}\\beta=\\dfrac{-1}{2}\\left(\\mathrm{cos}(\\alpha+\\beta)-\\mathrm{cos}(\\alpha-\\beta)\\right)\\;$ \u7b49\u3092\u7528\u3044\u308b
\n$\\displaystyle \\int \\mathrm{cos}7x\\,\\mathrm{cos}3x \\,dx=\\dfrac{1}{2}\\displaystyle \\int (\\mathrm{cos}10x+\\mathrm{cos}4x) \\,dx$
\n$=\\dfrac{1}{2}\\left(\\dfrac{1}{10}\\mathrm{sin}10x + \\dfrac{1}{4}\\mathrm{sin}4x \\right)+C=\\dfrac{1}{20}\\mathrm{sin}10x+\\dfrac{1}{8}\\mathrm{sin}4x +C$ (\u7b54)<\/span><\/div><\/div>\n

(4) $\\displaystyle \\int \\mathrm{sin}^2x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\;\\mathrm{sin}^2x=\\dfrac{1}{2}(1-\\mathrm{cos}2x),\\;\\mathrm{cos}^2x=\\dfrac{1}{2}(1+\\mathrm{cos}2x)\\;$ \u7b49\u3092\u7528\u3044\u308b
\n$\\displaystyle \\int \\mathrm{sin}^2x \\,dx=\\dfrac{1}{2}\\displaystyle \\int (1-\\mathrm{cos}2x) \\,dx=\\dfrac{1}{2}\\left(x-\\dfrac{1}{2}\\mathrm{sin}2x\\right)+C=\\dfrac{1}{2}x-\\dfrac{1}{4}\\mathrm{sin}2x+C$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u4e0d\u5b9a\u7a4d\u5206(\u90e8\u5206\u7a4d\u5206)<\/span><\/p>\n

1\uff0e\u6b21\u306e\u4e0d\u5b9a\u7a4d\u5206\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle \\int (3x+1)e^{2x+1} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u3000\"\"
\n$I=(3x+1)\\cdot\\dfrac{2}{4}e^{2x+1}-3\\cdot\\dfrac{1}{4}e^{2x+1}+C=\\dfrac{6x+2-3}{4}e^{2x+1}+C$
\n$\\quad =\\dfrac{(6x-1)e^{2x+1}}{4}+C$ (\u7b54)<\/span><\/div><\/div>\n

(2) $I=\\displaystyle \\int x^2\\mathrm{cos}(2x+3) \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u3000\"\"
\n$I=\\dfrac{1}{2}x^2\\mathrm{sin}(2x+3)+\\dfrac{1}{2}x\\mathrm{cos}(2x+3)-\\dfrac{1}{4}\\mathrm{sin}(2x+3)+C$ (\u7b54)<\/span><\/div><\/div>\n

(3) $I=\\displaystyle \\int (2x+1)^3\\mathrm{log}(2x+1) \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u3000\"\"
\n$I=\\dfrac{1}{8}(2x+4)^4\\mathrm{log}(2x+1)-\\dfrac{1}{4}\\displaystyle \\int (2x+1)^3 \\,dx$
\n$\\quad =\\dfrac{1}{8}(2x+4)^4\\mathrm{log}(2x+1)-\\dfrac{1}{4}\\cdot\\dfrac{1}{2\\cdot4}(2x+1)^4+C$
\n$\\quad =\\dfrac{1}{8}(2x+4)^4\\mathrm{log}(2x+1)-\\dfrac{1}{32}(2x+1)^4+C$ (\u7b54)<\/span><\/div><\/div>\n

(4) $I=\\displaystyle \\int e^{3x}\\mathrm{cos}2x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle \\int e^{ax}\\mathrm{cos}bx \\,dx=\\dfrac{e^{ax}}{a^2+b^2}(a\\mathrm{cos}bx+b\\mathrm{sin}bx)+C$
\n$\\displaystyle \\int e^{ax}\\mathrm{sin}bx \\,dx=\\dfrac{e^{ax}}{a^2+b^2}(a\\mathrm{sin}bx-b\\mathrm{cos}bx)+C$ \u7b49\u3092\u7528\u3044\u308b
\n$I=\\dfrac{e^{3x}}{3^2+2^2}(3\\mathrm{cos}2x+2\\mathrm{sin}2x)+C$ (\u7b54)<\/span><\/div><\/div>\n

\u4e0d\u5b9a\u7a4d\u5206(\u7f6e\u63db\u7a4d\u5206)<\/span><\/p>\n

2\uff0e\u6b21\u306e\u4e0d\u5b9a\u7a4d\u5206\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle \\int \\mathrm{sin}^5x\\mathrm{cos}x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$t=\\mathrm{sin}x\\;\\to\\;dt=\\mathrm{cos}xdx$ \u3088\u308a
\n$I=\\displaystyle \\int t^5 \\,\\underset{\\mathrm{cos}xdx}{dt}=\\dfrac{1}{6}t^6+C=\\dfrac{1}{6}\\mathrm{sin}^6x$ (\u7b54)<\/span><\/div><\/div>\n

(2) $I=\\displaystyle \\int \\dfrac{\\mathrm{cos}x}{\\mathrm{sin}^5x} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$t=\\mathrm{sin}x\\;\\to\\;dt=\\mathrm{cos}xdx$ \u3088\u308a
\n$I=\\displaystyle \\int \\dfrac{1}{t^5} \\,\\underset{\\mathrm{cos}xdx}{dt}=\\displaystyle \\int t^{-5} \\,dt=\\dfrac{1}{-4}t^{-4}+C=\\dfrac{-1}{4t^4}+C=\\dfrac{-1}{4\\mathrm{sin}^4x}+C$ (\u7b54)<\/span><\/div><\/div>\n

(3) $I=\\displaystyle \\int \\dfrac{\\mathrm{log}x}{x} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$t=\\mathrm{log}x\\;\\to\\;dt=\\dfrac{1}{x}dx$ \u3088\u308a
\n$I=\\displaystyle \\int t \\,\\underset{{\\scriptsize \\dfrac{1}{x}}dx}{dt}=\\dfrac{1}{2}t^2+C=\\dfrac{1}{2}(\\mathrm{log}x)^2+C$ (\u7b54)<\/span><\/div><\/div>\n

(4) $I=\\displaystyle \\int (2x+1)(x^2+x+1)^5 \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$t=x^2+x+1\\;\\to\\;dt=(2x+1)dx$ \u3088\u308a
\n$I=\\displaystyle \\int t^5 \\,\\underset{(2x+1)dx}{dt}=\\dfrac{1}{6}t^6+C=\\dfrac{1}{6}(x^2+x+1)^6+C$ (\u7b54)<\/span><\/div><\/div>\n

(5) $I=\\displaystyle \\int \\dfrac{x}{\\sqrt{x^2+1}} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$t=x^2+1\\;\\to\\;dt=2xdx$ \u3088\u308a
\n$I=\\dfrac{1}{2}\\displaystyle \\int \\dfrac{2x}{\\sqrt{x^2+1}} \\,dx=\\dfrac{1}{2}\\displaystyle \\int \\dfrac{1}{\\sqrt{t}} \\;\\underset{2xdx}{dt}=\\dfrac{1}{2}\\displaystyle \\int t^{\\tiny\\dfrac{-1}{2}} \\,dt$
\n$\\quad =\\dfrac{1}{2}\\cdot\\dfrac{2}{1}t^{\\tiny\\dfrac{1}{2}}+C=\\sqrt{t}+C=\\sqrt{x^2+1}+C$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u76f4\u7dda\u306e\u65b9\u7a0b\u5f0f<\/span><\/p>\n

1. \uff12\u70b9 ${\\mathrm{A}}(1,-5,2), {\\mathrm{B}}(3,-3,5)$ \u3092\u901a\u308b\u76f4\u7dda\u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u65b9\u5411\u30d9\u30af\u30c8\u30eb\u306f $\\vec{v}=\\overrightarrow{\\mathrm{AB}}=\\overset{B-A}{\\begin{pmatrix}3-1\\\\-3-(-5)\\\\5-2\\end{pmatrix}}=\\begin{pmatrix}2\\\\2\\\\3\\end{pmatrix}$ \u3067\u3001\u70b9 $\\mathrm{A}(1,-5,2)$ \u3092\u901a\u308b\u306e\u3067\u3001
\n$\\dfrac{x-1}{2}=\\dfrac{y+5}{2}=\\dfrac{z-2}{3}$(\u7b54)<\/span><\/div><\/div>\n

\u5e73\u9762\u306e\u65b9\u7a0b\u5f0f<\/span><\/p>\n

2. \u70b9 ${\\mathrm{A}}(1,-5,2)$ \u3092\u901a\u308a\u3001\u76f4\u7dda $\\dfrac{x+1}{3}=\\dfrac{y-2}{2}=\\dfrac{z-3}{-3}$ \u306b\u5782\u76f4\u306a\u5e73\u9762\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u76f4\u7dda\u306e\u65b9\u5411\u30d9\u30af\u30c8\u30eb\u304c\u6c42\u3081\u308b\u5e73\u9762\u306e\u6cd5\u7dda\u30d9\u30af\u30c8\u30eb $\\vec{n}=\\begin{pmatrix}3\\\\2\\\\-3\\end{pmatrix}$ \u3060\u304b\u3089\u3001\u70b9 ${\\mathrm{A}}(1,-1,3)$ \u3092\u901a\u308b\u306e\u3067\u3001
\n$3(x-1)+2(y+1)-3(z-3)=0$ \u3088\u308a\u3001\u5c55\u958b\u3057\u3066\u3001$3x+2y-3z+8=0$ (\u7b54)<\/span><\/div><\/div>\n

3\u70b9\u3092\u901a\u308b\u5e73\u9762\u306e\u65b9\u7a0b\u5f0f<\/span><\/p>\n

3. \uff13\u70b9 ${\\mathrm{A}}(2,2,-2), {\\mathrm{B}}(2,3,1), {\\mathrm{C}}(5,2,-1)$ \u3092\u901a\u308b\u5e73\u9762\u306e\u6cd5\u7dda\u30d9\u30af\u30c8\u30eb $\\vec{n}$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\overrightarrow{\\mathrm{AB}}=\\overset{B-A}{\\begin{pmatrix}2-2\\\\3-2\\\\1-(-2)\\end{pmatrix}}=\\begin{pmatrix}0\\\\1\\\\3\\end{pmatrix},\\;\\overrightarrow{\\mathrm{AC}}=\\overset{C-A}{\\begin{pmatrix}5-2\\\\2-2\\\\-1-(-2)\\end{pmatrix}}=\\begin{pmatrix}3\\\\0\\\\1\\end{pmatrix}$ \u307e\u305f\u3001$\\vec{n}=\\begin{pmatrix}1\\\\a\\\\b\\end{pmatrix}$ \u3068\u304a\u304f\u3068
\n$\\vec{n}\\cdot\\overrightarrow{\\mathrm{AB}}=a+3b=0,\\;\\vec{n}\\cdot\\overrightarrow{\\mathrm{AC}}=3+b=0$ \u3088\u308a $a=9,\\;b=-3$ \u3060\u304b\u3089\u3001$\\vec{n}=\\begin{pmatrix}1\\\\9\\\\-3\\end{pmatrix}$ (\u7b54)
\n(\u5225\u89e3) $\\vec{n}=\\overrightarrow{\\mathrm{AB}}\\times\\overrightarrow{\\mathrm{AC}}=\\begin{pmatrix}0\\\\1\\\\3\\end{pmatrix}\\times\\begin{pmatrix}3\\\\0\\\\1\\end{pmatrix}=\\begin{pmatrix}\\begin{vmatrix}
\n1 & 0 \\\\
\n3 & 1
\n\\end{vmatrix}\\\\-\\begin{vmatrix}
\n0 & 3 \\\\
\n3 & 1
\n\\end{vmatrix}\\\\\\begin{vmatrix}
\n0 & 3 \\\\
\n1 & 0
\n\\end{vmatrix}\\end{pmatrix}=\\begin{pmatrix}1\\\\-(-9)\\\\-3\\end{pmatrix}=\\begin{pmatrix}1\\\\9\\\\-3\\end{pmatrix} $ (\u7b54)<\/span><\/div><\/div>\n

\u70b9\u3068\u5e73\u9762\u306e\u8ddd\u96e2<\/span><\/p>\n

4\uff0e\u70b9 $(2,1,0)$ \u3068\u5e73\u9762 $2x-y+2z+1=0$ \u3068\u306e\u8ddd\u96e2 $h$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u70b9 $(x_0, y_0, z_0)$ \u3068\u5e73\u9762 $ax+by+cz+d=0$ \u3068\u306e\u8ddd\u96e2\u3092 $h$ \u3068\u3059\u308b\u3068\u304d
\n\u516c\u5f0f $h=\\dfrac{|ax_0+by_0+cz_0+d|}{\\sqrt{a^2+b^2+c^2}}$ \u3092\u7528\u3044\u308b
\n$h=\\dfrac{|2\\cdot2-1+2\\cdot0+1|}{\\sqrt{2^2+(-1)^2+2^2}}=\\dfrac{|4|}{\\sqrt{9}}=\\dfrac{4}{3}$ (\u7b54)<\/span><\/div><\/div>\n

\u76f4\u7dda\u3068\u5e73\u9762\u306e\u4ea4\u70b9<\/span><\/p>\n

5\uff0e\u76f4\u7dda $\\dfrac{x+1}{3}=\\dfrac{y-2}{2}=\\dfrac{z-3}{-3}$ \u3068\u5e73\u9762 $2x-y+2z+6=0$ \u306e\u4ea4\u70b9 $\\mathrm{P}$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\dfrac{x+1}{3}=\\dfrac{y-2}{2}=\\dfrac{z-3}{-3}=t$ \u3068\u304a\u304f\u3068
\n$x=3t-1, y=2t-2, z=2t+1\\cdots(*)$ \u3088\u308a\u3001\u5e73\u9762\u306e\u65b9\u7a0b\u5f0f\u306b\u4ee3\u5165\u3057\u3066
\n$2(3t-1)-(2t-2)+2(2t+1)+6=0\\;\\to\\;8t+8=0\\;\\to\\;t=-1$ \u3053\u308c\u3092 $(*)$ \u306b\u4ee3\u5165\u3057\u3066
\n$x=3(-1)-1=-4, y=2(-1)-2=-4, z=2(-1)+1=-1$ \u3088\u308a
\n\u4ea4\u70b9\u306f $\\mathrm{P}(-4, -4, -1)$ (\u7b54)<\/span><\/div><\/div>\n<\/div><\/div><\/div>\n","protected":false},"excerpt":{"rendered":"

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