高専の数学解説

第17章 17.36

(1)

∣S0−λE∣=∣cos⁡θ−λ0sin⁡θ0−λ0sin⁡θ0−cos⁡θ−λ∣=λ(cos⁡θ−λ)(cos⁡θ+λ)+λsin⁡2θ|S_0 - \lambda E| = \begin{vmatrix} \cos\theta - \lambda & 0 & \sin\theta \\ 0 & -\lambda & 0 \\ \sin\theta & 0 & -\cos\theta - \lambda \end{vmatrix} = \lambda(\cos\theta - \lambda)(\cos\theta + \lambda) + \lambda\sin^2\theta

=λcos⁡2θ−λ3+λsin⁡2θ=−λ3+λ=−λ(λ2−1)=0= \lambda\cos^2\theta - \lambda^3 + \lambda\sin^2\theta = -\lambda^3 + \lambda = -\lambda(\lambda^2 - 1) = 0

λ=0,±1\lambda = 0, \pm 1

λ=−1\lambda = -1のとき

(cos⁡θ+10sin⁡θ010sin⁡θ0−cos⁡θ+1)→(cos⁡2θ−10sin⁡θ(cos⁡θ−1)010sin⁡2θ0−sin⁡θ(cos⁡θ−1))→(−sin⁡2θ0sin⁡θ(cos⁡θ−1)010sin⁡2θ0−sin⁡θ(cos⁡θ−1))\left(\begin{array}{ccc} \cos\theta + 1 & 0 & \sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & -\cos\theta + 1 \end{array}\right) \to \left(\begin{array}{ccc} \cos^2\theta - 1 & 0 & \sin\theta(\cos\theta - 1) \\ 0 & 1 & 0 \\ \sin^2\theta & 0 & -\sin\theta(\cos\theta - 1) \end{array}\right) \to \left(\begin{array}{ccc} -\sin^2\theta & 0 & \sin\theta(\cos\theta - 1) \\ 0 & 1 & 0 \\ \sin^2\theta & 0 & -\sin\theta(\cos\theta - 1) \end{array}\right)

→(−sin⁡2θ0sin⁡θ(cos⁡θ−1)010000)→(10−cos⁡θ−1sin⁡θ010000)\to \left(\begin{array}{ccc} -\sin^2\theta & 0 & \sin\theta(\cos\theta - 1) \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) \to \left(\begin{array}{ccc} 1 & 0 & -\frac{\cos\theta - 1}{\sin\theta} \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right)

p1=C1(cos⁡θ−1sin⁡θ01)→(cos⁡θ−10sin⁡θ)→(−2sin⁡2θ20sin⁡θ)→(−sin⁡θ202sin⁡θ2cos⁡θ2)→(−sin⁡θ202sin⁡θ2cos⁡θ2)p_1 = C_1 \begin{pmatrix} \frac{\cos\theta - 1}{\sin\theta} \\ 0 \\ 1 \end{pmatrix} \to \left(\begin{array}{c} \cos\theta - 1 \\ 0 \\ \sin\theta \end{array}\right) \to \left(\begin{array}{c} -2\sin^2\frac{\theta}{2} \\ 0 \\ \sin\theta \end{array}\right) \to \left(\begin{array}{c} -\sin\frac{\theta}{2} \\ 0 \\ 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \end{array}\right) \to \left(\begin{array}{c} -\sin\frac{\theta}{2} \\ 0 \\ 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \end{array}\right)

λ=0\lambda = 0 のとき

(cos⁡θ0sin⁡θ000sin⁡θ0−cos⁡θ)→(cos⁡θsin⁡θ0sin⁡2θcos⁡θsin⁡θ0−cos⁡2θ000)→(cos⁡θsin⁡θ0sin⁡2θ00−1000)→(10sin⁡θcos⁡θ001000)→(100001000)\left(\begin{array}{ccc} \cos\theta & 0 & \sin\theta \\ 0 & 0 & 0 \\ \sin\theta & 0 & -\cos\theta \end{array}\right) \to \left(\begin{array}{ccc} \cos\theta\sin\theta & 0 & \sin^2\theta \\ \cos\theta\sin\theta & 0 & -\cos^2\theta \\ 0 & 0 & 0 \end{array}\right) \to \left(\begin{array}{ccc} \cos\theta\sin\theta & 0 & \sin^2\theta \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{array}\right) \to \left(\begin{array}{ccc} 1 & 0 & \frac{\sin\theta}{\cos\theta} \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right) \to \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right)

p2=C2(010)p_2 = C_2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}

λ=1\lambda = 1 のとき

(cos⁡θ−10sin⁡θ010sin⁡θ0−cos⁡θ−1)→(cos⁡2θ−10sin⁡θ(cos⁡θ+1)010sin⁡2θ0−sin⁡θ(cos⁡θ+1))→(−sin⁡2θ0sin⁡θ(cos⁡θ+1)010sin⁡2θ0−sin⁡θ(cos⁡θ+1))\left(\begin{array}{ccc} \cos\theta - 1 & 0 & \sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & -\cos\theta - 1 \end{array}\right) \to \left(\begin{array}{ccc} \cos^2\theta - 1 & 0 & \sin\theta(\cos\theta + 1) \\ 0 & 1 & 0 \\ \sin^2\theta & 0 & -\sin\theta(\cos\theta + 1) \end{array}\right) \to \left(\begin{array}{ccc} -\sin^2\theta & 0 & \sin\theta(\cos\theta + 1) \\ 0 & 1 & 0 \\ \sin^2\theta & 0 & -\sin\theta(\cos\theta + 1) \end{array}\right)

→(−sin⁡2θ0sin⁡θ(cos⁡θ+1)010000)→(10−cos⁡θ+1sin⁡θ010000)\to \left(\begin{array}{ccc} -\sin^2\theta & 0 & \sin\theta(\cos\theta + 1) \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) \to \left(\begin{array}{ccc} 1 & 0 & -\frac{\cos\theta + 1}{\sin\theta} \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right)

p1=C1(cos⁡θ+1sin⁡θ01)→(cos⁡θ+10sin⁡θ)→(2cos⁡2θ20sin⁡θ)→(cos⁡θ202sin⁡θ2cos⁡θ2)→(cos⁡θ202sin⁡θ2cos⁡θ2)→(cos⁡θ20sin⁡θ2)p_1 = C_1 \begin{pmatrix} \frac{\cos\theta + 1}{\sin\theta} \\ 0 \\ 1 \end{pmatrix} \to \left(\begin{array}{c} \cos\theta + 1 \\ 0 \\ \sin\theta \end{array}\right) \to \left(\begin{array}{c} 2\cos^2\frac{\theta}{2} \\ 0 \\ \sin\theta \end{array}\right) \to \left(\begin{array}{c} \cos\frac{\theta}{2} \\ 0 \\ 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \end{array}\right) \to \left(\begin{array}{c} \cos\frac{\theta}{2} \\ 0 \\ 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \end{array}\right) \to \left(\begin{array}{c} \cos\frac{\theta}{2} \\ 0 \\ \sin\frac{\theta}{2} \end{array}\right)

(2)

P=(cos⁡θ2−sin⁡θ20001sin⁡θ2cos⁡θ20)P = \begin{pmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2} & 0 \\ 0 & 0 & 1 \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2} & 0 \end{pmatrix}

とするとこれは直交行列であるので tP=P−1{}^t P = P^{-1}

v=Pv′v = Pv' となるような v′v' を考えると tPv=v′{}^t P v = v'

v′=(cos⁡θ20sin⁡θ2−sin⁡θ20cos⁡θ2010)(xyz)=(x⋅cos⁡θ2+z⋅sin⁡θ2−x⋅sin⁡θ2+z⋅cos⁡θ2y)v' = \begin{pmatrix} \cos\frac{\theta}{2} & 0 & \sin\frac{\theta}{2} \\ -\sin\frac{\theta}{2} & 0 & \cos\frac{\theta}{2} \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x \cdot \cos\frac{\theta}{2} + z \cdot \sin\frac{\theta}{2} \\ -x \cdot \sin\frac{\theta}{2} + z \cdot \cos\frac{\theta}{2} \\ y \end{pmatrix}

tSθv=tv′tSθPv′=tv′(1000−10000)v′=(x⋅cos⁡θ2+z⋅sin⁡θ2−x⋅sin⁡θ2+z⋅cos⁡θ2y)t(1000−10000)(x⋅cos⁡θ2+z⋅sin⁡θ2−x⋅sin⁡θ2+z⋅cos⁡θ2y){}^t S_\theta v = {}^t v' {}^t S_\theta P v' = {}^t v' \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} v' = \begin{pmatrix} x \cdot \cos\frac{\theta}{2} + z \cdot \sin\frac{\theta}{2} \\ -x \cdot \sin\frac{\theta}{2} + z \cdot \cos\frac{\theta}{2} \\ y \end{pmatrix}^{t} \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \cdot \cos\frac{\theta}{2} + z \cdot \sin\frac{\theta}{2} \\ -x \cdot \sin\frac{\theta}{2} + z \cdot \cos\frac{\theta}{2} \\ y \end{pmatrix}

=(x⋅cos⁡θ2+z⋅sin⁡θ2x⋅sin⁡θ2−z⋅cos⁡θ20)(x⋅cos⁡θ2+z⋅sin⁡θ2−x⋅sin⁡θ2+z⋅cos⁡θ2y)= \left( x \cdot \cos\frac{\theta}{2} + z \cdot \sin\frac{\theta}{2} \quad x \cdot \sin\frac{\theta}{2} - z \cdot \cos\frac{\theta}{2} \quad 0 \right) \begin{pmatrix} x \cdot \cos\frac{\theta}{2} + z \cdot \sin\frac{\theta}{2} \\ -x \cdot \sin\frac{\theta}{2} + z \cdot \cos\frac{\theta}{2} \\ y \end{pmatrix}

=(x⋅cos⁡θ2+z⋅sin⁡θ2)2−(x⋅sin⁡θ2−z⋅cos⁡θ2)2=x2(cos⁡2θ2−sin⁡2θ2)+4xzsin⁡θ2cos⁡θ2+z2(sin⁡2θ2−cos⁡2θ2)= \left( x \cdot \cos\frac{\theta}{2} + z \cdot \sin\frac{\theta}{2} \right)^2 - \left( x \cdot \sin\frac{\theta}{2} - z \cdot \cos\frac{\theta}{2} \right)^2 = x^2 \left( \cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2} \right) + 4xz \sin\frac{\theta}{2}\cos\frac{\theta}{2} + z^2 \left( \sin^2\frac{\theta}{2} - \cos^2\frac{\theta}{2} \right)

=x2⋅cos⁡θ+2xz⋅sin⁡θ−z2⋅cos⁡θ=c= x^2 \cdot \cos\theta + 2xz \cdot \sin\theta - z^2 \cdot \cos\theta = c

問題集 p.180 のように v′=(x′y′z′)v' = (x' \quad y' \quad z') と考えると上式は

x′2−y′2=cx'^2 - y'^2 = c

これは双曲柱面を表す。

(2) と同様に考えると、問題集 p.181 より v=Pv′=P(x′y′z′)v = Pv' = P \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} とおくと

v′=P−1v=(131313012−12−261616)(xyz)=(13(x+y+z)12(y+z)16(−2x+y+z))v' = P^{-1}v = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}}(x + y + z) \\ \frac{1}{\sqrt{2}}(y + z) \\ \frac{1}{\sqrt{6}}(-2x + y + z) \end{pmatrix}

S=vPAPv′+2=v(−200010001)v′+2=−2x′2+y′2+z′2+2=0S = \sqrt{v}PAP v' + 2 = \sqrt{v}\begin{pmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}v' + 2 = -2x'^2 + y'^2 + z'^2 + 2 = 0

v=p1v = p_1 とすると、vv に直交する平面は問題集 p.66 の平面の方程式から考えて、法線ベクトルが p1p_1 となるので

13x+13y+13z+d=0\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z + d = 0

13x+13y+13z=d\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z = d

x′=dx' = d となり、これを SS の式に代入すると

y′2+z′2=2a2−2y'^2 + z'^2 = 2a^2 - 2 の円となる

解説: r31bn1z