与三条直线相交的直线的轨迹

hbghlyj

书后的答案:

11. $4 x^{2}-9 y^{2}+10 x-45 y-84 z-120=0$
12. $x^{2}+y^{2}-z^{2}=1$,但要去掉$\left\{\begin{array} { l } { x = - 1 , } \\ { y = z , } \end{array} \left\{\begin{array}{l}x=1 \\ y=-z\end{array}\right.\right.$和$\frac{y}{-3}=\frac{y+\frac{5}{3}}{4}=\frac{z+\frac{4}{3}}{5}$三条直线.
13. $x^{2} \tan ^{2} \beta-y^{2}+\left(\tan ^{2} \beta-1\right) z^{2}=\left(\tan ^{2} \beta-1\right) c^{2}$.

hbghlyj

11.
In[1]:=
        U = {6 + 3 a, 2 a, 1 + a}; V = {3 b, 3 + 2 b, -4 - 2 b}; X = {x, y, z};
        GroebnerBasis[Cross[U - X, V - X]~Join~{Dot[U - X, {2, 3, 0}]}, {x, y, z}, {a, b}]
Out[1]:=
        $\left\{-120+10 x+4 x^2-45 y-9 y^2-84 z\right\}$


12.
In[2]:=
        U = {1, a, a}; V = {-1, b, -b}; W = {2 - 3 c, -1 + 4 c, -2 + 5 c}; X = {x, y, z};
        GroebnerBasis[Cross[U - X, V - X]~Join~Cross[U - X, W - X], {x, y, z}, {a, b, c}]
Out[2]:=
        $\left\{-1+x^2+y^2-z^2\right\}$


13.
In[3]:=
        GroebnerBasis[{Dot[Cross[{1, Tan[β], 0}, {x, y, z - c}], Cross[{1, -Tan[β], 0}, {x, y, z + c}]]}, {x, y, z}]
Out[3]:=
        $\left\{c^2-y^2-z^2-c^2 \text{Tan}[\beta ]^2+x^2 \text{Tan}[\beta ]^2+z^2 \text{Tan}[\beta]^2\right\}$

青青子衿

动直线产生的曲面方程
forum.php?mod=viewthread&tid=2897
【求助】与三条异面直线都相交的直线的轨迹。。。。