空间三直线

realnumber

l,m所成角为α，l,n所成角为β，
求m,n所成角的取值范围.----怀疑就是│α-β│~│α+β│（后者超过90度的话，最大就取90度）

realnumber

百度了下，是"三面角"

goft

就是以$l$为旋转轴的两个圆锥的母线位置关系么

乌贼

回复 3# goft
一针见血

其妙

回复 2# realnumber
定义
三面角：由三个面构成的多面角称为三面角，如图中三面角可记作∠O-ABC。
特别地，三个面角都是直角的三面角称为直三面角。
三面角的补三面角：由三条自已知三面角定点发出的垂直于已知三面角的三个平面的射线组成的三面角叫做已知三面角的补三面角。
性质
1、三面角的任意两个面角的和大于第三个面角。
2、三面角的三个二面角的和大于180°，小于540°。

三面角相关定理
设三面角∠O-ABC的三个面角∠AOB、∠BOC、∠AOC所对的二面角依次为∠OC，∠OA，∠OB。
1、三面角正弦定理：
sin∠OA/sin∠BOC=sin∠OB/sin∠AOC=sin∠OC/sin∠AOB。
证明过程如下：
三面角正弦定理
三面角正弦定理
2、三面角第一余弦定理：
cos∠BOC=cos∠OA×sin∠AOB×sin∠AOC+cos∠AOB×cos∠AOC。
证明过程如下：
三面角余弦定理证明
三面角余弦定理证明
3、三面角第二余弦定理：
cos∠OA=cos∠BOC×sin∠OB×sin∠OC-cos∠OB×cos∠OC。
从三面角第一余弦定理中消去∠AOB和∠AOC即可得
123.125.115.53/view/2692921.htm

hbghlyj

其妙 发表于 2014-1-15 05:00
123.125.115.53/view/2692921.htm

链接已过期 😢

hbghlyj

math.ucr.edu/~res/math133/winter09/polyangles.pdf
THEOREM 1 (Triangle Inequality). In the trihedral angle $\angle A\text{-}BCD$, one has
\[
\angle BAC + \angle CAD > \angle BAD.
\]
In the planar case where $C$ lies in the interior of $\angle BAD$, one has equality.

THEOREM 2 (Angle Sum Inequality). In the trihedral angle $\angle A\text{-}BCD$, one has
\[
\angle BAC + \angle CAD + \angle BAD < 360^\circ.
\]
These theorems reflect a basic geometrical fact: a set of coplanar points cannot be isometric to a set of noncoplanar points (compare the discussion in Section 8.5). Physically, this means that a tripod whose legs are locked into rigid positions with respect to each other cannot be moved so that the three feet and the top all lie on a flat surface.

PROOF OF THEOREM 1. If $\angle DAB \le \angle CAD$ or $\angle DAB \le \angle BAC$, the inequality is immediate, so we may assume that $\angle DAB > \angle CAD$ and $\angle DAB > \angle BAC$. Choose $E \in AB$ and $G \in AD$, and let $K \in \operatorname{Int} \angle DAB$ be a point such that
\[
\lvert \angle KAB \rvert = \lvert \angle BAC \rvert \quad (<\lvert \angle BAD \rvert).
\]
By the Crossbar Theorem there is a point $F \in EG \cap AC$. Choose $H \in AC$ so that $|AF|=|AF|$. Then $\triangle EAH \cong \triangle EAF$ by S.A.S., and therefore $|EH|=|EF|$.

By the Triangle Inequality (for ordinary plane triangles) and $E-F-G$ we have
\[|E H|+|H G|>|E G|=|E F|+|FG| ;
\]
since $|E H|=|E F|$, we conclude that $|H G|>|F G|$.
Since $|H A|=|FA|$ and $|H G|>|F G|$, the Hinge Theorem implies that $|\angle H A G|>|\angle F A G|$. On the other hand,
\[
|\angle B A D|=|\angle B A F|+|\angle F A G|<|\angle B A F|+|\angle E A G| .
\]
Since $\angle B A F=\angle B A C|$ and $\angle B A G=\angle C A D$, the inequality above reduces to
\[
|\angle B A D|<|\angle B A C|+|\angle C A D|
\]
PROOF OF THEOREM 2. The two main tools are Theorem 1 and the angle sum theorem for Euclidean triangles.
Consider the trihedral angles $\angle B\text-A C D, \angle C\text-A B D, \angle D\text-A B C$.
Applying Theorem 1 to each of them, we obtain the following inequalities:
(i) $|\angle B D C|<|\angle B D A|+|\angle A D C|$
(ii) $|\angle D C B|<|\angle D C A|+|\angle B C A|$
(iii) $|\angle D B C|<|\angle D B A|+|\angle C B A|$.

Since the angle-sum of a triangle is $180^{\circ}$ we have the following equalities:
\[
\begin{aligned}
& \text { (iv) }|\angle B D C|+|\angle D C B|+|\angle D B C|=180^{\circ} \\
& \text { (v) }|\angle B A D|=180^{\circ}-|\angle A D B|-|\angle A B D| \\
& \text { (vi) }|\angle B A C|=180^{\circ}-|\angle A C B|-|\angle A B C| \\
& \text { (vii) }|\angle C A D|=180^{\circ}-|\angle A D C|-|\angle A C D|
\end{aligned}
\]
Adding (v) -(vii) together, we obtain
\[
\begin{aligned}
& |\angle B A D|+|\angle B A C|+|\angle C A D|= \\
& 3 \cdot 180-|\angle A D B|-|\angle A B D|-|\angle A C B|-|\angle A B C|-|\angle A D C|-|\angle A C D| \\
= & 3 \cdot 180-(|\angle A D B|-|\angle A D C|)-(|\angle B C A|+|\angle D C A|) \\
& -(|\angle D B A|+|\angle C B A|) .
\end{aligned}
\]
Substitution of inequalities (i)-(iii) in the latter expression yield
\[
|\angle B A D|+|\angle B A C|+|\angle C A D|<3\cdot180-|\angle B D C|-|\angle D B C|-|\angle B C D|
\]
and by (iv) the right hand side is equal to $3\cdot 180-180=360$, as claimed. $\blacksquare$