两异面直线L1: x-1_y-y1_3-列 x-x3 卩-12=22间的距离 d=Prja, xa, M1M2 (a1xa,\、M1 n2 p2 2-X1y2-yz2-1 a1×c 2 k
: , 1 1 1 1 1 1 1 p z z n y y m x x L − = − = − 两异面直线 : : 2 2 2 2 2 2 2 间的距离 p z z n y y m x x L − = − = − → = Pr 1 2 M1M2 d j 1 2 1 2 1 2 ( ) = → M M . 2 2 2 1 1 1 2 1 2 1 2 1 2 2 2 1 1 1 m n p m n p i j k x x y y z m n p m n p − − − =
(5)平面及直线间的位置关系 平面与平面: 丌1:A1x+B1y+C1z+D1=0, 72:A2x+B2y+C2z+D2=0, [A1A2+B,B2+Cc2 cOs 2 +B12+C1√42+B2+C2 丌1⊥丌2分A142+B1B2+C1C2=0 丌12分= CI 2 K图心
(5) 平面及直线间的位置关系 平面与平面: : 0, 1 A1x + B1 y + C1 z + D1 = : 0, 2 A2 x + B2 y + C2 z + D2 = 2 2 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1 2 | | cos A B C A B C A A B B C C + + + + + + = = 1 ⊥ 2 0; A1A2 + B1B2 + C1C2 = . 2 1 2 1 2 1 C C B B A A 1 // 2 = =
直线与直线: L x-1y-y_z-列 ,L2 x-12y-J2- ni COS(L1,L,)= Im, m,+n, "2+PiPal m2+n2+p2·、m2+n2+n2 L1⊥L2<→mm2+n1m2+p1D2=0, L1∥/L2∈→ m n p p2 1y2-y2-列1 L1与L2共面∈→m1 P1|=0 K图心
直线与直线: : , 1 1 1 1 1 1 1 p z z n y y m x x L − = − = − : , 2 2 2 2 2 2 2 p z z n y y m x x L − = − = − 2 2 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 2 1 2 | | cos( , ) m n p m n p m m n n p p L L + + + + + + ^ = L1 ⊥ L2 0, m1m2 + n1n2 + p1 p2 = , 2 1 2 1 2 1 p p n n m m L1 // L2 = = L1与L2共面 0 2 2 2 1 1 1 2 1 2 1 2 1 = − − − m n p m n p x x y y z z
平面与直线: L -o-y-yo i-10, 1: Ax+ By+Cz+D=U, n Am+ Bn+ Cp sin p A2+B2+C2.√m2+n2+p AB C L⊥丌今 L∥冷Am+Bn+Cp=0 Lc丌分Am+Bn+(p=0,且Ax+B+C+D=0 x= xo t mt 已知与L,求交点:令{y=y+nt, z=Z0 t pt 代入4x+B+C+D=0得,从而可得交点.网国圆
平面与直线: : , 0 0 0 p z z n y y m x x L − = − = − : Ax + By +Cz + D = 0, 2 2 2 2 2 2 | | sin A B C m n p Am Bn Cp + + + + + + = L ⊥ . p C n B m A = = L// Am+ Bn+Cp = 0. L Am+ Bn+Cp = 0,且Ax0 + By0 +Cz0 + D = 0 已知与L, 求交点: , 0 0 0 = + = + = + z z pt y y nt x x mt 令 代入Ax + By +Cz + D = 0得t,从而可得交点