证明略
设
=a,
=b,
=c
则a、b、c两两垂直且模相等.
∴a·b=b·c=a·c=0,
又∵
=
NB 1
∴
=
=
b,
=
+
=
a+
b,
=
+
+
=-
a+b+c,
∴
·
=(
a+
b)·(b+c-
a)
=
-
=0.
∴MN⊥MC,
又
=
+
=
+
(b+c)=
(a+b+c),
=
+
=-a+c.
∴
·
=
(a+b+c)(c-a)=0.∴MP⊥B 1C.
证明略
设
=a,
=b,
=c
则a、b、c两两垂直且模相等.
∴a·b=b·c=a·c=0,
又∵
=
NB 1
∴
=
=
b,
=
+
=
a+
b,
=
+
+
=-
a+b+c,
∴
·
=(
a+
b)·(b+c-
a)
=
-
=0.
∴MN⊥MC,
又
=
+
=
+
(b+c)=
(a+b+c),
=
+
=-a+c.
∴
·
=
(a+b+c)(c-a)=0.∴MP⊥B 1C.