It is proved by vector method that the diagonal of a quadrilateral in space is perpendicular if and only if the sum of squares of two opposite sides is equal

For convenience, the following is the representative vector of the following, which is the representative vector of the following. CD = # (CD = # # # # # # # # # 35; # # # # # # 3535; 35; 35; 35\35\3535\3535\35; # # # # 3535; \35; # \\\\\\\\results: AB2 + CD2 = AB2 + (# BD - # BC) 2 = AB2 + BD2 + bc2-2 # BD ·# bcad2 + BC2 = (# BD - # BA) 2

It is proved by the method of space vector that the diagonal of a quadrilateral in space is perpendicular if and only if the sum of squares of two groups of opposite sides is equal

Let four vertices correspond to vectors a, B, C and D
Diagonal vertical
(A-C) * (B-D) = 0 (* denotes dot product)
a*b+c*d=b*c+d*a
(a-b)*(a-b)+(c-d)*(c-d)=(b-c)*(b-c)+(d-a)*(d-a)
The sum of squares of two sets of opposite sides is equal

Help prove that "any two vectors in space are coplanar"
If there are two vectors that are not in the same plane but perpendicular to each other, how can we prove that they are coplanar?

Vector only has direction, regardless of the starting point, we can move vector arbitrarily. As long as we move two vectors to the common starting point, they must be coplanar. First of all, this sentence is right. Vector has both size and direction. There is no space vector in the two vectors. So, there are only two kinds of relations between two vectors: parallel and non parallel. The difference between vector and straight line lies in

It is known that the vectors a and B are unit vectors perpendicular to each other in the plane, if the vector C with (3a + C) (4b-c) = 0 satisfies | c-b|

Let a = (1,0), B = (0,1), C = (x, y), then x & sup2; - 4x + Y & sup2; + 3Y = 0 from (3a + C) × (4b-c) = 0, so (X-2) & sup2; + (y + 3 / 2) & sup2; = 25 / 4
K & sup2; ≥ | C-B | & sup2; = (x-1) & sup2; + Y & sup2; its geometric meaning is the square of the distance from point to point (1,0) on the circle (X-2) & sup2; + (y + 3 / 2) & sup2; = 25 / 4
The maximum distance from point to point (1,0) on the circle (X-2) & sup2; + (y + 3 / 2) & sup2; = 25 / 4 = {√ [(2-1) & sup2; + (3 / 2 - 0) & sup2;]} + 5 / 2 = (√ 13 + 5) / 2
The minimum value of K is (√ 13 + 5) / 2

It is proved by vector method that the diagonal of a quadrilateral in space is perpendicular if and only if the sum of squares of two opposite sides is equal