In the regular tetrahedron ABCD with edge length of 1, e and F are the midpoint of AD and BC, respectively RT

In the regular tetrahedron ABCD with edge length of 1, e and F are the midpoint of AD and BC, respectively RT

Relatively basic, it is easy to understand the definition of the distance between spatial lines

If the triangular pyramid p-abc, PA, Pb and PC are perpendicular, PA = 1, Pb = PC = √ 2, then the length of the distance from a point O to P, a, B and C is

Establishment of three dimensional rectangular coordinate system
P(0,0,0) A(0,0,1) B(√2,0,0) C(0,√2,0)
Let o (x, y, z)
D = root sign (x ^ 2 + y ^ 2 + Z ^ 2) = root sign [(x - √ 2) ^ 2 + y ^ 2 + Z ^ 2]=
Root sign [x ^ 2 + (Y - √ 2) ^ 2 + Z ^ 2] = root sign [x ^ 2 + y ^ 2 + (Z-1) ^ 2]
The solution is d = root 5 / 2

If PA, Pb and PC are perpendicular and PA = Pb = PC = a, then the distance from the center O to the plane ABC is 0______ ．

If the four points P, a, B and C in space are on the same sphere, PA, Pb and PC are perpendicular, and PA = Pb = PC = a, then PA, Pb and PC can be regarded as three edges from one vertex of the cube. Therefore, the sphere passing through the four points P, a, B and C in space is the circumscribed sphere of the cube with edge length a, and the diameter of the sphere is the diagonal of the cube with length a 3a, so the radius of the sphere is 123A, and the distance from the center O to the plane ABC is 16 times of the volume diagonal, that is, the distance from the center O to the plane ABC is 36a

If PA, Pb and PC are perpendicular to each other and PA = Pb = PC
A regular triangular pyramid p-abc is inscribed in a sphere o, r = 2. PA, Pb, PC are perpendicular to each other, PA = Pb = PC
Find the side length (i.e. PA, Pb, PC)

PA, Pb and PC are perpendicular to each other, and PA = Pb = PC;
Then 2R = √ 3 Pa; PA = 4 √ 3 / 3 = Pb = PC

How to find the inverse function of a function, for example, y = √ X-1

Finding the inverse function of function y = 1-x / 1 + X

Let f (x) = y, y = (1-x) / (1 + x) (x + 1 cannot be equal to 0 = > x is not equal to - 1) y (1 + x) = 1-x, y + x = 1-x, y + x = 1-y, x = 1-y, X = (1-y) / (1 + y) and X is not equal to - 1 (1-y) / (1 + y) is not equal to - 1, 1-y is not equal to - 1-y. therefore, f ^ - 1 (x) = (1-x) / (1 + x)

Inverse function of function y = arccosx (- 1 ≤ x ≤ 0)

Because - 1 ≤ x ≤ 0
therefore
y=π-arccos(-x)
arccos(-x)=π-y
-x=cos(π-y)
So the inverse function is:
y=-cos(π-x)

If f (x) is the inverse of y = 2x, then f (x) is always over a fixed point______ ．

Since f (x) is the inverse function of y = 2x, the image of F (x) and y = 2x is symmetric with respect to the straight line y = x, so that the image of F (x) is always over the fixed point (1,0), so the answer is: (1,0)

The image of the inverse function of the function y = a Λ (X-2) + 4 (a > 0, a ≠ 1) is always over a fixed point independent of A

The image of the inverse function of the function y = a Λ (X-2) + 4 (a > 0, a ≠ 1) is always over a fixed point independent of A
A: first of all, the function y = a Λ (X-2) + 4 (a > 0, a ≠ 1) must have an inverse function, because the function is monotone on the real number set R, when a > 1 is monotone increasing, when 0

Y = f (x) is constant over (0,1), and its inverse function is g (x)?

The inverse function g (x) is always over the fixed point (1,0)
So y = g (x) + 1 is constant over the fixed point (1,1)