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Given that the coordinates of three vertices of triangle ABC are a (0,0,2) B (4,2,0) C (2,4,0), the unit normal vector of plane ABC is obtained We just learned this, but I don't know much about it
Specific steps of plane normal vector: (undetermined coefficient method)
1. Establishing rectangular coordinate system
2. Let n = (x, y, z)
3. Find two non collinear vectors in the plane, denoted as a = (A1, A2, A3) B = (B1, B2, B3)
For example, ab = (4,2, - 2), BC = (- 2,2,0)
4. According to the definition of normal vector, the system of equations is established: ① n · a = 0; ② n · B = 0
That is 4x + 2y-2z = 0, - 2x + 2Y = 0
5. Solve the equations, take one of the solutions
The solution is x = 1, y = 1, z = 3
6. The unit is (V11 / 11, V11 / 11, 3v11 / 11) (V is the root ~)
Of course, you can also use the cross product to calculate directly, abxbc = (4,4,12), even if you don't learn
In space rectangular coordinate system, if a (1,0,0), B (0,1,0), C (0,0,1), P (x, y, z) is any point in plane ABC, then x, y, Z satisfy equation ()
P (x, y, z) is any point in the plane ABC
Then p, a, B and C are coplanar
OP=mOA+nOB+zOC,m+n+z=1
(x,y,z)=m(1,0,0)+n(0,1,0)+z(0,0,1)=(m,n,z)
x+y+z=m+n+z=1
For nonzero vectors a, B, C, if a · B = a · C, then B = C
Not equal to, possibly perpendicular to, multiplied by zero
As shown in the figure, in trapezoidal ABCD, AD / / BC, diagonal AC and BD intersect at O. (2) if triangle ACD = 6, s triangle OBC = 8, calculate the area of trapezoidal ABCD
If s triangle AOD = 2, s triangle cod = 3, find the area of ladder ABCD
If ACD = 6, OBC = 8
Ad / BC = 3 / 4 (the area ratio of the same height is equal to the ratio of the bottom edge)
AO / OC = 3 / 4 (the corresponding sides of similar triangles are proportional)
S triangle AOD / s triangle ODC = 3 / 4
The ODC of s triangle is 24 / 7
S trapezoid = 6 + 8 + 24 / 7 = 122 / 7
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