Why is the normal vector of a plane equal to the outer product of two nonparallel vectors? Please explain the reason in detail

Why is the normal vector of a plane equal to the outer product of two nonparallel vectors? Please explain the reason in detail

The direction of the outer product (or cross product) of two nonparallel vectors is perpendicular to the plane determined by the two vectors, and its direction conforms to the right-hand rule, which is explained in general higher mathematics books (such as the second volume of the special textbook for Physics Department of Sichuan University Edition)

Is the normal vector parallel? Is it true that the plane must be parallel and vice versa?

If we regard coincidence as a special form of parallelism, yes, this is a necessary and sufficient condition

The area of the triangle is 21.6 hectares, and the bottom is 800 meters high

1 ha = 10000 M2
21.6 ha = 216000 M2
Height × bottom / 2 = 216000
Height = 216000 / (800 / 2) = 216000 / 400 = 2160 / 4 = 540M

Solve the equation: X divided by two seventh, x = one fourteenth

There is no solution to this problem

17. The diameter of a semicircle is 6 decimeters, its perimeter is () decimeters, and its area is () square decimeters
17. The diameter of a semicircle is 6 decimeters, its perimeter is () decimeters, and its area is () square decimeters

If the diameter is 6dm, then the radius is 3DM
Therefore, the circumference of the circle is 2 × 3.14 × 3 = 18.84dm, and the area is 3.14 × 3 × 3 = 28.26
Then, the arc length of the semicircle is 18.84 △ 2 = 9.42dm
Therefore, the circumference of the semicircle is 9.42 + 6 = 15.42dm; the area is 28.26 △ 2 = 14.13dm ^ 2

Simple mathematical operation: 3.64 × 0.8 + 6.36 × 0.8-3.8

3.64×0.8+6.36×0.8-3.8
=0.8×（3.64+6.36）-3.8
=0.8×10-3.8
=8-3.8
=4.2
I wish you progress in your study

In the angle ABC, BD = DC, de vertical AB, DF vertical AC, e, F, be = CF, it is proved that ad is the bisector of the angle BAC

∵DE⊥AB,DF⊥AC
Ψ△ BDE and △ CDF are right triangles
∵BE=CF,BD=DC
∴Rt△BDE≌Rt△CDF(HL)
∴DE=DF
∵ in RT △ ade and RT △ ADF
DE=DF,AD=AD
∴Rt△ADE≌Rt△ADF(HL)
∴∠EAD=∠FAD
That is, bad = CAD
Ψ ad is the bisector of ∠ BAC

Given a series of numbers A1, A2, A3, A4, A5, A6, a7, and A1 = 1, a7 = 729, A1 / A2 = A2 / A3 = A3 / A4 = A4 / A5 = A5 / A6 = A6 / A7, can you work out this series of numbers

From A1 / A2 = A2 / A3 = A3 / A4 = A4 / A5 = A5 / A6 = A6 / A7, we can see that it is an equal ratio sequence
an=a1*q^(n-1)=q^(n-1)
a7=q^6=729=27^2
q=3
an=3^(n-1)

If the sum of the top and bottom of the isosceles trapezoid ABCD is 4, and the acute angle between the two diagonals is 60 degrees, what is the area of the isosceles trapezoid?
You'd better tell me how to get high. Thank you

Extend BC to e, make CE = ad, easy to get quadrilateral aced is parallelogram
Diagonal lines AC and BD intersect at point O, angle doc = 60, easy angle AOD = 120, because trapezoid isosceles, easy angle oad = angle ODA = 30
Because the quadrilateral aced is a parallelogram, the angle e = angle oad = 30
Make DM perpendicular to BC, intersect in M, RT △ DME, ∠ e = 30
So DM = (1 / 2) de
DE^2-DM^2=(4/2)^2=4
DM=2√3/3
S=(1/2)DM(AD+BC)=(1/2)*(2√3/3)*4=4√3/3

Let X and Y belong to R, and lgx + lgY = LG (x + y), find the minimum value of X + 4Y

xy=x+y
x(y-1)=y
x=y/(y-1)
x>0,y>0
x+4y=y/(y-1)+4y=1+1/(y-1)+4y=1/(y-1)+4(y-1)+5>=9