How to find the normal vector of the high number plane is as follows The normal vector of the plane ax + by + CZ + D = 0 should be n = (a, B, c), but I don't know why it is this vector, how it comes from, and what should I pay attention to?

Let a (x1, Y1, z1), B (X2, Y2, Z2) be on this plane
Then the vector AB (arrow cannot be typed, sorry) = (x1-x2, y1-y2, z1-z2)
By substituting these two points into the plane equation, a (x1-x2) + B (y1-y2) + C (z1-z2) = 0
That is, (a, B, c) (x1-x2, y1-y2, z1-z2) = 0. This equation is true for all points a and B in the plane
So the normal vector is (a, B, c)
If you don't know, ask again,

Given the set a = X-1 ≥ a, B = x + 1 ≤ 4, and a ∩ B = &;. Find the range of real number a

A: X > = 1 + A or X

Given the ellipse equation y ^ 2 + 2x ^ 2 = 4, the intersection of line L and Y axis and the intersection of P (0, m) and ellipse are different points A.B, and the AP vector is equal to twice the Pb vector, the value range of M is calculated

This paper mainly examines the application of the positional relationship between straight line and ellipse. Straight line and curve are simultaneous. According to the relationship between the root and coefficient of equation, this is the most commonly used method to deal with this kind of problem. However, conic curve is characterized by large amount of calculation, which requires strong ability of operation and reasoning. The key is to see clearly the conditions given in the paper, To solve this kind of problem, one must be familiar with the definition of curve and the graphic characteristics of curve, which is also the common knowledge of college entrance examination

Given x2 + X + 1 = 0, find the value of 1 + X + x2 + X3 + X4. + x2013
The result is 1 + X (1 + X + x2) +. + x2011 (1 + X + x2) = 1 + 0 +. + 0 = 1
But the original formula = (1 + X + x3) + X4 + X5 + X6 +... + x2011 + x2012 + x2013 = 0 + X4 (1 + x) + X6 +. + x2010 (1 + x) + x2012 + x2013 = - X6 + X6 +. - x2012 + x2012 + x2013 (a total of 2014 numbers, no three terms sum is 0, there will be one x2013) = x2013
So 1 = x2013  x = 1, then 1 + X + X2 is not zero

There is nothing wrong with it
X2013 = 1, the conclusion is not x = 1, but a complex value, which satisfies 1 + X + x2 = 0
You can change your mind and solve the equation 1 + X + x2 = 0 to get two complex solutions. Then you can substitute any of the two complex solutions into the following equation to get (complex solution) ^ 2013 = 1
However, it is obvious that the result gives a simpler and clearer answer

It is known that the three edges of the parallelepiped abcd-a1b1c1d1 with the same vertex a as the end point are all equal to 1 and the angle between them is 60 degrees, and the length of the ball AC1
The answer is √ 6. Can you tell me how to do it

∵∠DAB=60°,∴∠ABC=120˚；
So AC & sup2; = AB & sup2; + BC & sup2; - 2Ab * bccos120 & # 730; = 2 + 2cos60 & # 730; = 3
If a1e ⊥ AB is made in plane aa1b1b, then a1e = aa1sin60 & # 730; = √ 3 / 2
Make a1f ⊥ AC in the diagonal plane acc1a1 and ABCD on the bottom of the diagonal plane acc1a1,
AC is their intersection, so a1f ⊥ bottom ABCD, even EF, according to the three perpendicular theorem,
AF⊥EF.
A1E=A1Asin60°=√3/2.
AE=A1Acos60°=1/2.
EF=AEtan30°=(1/2)(√3/3)=√3/6.
∴A1F=√(A1E²-EF²)=√[(√3/2)²-(√3/6)²]=√(2/3).
So sin ∠ a1af = a1f / A1A = √ (2 / 3)
cos∠A1AF=√(1-2/3)=√(1/3).
cos∠ACC1=cos(180˚-∠ACC1)=-cos∠ACC1=-√(1/3).
∴AC1=√[AC²+C1C²-2AC*C1C*cos∠ACC1]
=√[3+1+2(√3)√(1/3)]=√6.

1001+1003+1005+1007...+2007+2009=?
Answer 200 by 10:00 tonight

The answer is 760025, number of items = (last first) / tolerance + 1, and = (first + last) * number of items / 2

Given the set a = {x 2 + 2 (a + 1) x + A2-1 = 0} B = {x 2 + 4x = 0} AUB = a, find the value range of real number a

x²+4x=0
x(x+4)=0
x1=0,x2=-4
So B = {0, - 4}
Because a ∪ B = a, and B has only two elements, a has at most two elements
So a = b = {0, - 4}
So 0 and - 4 are two roots of the equation x & # 178; + 2 (a + 1) x + A & # 178; - 1 = 0
Weida theorem: 0-4 = - 2 (a + 1)
0×(-4)=a²-1
The solution is a = 1
Answer: a = 1

5x (7 + x) = 52.5

5*（7+x）=52.5
35+5x=52.5
5x=52.5-35
5x=17.5
x=17.5÷5
x=3.5

In a parallelogram, if the length of one side is 6.5 and the diagonal length is 5 and 12, the area is ()
A. 23.5B. 39C. 60D. 30

As shown in the figure, ▱ ABCD's diagonal AC and BD intersect at O, AC = 5, BD = 12, BC = 6.5. ∵ quadrilateral ABCD is a parallelogram, ▱ ob = 12bd = 6, OC = 12ac = 2.5. In △ BOC, ∵ ob2 + oc2 = 36 + 6.25 = 42.25 = BC2, ▱ BOC = 90 °, ▱ ABCD is a diamond, ∵ its area is: 12 × AC ×

12 + X / (5 + x) = 12 / 7

12+x/(5+x)=12/7
12*7(5+x)+7x=12(5+x)
72*(5+x)+7x=0
360+79x=0
x=-360/79
The results showed that x = - 360 / 79
Is the root of the original equation

How to find the normal vector of the high number plane is as follows The normal vector of the plane ax + by + CZ + D = 0 should be n = (a, B, c), but I don't know why it is this vector, how it comes from, and what should I pay attention to?