Let vector a and B be two vectors. When vector a and B satisfy what conditions, vector a + vector b = vector 0

Let vector a and B be two vectors. When vector a and B satisfy what conditions, vector a + vector b = vector 0

Equal in size, opposite in direction

Let the vector | a | ·| B | ≠ 0, then the vector A / / B is the integral of the vector a = B______ Conditions

Answer: necessary
A / / B may not have a = B, for example, a = 1, B = 2
A = B must have a / / b. for any nonzero vector a, there must be a / / A

Why is the product of two vectors a and B perpendicular to each other zero
Isn't it (π / 2)
Isn't it (π / 2)? Why is the angle 90 degrees? Is π 180 degrees?

Yeah
The scalar product of vector AB is the angle of abcos
Because it's vertical, the angle is 90 degrees, and the cos value of 90 degrees is 0
So the scalar product of the vertical vector is zero
PI is 180 degrees
So π / 2 is 90 degrees
PI is 180 degrees
Two PI is 360 degrees

If vector | a | = 1, | B | = 2, vector C = a + B, and C is perpendicular to a, then the angle between vector a and B is ()

Given that | a | = 1, | B | = 2, vector C = a + B, vector C ⊥ a, then:
Vector C · a = (a + b) · a = 0
That is: |a| & #178; + B · a = 0
So: B · a = - |a | & #178; = - 1
That is: | B | * | a | * cos = - 1
2*1*cos=-1
So: cos = - 1 / 2
So: = 120 degree
The angle between vector a and B is 120 degrees

(1) Solve the equation about X: log5 (x + 1) - Log & nbsp; 15 (x-3) = 1 (2) the equation about X (34) x = 3A + 25-A has negative roots, find the value range of A

(1) The original equation is transformed into & nbsp; log5 (x + 1) + log5 (x-3) = log55, thus (x + 1) (x-3) = 5, and the solution is & nbsp; X = - 2 or x = 4 ∵ the original equation must satisfy the condition of X + 1 ﹥ x-3 ﹥ 0 ﹥ x = - 2, so the solution of the equation is x = 4 (2). If the negative root of the equation is x0 (x0 ﹤ 0), then (34) x0 = 3A + 25-A ﹥ 1 ﹥ a satisfies the condition 3A + 25-A ﹥ 03A + 25-A ﹥ 1, and the solution is 34 ﹤ a ﹤ 5 & nbsp; So the value range of real number a is (34,5)

The equation about X is known: the square of X - 2x = (3a-2) / (5-a) has real root in (1 / 2,2), and the range of a is obtained

x^2-2x=(3a-2)/(5-a)
（x-1）^2-1-(3a-2)/(5-a)=0
When x = 1, there is a minimum value of - 1 - (3a-2) / (5-a)
There are real roots in (1 / 2,2)
So - 1 - (3a-2) / (5-a) ≤ 0
-3/2≤a＜5
When x = 2, y ≥ 0
That is, (3a-2) / (5-a) ≥ 0
2/3≤a

A is the root of the equation x-square-x-1 = 0. Find the direct solution of the fourth power-3a-2 of A

0
X-1 = 0
X square = x + 1
The fourth power of x = x square + 1 + 2x,
Because a is the root of the equation, we substitute "the 4th power of x = x square + 1 + 2x" into "the 4th power of a-3a-2" to get
X squared + 1 + 2x-3x-2
Xsquare - X-1 = 0

The solution set of the equation solution 2x-3y + 7 = 0

This is a set of points, represented by a set of points

If the equation 2x ^ (- 2m + 3) + 3Y ^ (5n-7) = 4 is a quadratic equation with respect to X and y, try to find the value of m ^ 2 + 3N

For the binary linear equations of X and y, then the degree of X and Y is 1
So - 2m + 3 = 1
5n-7=1
m=1,n=8/5
So M & sup2; + 3N = 1 + 24 / 5 = 29 / 5

If the solution of equation 2aX + 6 = 3 (x + 1) is the same as that of equation 2x + 5 = 10x + 13, find the value of A

2x+5=10x+13
10x-2x=5-13
8x=-8
x=-1
Substituting 2aX + 6 = 3 (x + 1)
We get - 2A + 6 = 0
a=3