What is the sum of unit vectors? How much is the sum of two unit vectors? 1 or 0 or no specific number? How much is the multiplication of two unit vectors? 1 or 0 or no specific number?

What is the sum of unit vectors? How much is the sum of two unit vectors? 1 or 0 or no specific number? How much is the multiplication of two unit vectors? 1 or 0 or no specific number?

The sum of unit vectors is still a vector, not a number
The scalar product of two unit vectors is a number, which depends on the angle between the two vectors

Why is the unit vector x plus y equal to one

You are talking about the two-dimensional unit vector. The definition of the two-dimensional unit vector is the vector formed by taking the origin as the center of the circle, taking a point on the circle with radius of 1, connecting the center of the circle. Similarly, in the case of three-dimensional space, it is to take a unit ball. This is the definition. So the above x ^ 2 + y ^ 2 = 1

Because a and B are unit vectors perpendicular to each other, why is a equal to 1 and B equal to 1?

The unit vector is that the length of the vector is one, that is, the length of the module is 1, that is, square a equals 1

It is known that the length of the major axis of an ellipse is twice that of the minor axis, and the right quasilinear equation is that x equals 4 / 3 times root 3
If I can't work out the equation for an ellipse, I'll have to deal with it

`a=2b
c²=a²-b²=4b²-b²=3b²,∴c=√3b
The right quasilinear equation is as follows: A & sup2 / / C = 4B & sup2; / √ 3B = 4 √ 3 / 3, ｜ B = 1
∴a=2,
The equation of ellipse is X & sup2 / 4 + Y & sup2; = 1

Let L: y be equal to x plus 1 and the square of x plus B of ellipse a, and the square of Y is equal to 1, which is called the intersection of L and X axis at two different points a and B
(1) It is proved that the square of a plus the square of B is greater than 1 (2). If f is a focal point of the ellipse and the phasor AF is equal to 2fb, the elliptic equation is solved

If 2 / 2 of 1-A + 1 / 2 of a = 0 (a ≠ 0), then the - 1 power of a is equal to () a - 1 B 1 C 2 D - 1 or 2

1-2/a+1/a²=0
Using the complete square formula
(1-1/a)²=0
1-1/a=0
a=1
B!
If you still have doubts, please ask
I wish: progress in learning!

What is the square of (a + B + C + D),

a2+b2+c2+d2+2ab+2ac+2ad+2bc+2bd+2cd

If the square of a plus the square of B equals 1, and the square of C plus the square of D equals 1, find the range of AC + BD

Given that: A, B, C, D ∈ R, prove: (AC + BD) 2 ≤ (A2 + B2) (C2 + D2)

It is proved that solution 1 (analytical method) needs to prove (AC + BD) 2 ≤ (A2 + B2) (C2 + D2), (2 points) that is, a2c2 + b2d2 + 2abcd ≤ a2c2 + a2d2 + b2c2 + b2d2, (4 points) that is, 2abcd ≤ a2d2 + b2c2, (6 points) that is, 0 ≤ a2d2 + b2c2-2abcd = (AD + BC) 2, (8 points)

It is known that the square of a plus the square of B is equal to 1, the square of C plus the square of D is equal to 1, AC + BD = 0

a²+b²=1 c²+d²=1
Let a = sin α, B = cos α, C = sin β, d = cos β
therefore
From AC + BD = sin α sin β + cos α cos β = cos (α - β) = 0
AB + CD = sin α cos α + sin β cod β = 1 / 2 (sin2 α + sin2 β) = 1 / 2 [2Sin [α + β] cos [α - β)] = 0