The vectors a, b, satisfying modulus a modulo b 1, and modulo a kb 3 modulo ka b, where k 0

a-kb|=√3|ka+b|

then (a-kb) 2=3(ka+b) 2

Because a 2=|a|^2=1,b^2=|b|2=1 so: 1+k 2-2ka*b=3(k 2+1+2ka*b)a*b=-(k 2+1) 4k -2k 4k=-1 2 take the equal sign if and only if k=1.

That is, when k=1, a*b obtains the maximum value of -1 2.

In this case: a*b=|a||b|cos = cos = -1 2, =120 degrees a+ b|^2=1+λ^2+2λa*b=λ^2-λ+1=(λ-1/2)^2+3/4

So when =1 2, |a+λb|The minimum value is 3 2

The geometric meaning is that the end point of vector A is connected to the point on the line where vector B is located, and the distance from the end point of a to the line is the shortest.

a*b=-(k 2+1) 4k =-1 4*(k+1 k) -1 4*2 (if and only if k=1 k, i.e., k=1 is equal sign).

modulo a+ b= ( 2+2 a*b+1)= ( 2- +1), when =1 2 the value of modulo a+ b is the smallest.

Summary. Hello, happy to answer your <>

It is known that the vector a, b satisfies that the a modulo is equal to 2, the b modulus is equal to 1, the angle between a and b is 60°, c = 2a + 3b, d = a + kb, when the real number k is what is the value. (1) c d (2) d module is equal to 2 times the root number 13: First, we can get the following relationship according to the modulus and angle formulas of the vector:

a| =2|b|=1 Angle =60°Next, let's solve the requirements of the problem one by one. Requirement 1: C d (vertical) Vectors c are perpendicular to vector d meaning that their inner product is zero.

We can calculate the inner product of vector c and vector d: c · d = 2a + 3b) ·a + kb) = 2a · a + 2a · kb + 3b · a + 3b · kb= 2|a|^2 + 2k + 3)|a||b|cosθ +3|b|2 substitutions for known values: c · d = 4 + 2k + 3) (2) (1) ( 3 = 4 + 2k + 3) +3 = 2k + 10 In order for c d to hold, we need 2k + 10 = 0.

Solving this equation gives k = 5. So, when k = 5, c d holds. Requirement 2:

d|=2 The modulo of the vector d of 13 is equal to 2 times the root number 13 can be expressed as: |d| =a + kb| =a|)^2 + k|b|)^2 + 2k|a||b|cos] substitution of known values: [2) 2 + k(1)) 2 + 2k(2)(1)( 2 13 Simplification:

4 + k 2 + 2k] = 2 13 squared sides: 4 + k 2 + 2k = 52 shifts and collated into the standard form of a quadratic equation: k 2 + 2k - 48 = 0 We can factor it or use the root finding formula to solve this quadratic equation and get k = 6 or k = 8.

So, when k is equal to 6 or -8, |d|=2 13 holds. When the real number k is equal to or -8, the requirement (1) and the requirement (2) are satisfied.

2) Modulo d is equal to 2 times the root number 13

It is known that the vectors a, b satisfy the equivalence of a modulus and so on in 2, b modulo which Sun argues is equal to 1, the angle between a and b is 60°, c = 2a + 3b, d = a + kb, when the real number k is the value of the keygen.

1) c d known vector a, b satisfies a modulo and so on is missing 2, b modulo which Sun argues is equal to 1, the angle between a and b is 60°, c = 2a + 3b, d = a + kb, when the real number k is the value of the key.

2) Modulo d is equal to 2 times the root number 13

1) c d known vector a, b satisfies a modulo and so on is missing 2, b modulo which Sun argues is equal to 1, the angle between a and b is 60°, c = 2a + 3b, d = a + kb, when the real number k is the value of the key.

Summary. B square is equal to 9 times, A square A is equal to 25, so B square is equal to 9*25 = 2258Knowing that the modulo of vector a is 5, and b=-3a then the modulo of b is.

b square is equal to 9 times, a square a is equal to 25, so b square is equal to 9*25 = 225, b is equal to 5 times, root number 5

There are two kinds of modulus in mathematics: 1. The modulo of complex numbers in mathematics. The value of the square root of the square of the sum of the real and imaginary parts of the complex number is called the modulo of the complex number.

The problem is that we can square the square first, and then we square it and then calculate this mold will lose a lot of pants and bright tan, because the key Qi mold Hutong style has a size and direction, that is, there is a plus or minus sign, and the plus and minus signs are gone after we are squared.

Because of the input, I use uppercase letters for the lower volume and lowercase letters for the angles!

ka+b=(kcosa+cos, ksina+sin) a-kb = (cosa-kcos, sina-ksin) so (kcosa+cos) ksina +sin ) = 3(cosa-kcos) 3(sina-ksin )

The simplification is: cosacos +sinasin = (k +1) 4k, so f(k)=(k +1) 4k = k 4 + 1 4k(2) is made by (1).

f(k) minimum = (mean inequality).

If and only if k 4 = 1 4k i.e. k = 1, then the product of vector a and vector b =

The product of the modulus = 1

So the cosine value of the included angle = =

So the included angle = 60°

a*b=-(k^2+1)/4k

1 4*(k+1 k) -1 4*2 (if and only if k=1 k, i.e., k=1 is equal sign).

modulo a+ b= ( 2+2 a*b+1)= ( 2- +1), when =1 2 the value of modulo a+ b is the smallest.

Modulo of ka+b = 3

Square on both sides. k^2+1+2kcosα=3

i.e. k+2cos = 2 k

Find the minimum value of vector a point multiplied by vector b, i.e., find the minimum cos by the above equation, cos = 1 k-k 2

The derivative on the right side is not difficult (note that k is greater than zero, and treat k as a variable.)

|a|=|b|=1

ka+b|=|3a-kb|

Square on both sides: k |a|²+2ka●b+|b|²=3|a|²-2√3ka●b+k²|b|or collapsed.

k²+2ka●b+1=3-2√3ka●b+k²2(1+√3)ka●b=1

k>0a●b=1/[1(√3+1)k]

a●b=(√3-1)/(2k)

If a b, =0 or shout = 180

a●b=|a||b|cos=±1

3-1)/(2k)=±1

k = ( 3-1) 2 or k = (1-3) 2