The angle between vectors a and B is 45 degrees, and the modulus of a is 1, | 2a-b | = √ 10, | B | =?

The angle between vectors a and B is 45 degrees, and the modulus of a is 1, | 2a-b | = √ 10, | B | =?

It is known that vector a = 1, vector b = 2, the angle between vectors a and B is 120 °, vector C = 2A + 3b, vector D = pa-5b, and C is perpendicular to D to find the real number P

Vector c * d = (2a + 3b) (pa-5b) = 2p|a| & #178; + (3p-10) a * b-15|b| & #178;
∵|a|=1 |b|=2 (a,b)=120°
∴a*b=|a||b|cos120°=-1
∵c⊥d
∴c*d=0
∴2p-(3p-10)-60=0
∴p=-50

Vector a = 3, vector b = 2, the angle between vector a and vector B is 90 degrees (1) if ka-2b is perpendicular to 4A + 3B
Finding the value of K
(2) If the vector ka-2b and 4A + 3
The angle between B and K is an acute angle

|a|=3,|b|=2,ab=0
one
(ka-2b)(4a+3b)
=4ka^2-6b^2
=36k-24
=0
Therefore, k = 24 / 36 = 2 / 3
two
(ka-2b)(4a+3b)
=4ka^2-6b^2
=36k-24
>0
Therefore, k > 2 / 3
If you don't understand, please ask

Given that the angle between vectors a and B is 60, | a | = 2, | B | = 1, and (KA + b) ⊥ (2a-b), then the real number k =

Because (KA + b) ⊥ (2a-b),
therefore
(ka＋b)(2a－b) = 0
2k|a|² + （2-k）|a||b|cost -|b|² = 0
8k + (2-k)2cos60 -1 = 0
7k = -1
k = -(1/7)

Vector b = (- 3,4), vector C = (- 1,1) and its relation with vector a is a = b-2c
1 to find the coordinates of vector a
2 find (vector a + vector C) times (vector a-vector C)

Because C = (- 1,1)
So - 2C = (2, - 2)
So a = B – 2C = (- 1,2)
(a + c)(a - c) =｜a｜^2*｜c｜^2 = 5 – 2 = 3

a. B. C is three vectors, A-B + 2C = 0 vector. The angle between a and C is 60 degrees, | a | = 2, | C | = 1, find | B | =?

From A-B + 2C = 0: B = a + 2c, so | B | = | a + 2C | = √ (a + 2C) ^ 2 = √ a ^ 2 + 4ac + 4C ^ 2 = √ (4 + 4 * 2 * 1 * cos60 + 4 * 1)
=√12=2√3

a-1/2（a+b）-1/2（c+a-b）=____ (vector computing)

a-1/2（a+b）-1/2（c+a-b）
= 2a-a-b-c-a+b )/2
= -c/2

Let a = (3, - 2) B = (- 1,2), then the cosine of the angle between A-B and B is

Well, there is a formula. The product of vector A-B and vector b divided by the absolute value of vector A-B and the absolute value of vector B is the cosine value of the angle between vector A-B and vector B. therefore, in this problem, vector A-B = (4, - 4), so the product of vector A-B and vector b = - 12, so the absolute value of vector A-B = 4, root 2, the absolute value of vector B

Why is the value of the second-order determinant the area of the parallelogram composed of the two vectors?
How to prove it?

First of all, we prove that the first vector of the formula cosx = a. * B / |||||||||||||||||124;|||||\\124\124\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\it's level 2

Prove by definition that the module of cross product of vector a and B is equal to the area of parallelogram composed of a and B (the content of tensor)

Let the parallelogram composed of a and B be ABCD