Vector problem Let o be a regular pentagon ABCDE Vector AB + CB + CD + ed + EA = 2 [DB + CO + od + ed]

Vector problem Let o be a regular pentagon ABCDE Vector AB + CB + CD + ed + EA = 2 [DB + CO + od + ed]

The left side of the equal sign = (AB + EA) + CB + CD + ed = EB + CB + CD + ed, the right side of the equal sign = 2 [DB + CO + od + ed] = 2 [CD + EB], that is, CB + ed = CD + EB, that is, cb-eb = cd-ed, that is, CE = CE
It is proved that this is analytical method

1) We know that | → a | = 1, | → B | = 2, and the angle β between | → A and | → B is 60 degrees
(1) Find → a * → B, (→ A-2 → b) & # 178;, | → a + → 3B|
(2) It is proved that: → a - → B is perpendicular to → a
2) Known → AB = → a = (1,2), → BC = → B = (- 3,2), → CD - (6,4)
(1) prove that three points of abd are collinear
(2) What is the value of K when (1) wolf like k → a + → B is parallel to → a - → 3B; (2) vector → Ka + → B is perpendicular to → A-3 → B
3) It is known that → a, → B, → C are three vectors in the same plane, temperament → a = (1,2)
(1) | → C | = 2 √ 5, and → C / / → a, find the coordinates of → C
(2) If | → B | = √ 5 / 2, and → a + → 2b is perpendicular to 2 → a - → B, find the angle θ between → A and → B

1) Known as the title set
(1) Vector A. vector b = | vector a | vector B | * cos
a.b=|a||b|cos60°.
=1*2*(1/2).
∴a.b=1.
(a-2b)^2=a^2-4ab+b^2.
=|a|^2-4|a||b|cos+4|b|^2.
=1-4*1+16.
=13.
|a+3b|^2=(a+3b)^2.
=a^2+6ab+9b^2.
=1+6*1+9*4.
=43.
∴|a+3b|=√43.
(2) Prove: ab ⊥ a (is the original question like this? If so, the question is wrong, A.B (the scalar product of L two vectors is scalar, scalar and vector are not perpendicular to each other!). Please correct it!
2) The known vectors AB = (1,2), BC = (- 3,2), CD = (6,4)
(1) Prove that a, B and D are collinear
To prove that a, B and D are collinear, we only need to prove that AD and ab are collinear
∵AD=AB+BC+CD.
=(1-3+6,2+2+4).
=(4.8).
AD=4(1,2).
AD=4AB.
The vector AD and the vector AB are collinear,
A, B, D are collinear
(2) ka+b=k(1,2)+(-3,2).
=(k-3,2k+2).
a-3b=((1,2)-3(-3,2)).
=(1+9,2-6).
=(10,-4).
∵(ka+b)∥(a-3b)
∴ (k-3)*(-4)-(2k+2)*10=0.
-4k+12-20k-20=0.
-24k=8.
That is to say, k = - 1 / 3
3) Known as the title set
(1) Let the coordinate of direction C be C (x, y). [a = (1,2)]
|c|=√(x^2+y^2)=2√5.
x^2+y^2=20.
∵c∥a,∴ 2x-y=0.
y=2x.
x^2+(2x)^2=20.
5x^2=20.
x^2=4.
x=±2,
y=±4.
The coordinates of vector C are: vector C = (2,4) or C '= (- 2, - 4)
(2) }b}=√5/2,
∵(a+2b)⊥(2a-b).
∴(a+2b).(2a-b)=0.
2a^2+3ab-2b^2=0.
2*5+3|a||b|cos-2*5/4=0.
10+3√5*√5/2cos-5/2=0.
cos=(5/2-10)/(15/2).
=(-15/2)/(15/2).
cos=-1.
∴=180°
In other words, under the condition of question setting, vector a and vector B are inversely collinear