It is known that | AC vector | = 5, | AB vector | = 8, Ad vector = 5 / 11, DB vector, CD vector * AB vector = 0, and the angle BAC = a, cos (a + x) = - π The answer is specific numbers. Is - (3 + 4 radical 3) / 10 You can help me, my friend

It is known that | AC vector | = 5, | AB vector | = 8, Ad vector = 5 / 11, DB vector, CD vector * AB vector = 0, and the angle BAC = a, cos (a + x) = - π The answer is specific numbers. Is - (3 + 4 radical 3) / 10 You can help me, my friend

Taking D as the origin, AB as the x-axis, the vector CD * vector AB = 0, then CD is perpendicular to the AB vector ad = 5 / 11, the vector dB, | vector ab | = 8, then ad = 2.5, DB = 5.5, that is, a (- 2.5,0) B (5.5,0) is known from the | vector AC | = 5, AC = 5, then cos θ = 0.5, so θ = 60 degrees, cos (θ + x) = 4 / 5, that is cos (60 degrees + x) = 4 / 5, which is expanded to 0.5 * cos

A vector problem in Senior High School
Let the vectors a and B satisfy: | a | = 3, | B | = 4, a · B = 0. If a triangle is formed with the module of a, B, A-B as the side length, then the maximum number of common points between its side and the circle with radius 1 is?

It can be seen that the module of a, B, A-B is a triangle with side length of 3,4,5
And its inscribed circle radius r = (3 + 4-5) / 2 = 1
So the most common point is 4

It is known that the three vertices of a, B and C are (2,6), (6,5) (- 3, - 6) and the weights of 3kg, 2kg and 1kg are placed at a, B and C respectively, then the coordinates of the center of gravity of ABC are

If it is an ordinary triangle, the coordinates of the center of gravity are ((2 + 6-3) / 3, (6 + 5-6) / 3) = (5 / 3,5 / 3). In this problem, if a heavy object is placed at the vertex, the weighted center of gravity is obtained. The weighted average value should be used to express the weighted center of gravity, that is: ((3 * 2 + 2 * 6-1 * 3) / (1 + 2 + 3), (3 * 6 + 2 * 5-1 * 6) / (1 + 2 + 3)) =

Four vector problems in Senior High School
Given that the area of triangle ABC is 100, points D and E are the points on edge AB and BC respectively, and ad: DB = Ce: EB = 2:1, AE and CD intersect at point P, the area of triangle APC is calculated?

From the title: De / / AC, and de: AC = 1:2, from the similar triangle (triangle DPE and triangle CPA), we know that AP: PE = DP: PC = 1:2, so point P is the area of triangle ABC. Therefore, the area of triangle APC is 1 / 3 of the area of triangle ABC