(a + 3b) ⊥ (7a-5b), (a-4b) ⊥ (7a-2b) a and B are vectors to find the angle between a and B

(a + 3b) ⊥ (7a-5b), (a-4b) ⊥ (7a-2b) a and B are vectors to find the angle between a and B

(a + 3b) ⊥ (7a-5b), (a-4b) ⊥ (7a-2b) means (a + 3b) * (7a-5b) = 0, (a-4b) * (7a-2b) = 0. The solution of the simultaneous equations is 2A = B, and 1 / 2 is obtained by applying the cosine angle formula, so the included angle is 60 degrees

For non-zero vector a, B satisfies a + 3b ⊥ 7a-5b, a-4b ⊥ 7a-2b, then the included angle of vectors a and B is () For non-zero vector a, B satisfies a + 3b ⊥ 7a-5b, a-4b ⊥ 7a-2b, then the included angle of vectors a and B is () There should be a process, thank you

(a+3b)·(7a-5b)=0
7a·a +16a·b -15 b·b =0
(a-4b)·(7a-2b)=0
7a·a -30 a·b +8 b·b =0
a·b =0.5 b·b =0.5 a·a
Cosine of angle between vectors a and B = 0.5
The included angle of vectors a and B is 60 °
There is no checking calculation. Check it yourself

Given the vector a = (COSA, Sina) B = (root 3,1), find the maximum value of a vector + B vector

Cosa = root 3 / 2 Sina = 1 / 2 in the same direction
Max. 3

Let vectors a and B satisfy: a = 1, a * b = 3 / 2, a + B = 2, root sign 2, then B = how many calculation processes!

Because a + B = 2 root sign 2
(a+b)^2=8
b^2=8-a^2-2ab=8-1-2*3/2=4
So B = 2

If the plane vector B is parallel to the vector a = (2,1), and B = 2, radical 5, then B= |B| = root sign (x ^ 2 + y ^ 2) = 2 root signs 5 x^2+y^2=20 Because B / / A, B = (x, y) - "B = (2x, x) So 4x ^ 2 + x ^ 2 = 20 5x^2=20 x^2=4 X = plus or minus 2 So vector b = (2,1) or (- 2, - 1) But the correct answer seems to be (4,2) (- 4, - 2) Want to know why my solution is wrong

b=(2x,x)
And x = 2 or - 2
So B = (4,2) or (- 4, - 2)

Vector a = (1.1) a × B = 3, a + B = root 13. Then B =?

It should be a times B
Square a + B to get a ^ 2 + B ^ 2 + 2A * b = 13
Then a ^ 2 = 2 and a * b = 3 can get B ^ 2 = 5
So B = root 5