|A|=1,|b|=2, and |a+b|=root number 7, find the included angle of vector ab

|A|=1,|b|=2, and |a+b|=root number 7, find the included angle of vector ab

|A+b|^2=|a|^2 b|^2+2ab=1+4+2ab=7,
Ab=1,
Cos=ab/|a||b|=1/2(a, b interangle

|A+b|^2=|a|^2 b|^2+2ab=1+4+2ab=7,
Ab=1,
Cos=ab/|a||b|=1/2(a, b angle

A vector + b vector + c vector =0 vector, module of vector a =3, module of vector b =5, module of vector c =7. Find the angle between vector a and b

A+b+c=0
A+b=-c
(A+b)^2=(-c)^2
A^2+b^2+2ab=c^2
3^2+5^2+2Ab=7^2
Ab =15/2
Cos=ab/(|a||b|)=(15/2)/(3*5)=1/2
So =60 degrees

Given a, b is a space vector,|a|=3,|b|=2,|a-b|=root 7, then the angle between a and b is

Cos=(|^a2+b |^2-||a-b |^2)/(2 a b|)
=(9+4-7)/(2*3*2)
=1/2
=60°
The angle between a and b is 60°

Cos=(|^a2+b|^2-||a-b|^2)/(2 a b||)
=(9+4-7)/(2*3*2)
=1/2
=60°
The angle between a and b is 60°

The two vectors are parallel. About the mold?

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A vector =(3, m), b vector =(2,3), and a vector is perpendicular to b vector, find a Ask for detailed process, can not do.

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Given points A (-2,3), B (3,5), find the coordinates of points A, B in relation to the center A', B' of point M (1,1), and show that vector A' B'= negative vector A Given points A (-2,3), B (3,5), find the coordinates of points A, B in relation to the center A', B' of point M (1,1), and prove that vector A' B'= negative vector A

Let A'=(x1, y1) vector AM=vector MA'(1+2,1-3)=(x1-1, y1-1)3=x1-1;-2=y1-1 solution give x1=4; y1=-1 Let B'=(x2, y2) vector BM=vector MB'(1-3,1-5)=(x2-1, y2-1)-2=x2-1;-4=y2-1 solution give x2=-1; y2=-3 vector A'B'=(-1-4,-3+1)=(-5,-2) vector AB=(3+2,5-2)...