Vector a, b, c has the same starting point,|a|=|b|=2, ab=-1,(a-c, b-c)=60 degrees,|a-c|=Root 7, then the included angle of vector a, c is The vector a, b, c has the same starting point,|a|=|b|=2, ab=-1,(a-c, b-c)=60 degrees,|a-c|=Root 7, then the included angle of vector a, c is The vector a, b, c the same starting point,|a|=|b|=2, ab=-1,(a-c, b-c)=60 degrees,|a-c|=Root 7, then the included angle of vector a, c is

Vector a, b, c has the same starting point,|a|=|b|=2, ab=-1,(a-c, b-c)=60 degrees,|a-c|=Root 7, then the included angle of vector a, c is The vector a, b, c has the same starting point,|a|=|b|=2, ab=-1,(a-c, b-c)=60 degrees,|a-c|=Root 7, then the included angle of vector a, c is The vector a, b, c the same starting point,|a|=|b|=2, ab=-1,(a-c, b-c)=60 degrees,|a-c|=Root 7, then the included angle of vector a, c is

60 Degrees

Let |a|=|b|=1 and |3a-2b|=Root number 7, then |3a+b| is

|3A-2b|^2=9||a|^2-12ab+4|^2=7, ab=|a||b||cos=1/2
Cos=a*b/|a||b|=1/2
I.e.=60°
Similarly |3a+b|=sqrt (9|a|^2+6ab|^2) sqrt represents the square root
The calculation result is sqrt13.

Given vector a+b+c=0, and the module of vector a is 3, the module of vector b is 5, and the module of vector c is 7, (1) Angle between vectors a and b (2) Is there a real number k such that ka+b is perpendicular to a-2b?

According to cosine theorem, cos (180°-θ)=(3^2+5^2-7^)/(2*3*5)=-1/2θ=60°, if there is a real number k, then there is (ka+b).(a-2b)=0ka^2-2kab+.

If the module of a vector is 5, the module of b vector is 3, and the module of (a vector-b vector) is 7, then what is the module of a vector*b vector =?

Square of module of (a vector-b vector)= square of module of a vector + square of module of b vector-2a vector*b vector
2A vector * b vector = square of module of a vector + square of module of b vector -(a vector - b vector) square of module =5^2+3^2-7^2=34-49=-15

Square of module of (a vector-b vector)= square of module of a vector + square of module of b vector-2a vector *b vector
2A vector * b vector = square of module of a vector + square of module of b vector -(a vector - b vector) square of module =5^2+3^2-7^2=34-49=-15

AB C is known as three internal angles of triangle ABC, vector a=(sinB+cosB, cosC) vector b=(sinC, sinB-cosB) If a·b=-1/5, calculate tan2A (please specify the value of cos2A)

A·b=(sinB+cosB)(sinC)+(cosC)(sinB-cosB)
= SinB sinB + cosB sinC + cosC sinB - cosC cosB
=-CosCcosB+sinBsinB+cosBsinC+cosCsinB
=-Cos (B+C)+ sin (B+C)=-1/5
Also [sin (B+C)]^2+[ cos (B+c)]^2=1
Solution gives sin (B+C)=3/5 cos (B+C)=4/5(where a set of solutions is rounded off by sin (B+C)>0 because B, C is the inner angle of the triangle)
SinA = sin (180-A)= sin (B+C)=3/5
CosA=-cos (180-A)=-cos (B+C)=-4/5
TanA=sinA/cosA=-3/4
Tan 2A =2tanA /[1-(tanA)^2]=2*(-3/4)/(1-9/16)=-24/7
Cos2A =2(cosA)^2-1=2(-4/5)^2-1=7/25

Given that the point G is the center of gravity of the triangle ABC passing through G as a straight line intersects both sides of ABAC and MN respectively, and the vector AM=xABAN=yAC, then xy/x+y=? AG=mAM+nAN, m+n=1 under collinear condition AG=(1/3) AB +(1/3) AC AM=xAB, AN=yAC So mx =1/3, ny =1/3 M =1/(3x), n =1/(3y) Then 1/(3x)+1/(3y)=1 3=(X+y)/xy Xy/(x+y)=1/3 Q: Why m+n=1, and why 1/3 AM+1/3 AN=AG?

First, answer the first question: This is a vector collinear basic problem: If the vector satisfies OA=mOB+nOC relation (where m, n are non-zero real numbers), and A, B, C point collinear, there must be m+n=1; Conversely, if the vector satisfies OA=mOB+nOC relation (where m, n are non-zero real numbers), and m+n=1, there must be...

First, answer the first question: This is a basic problem of vector collinearity. If the vector satisfies the relation OA=mOB+nOC (where m, n are nonzero real numbers), and A, B, C are collinear, then there must be m+n=1. Conversely, if the vector satisfies the relation OA=mOB+nOC (where m, n are nonzero real numbers), then there must be m+n=1.