9. Given two points A (2,0,3), B (1,2,4), calculate the module, direction cosine and direction angle of vector AB.

9. Given two points A (2,0,3), B (1,2,4), calculate the module, direction cosine and direction angle of vector AB.

| AB |=√[(1-2)^2+(-√2-0)^2+(4-3)^2]=2. Let the angle between vector AB and X axis, Y axis, Z axis be α,β, and γ respectively. Then cosα/((1-2)= cosβ/(-√2-0)= cosγ/(4-3)=1/|AB |=1/2.cosα=-1/2,α=120°; cosβ=-√2/2,β=135°; cos γ=1/2.γ=60...

Given four points A, B, C, D in coordinate plane, and vector AB=(6,1), vector BC=(x, y), vector CD=(-2,-3) Find: if vector BC is parallel vector DA, find the relation between x and y At the same time, the vector AC is vertical BD, and the value of x, y is obtained.

1) AD=AB+BC+CD=(4+x,-2+y), so BC//DA, so x (-2+y)-y (4+x)=0, reduced to x+2y=0.2) AC=AB+BC=(6+x,1+y), BD=BC+CD=(-2+x,-3+y), because AC, BD, so,(6+x)(-2+x)+(1+y)(-3+y)=0, reduced to x^2+y^2+4x-2y-15=0,...

Vector a=(0,1,2), b=(1,0, negative 1), then product of quantities a times b=

Is it an inner product? The product of the vector is defined in many ways. If the Euler inner product is the multiplication of the corresponding elements, and then the sum =0*1+1*0+2*(-1)=-2

Is it an internal product? The product of the vector is defined in many ways. If the Euler inner product is the multiplication of the corresponding elements, and then the sum =0*1+1*0+2*(-1)=-2

Is it the inner product? The product of the vector is defined in many ways. If the Euler inner product is the multiplication of the corresponding elements, and then the sum =0*1+1*0+2*(-1)=-2

We the vector OA=(1,1), the vector OB=(3,-1), the vector OC=(a, b).(1) If ABC three points are collinear, find the relation of a, b. We the vector OA=(1,1), the vector OB=(3,-1), the vector OC=(a, b).(1) If ABC three points are collinear, find the relation between a, b.

Vector AB=OB-OA=(3,-1)-(1,1)=(2,-2)
Vector BC=OC-OB=(a, b)-(3,-1)=(a-3, b+1)
If A, B and C are collinear, then AB//BC gives:(a-3)/2=(b+1)/(-2)
A-3=-(b+1)
Get: a+b=2

Vector AB=OB-OA=(3,-1)-(1,1)=(2,-2)
Vector BC=OC-OB=(a, b)-(3,-1)=(a-3, b+1)
A, B, C are collinear, then AB//BC, get:(a-3)/2=(b+1)/(-2)
A-3=-(b+1)
Get: a+b=2

Given vector OB=λOA OC If ABC is collinear at three points, prove =1 The above data are all vectors Given the vector OB=λOA OC, if ABC is three-point collinear, prove =1 The above data are all vectors

ABC three-point collinear
Set vector AB=x vector BC
Vector AB = Vector OB - Vector OA
Vector BC = Vector OC - Vector OB
Vector OB-vector OA=x (vector OC-vector OB)
Vector OB + x Vector OB = Vector OA + x Vector OC
Vector OB=1/(1+x) Vector OA+x/(1+x) Vector OC
Vector OB=λOA OC
∴λ=1/(1+X)
μ= X/(1+x)
∴λ+μ
=(X+1)/(x+1)
=1
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Given vector OA, OB, OC and vector OC=λ vector OA +μ vector OB If given, prove ABC three-point collinear

If =1, then λ=1-μ
So OC =λOA +μOB
I.e. OC=(1-μ) OA OB
So OC-OA=μ(OB-OA)
I.e. AC=μAB
So AC∥AB,
So A, B and C are collinear;