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Given three points A (-2,0,2), B (-1,1,2), C (-3,0,4), let a=AB, b=AC (1), find the angle between vector a and vector b.(2) If... A (-2,0,2), B (-1,1,2), C (-3,0,4) Let a=AB, b=AC (1) Find the angle between vector a and vector b.(2) If vector ka+b and vector ka-2b are perpendicular to each other, find the value of k.
0
Let A (-2,0,2), B (-1,1,0), C (-3,0,4)-3,0,4). Let a=AB, b=AC, if λ(a+b)(a-b) is perpendicular to the z-axis. Find the relation between λ and μ
By the title
A =(1,1,2), b =(-1,0,2)
A+b=(0,1,0), a-b=(2,1,4)
Let the vector on the z-axis be (0,0,1)
If λ(a+b)(a-b) is perpendicular to the z-axis
Then [λ(a+b)(a-b)]*(0,0,1)=0
I.e.-4μ=0
So μ=0,λ is any value.
Answers are for reference only
How to prove the vertical formula of vector?
Let β1=(x1, y1).β2=(x2, y2).
The angle from x axis to β1 is α1, the angle from x axis to β2 is α2,
Then: sinα1= y1/√(x12+y12), cosα1= x1/√(x12+y12),
Sinα2= y2/√(x22+y22), cosα2= x2/√(x22+y22),
X1x2+y1y2=0(x1x2+y1y2)/[√(x12+y12)√(x22+y22)]=0
Cosα1 cosα2+sinα1 sinα2=0 cos (α1-α2)=0α1-α2=/2
↔β1⊥β2.
In triangle ABC, A (7,8).B (3,5), c (4,3), M, N, D are AB, AC, ADj and MN at the midpoint of BC are intersected by F for vector DF
In the triangle ABC, A (7,8), B (3,5), c (4,3), M, N, D are the midpoint of AB, AC, BC respectively. MN and AD intersect at F to find the vector DF.
M (5,6.5), N (5.5,5.5), D (3.5,4), F (5.25,6) are known from the problem, so the vector DF=(1.75,2).
In the triangle ABC, A (7,8), B (3,5), c (4,3), M, N, D are the midpoint of AB, AC, BC, respectively. MN and AD intersect at F to find the vector DF.
M (5,6.5), N (5.5,5.5), D (3.5,4), F (5.25,6) are known from the problem, so the vector DF=(1.75,2).
In triangle ABC, A (7,8), B (3,5), c (4,3), M, N, D are the middle points of AB, AC, BC. MN and AD intersect at F to find the vector DF.
M (5,6.5), N (5.5,5.5), D (3.5,4), F (5.25,6) are known from the problem, so the vector DF=(1.75,2).
Given the vector AB=a+5b, BC=-2a+8b, CD=3(a-b), A, B, D are collinear, it is proved that CA=xCB+yCD (where x+y=1) AB and a, b, BC, CD are vectors,
CA=-(AB+BC)=a-13b, CB=-BC=2a-8b, CD=3a-3b
CA=xCB+yCD
That is, a-13b=x (2a-8b)+y (3a-3b)
So 1=2x+3y-13=-8x-3y
X=2 y=-1
So x+y=1
Vector a=(1,2), vector b=(-1,2), then the number product of vector a and vector b is ()
Vector p=(x1, y1), vector q=(x2, y2)
Then vector p, the number product of vector q is r=x1*x2+y1*y2
A*b=-1*1+2*2=3
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