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In parallelogram ABCD, AB vector=a vector, AD vector=b vector, AC vector=3NC vector, M is the midpoint of BC, then MN vector=
1/3A-1/6b
If you want process
1/3A-1/6b
Tell me the process.
Let A, B and C be nonzero vectors, where any two vectors are not collinear, but A+B and C are collinear, B+C and A are collinear, and A+B+C=0 is proved.
Because A+B is collinear with C, there is a real number x such that A+B=xC,
Similarly, because B+C is collinear with A, there is a real number y such that B+C=yA,
Change the second expression into: B=-C+yA, substitute the first expression to obtain: A-C+yA=xC,
If (1+y) A-(1+x) C=0, then both A and C are non-zero vectors and are not collinear,
Then from the fundamental theorem of plane vector or the necessary and sufficient condition of vector collinearity, we can get:1+y=0,1+x=0, so x=y=-1,
So A+B=xC=-C, so A+B+C=0.
If abc is a non-zero non-linear vector, why is (bc) a-(ca) b not perpendicular to c wrong? Such as a question If abc is a non-zero non-linear vector, why is (bc) a-(ca) b not perpendicular to c wrong? E.g.
(Bc) a-(ca) b multiplied by c
(Bc)(ac)-(ca)(bc)=0
So (bc) a-(ca) b is perpendicular to c
How to find the unit vector parallel to the vector AB by knowing the coordinates of A and B points M how to ask How to find the unit vector parallel to the vector AB by knowing the coordinates of A and B points M how to seek
Let AB (a, b) be the vector and let x, y be the unit vector, then (x, y)=m (a, b) and the sum of x, y is 1. Solve the equation
X=ma, y=mb, substitute m, and then (x, y)
The sum of the squares of ma, mb is 1, ab is known
Let AB (a, b) be the vector, and let (x, y) be the unit vector, then (x, y)=m (a, b), and the sum of x, y is 1. Solve the equation
X=ma, y=mb, substitute m, and then (x, y)
The sum of the squares of ma, mb is 1, ab is known
Let AB (a, b) be the vector, then let (x, y) be the unit vector, then (x, y)=m (a, b), and the sum of x, y is 1. Solve the equation
X=ma, y=mb, substitute m, and then (x, y)
The sum of the squares of ma, mb is 1, ab is known
Let a.b.c be three vectors in the same plane, where a =(-1,2). Let b be the unit vector and B be parallel to a, and find the coordinates of b Let a.b.c be three vectors in the same plane, where a =(-1,2). Let b be the unit vector and B be parallel to a, and find the coordinates of b.
B//a
Set b=(-x,2x)
B is the unit vector
X^2+4x^2=1
B =(-root number 5/5,2root number 5/5) or (root number 5/5,-2root number 5/5)
B//a
Set b =(-x,2x)
B is the unit vector
X^2+4x^2=1
B =(- root number 5/5,2 root number 5/5) or (root number 5/5,-2 root number 5/5)
B//a
Set b=(-x,2x)
B is the unit vector
X^2+4x^2=1
B =(- root number 5/5,2 root number 5/5) or (root number 5/5,-2 root number 5/5)
A vector a=(1,1), b=(1,0), c satisfies a·c=0, and |a|=|c|, b·c >0. (1) Find vector c (2) If the mapping f:(x, y)→(x `, y `)=xa+yb, find the primitive image of the points (1,2) under the mapping f. Above abc are vectors! A vector a=(1,1), b=(1,0), c satisfies a·c=0, and |a|=|c|, b·c >0. (1) Find vector c (2) If the mapping f:(x, y)→(x `, y `)=xa+yb, find the prime image of the points (1,2) under the mapping f. Above abc are vectors!
(1)|A|=√2let c=(x, y)=> x^2+y^2=2(1) a.c=0(1,1)(x, y)=0=> x+y=0(2) b.c >0(1,0)(x, y)>0=> x >0 sub (2) into (1)2x^2=2x=1 or -1( rejected, x >0) y=-1c (1,-1)(2) f...
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