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Let X be the normal vector of X plane through three points A (1,2,3) B (2,0,-1) C (3,-2,0) Given X be the normal vector of the X plane through three points A (1,2,3) B (2,0,-1) C (3,-2,0) Given the plane X through three points A (1,2,3) B (2,0,-1) C (3,-2,0) to find the normal vector of X-plane
AB=(1,-2,-4), AC=(2,-4,-3)
N=AB×AC=(-10,-5,0)
Unified n0=n/|n|=(2,1,0)/√5
How to find the normal vector for 3 points (0,0,0)(1.2.3)(4.5.6) in a given plane
There's already 3 points, and you can construct two vectors.
The normal vector is perpendicular to these two vectors, which can be expressed by cross multiplication of the two vectors.
And n=AB×AC. or n=AC×AB
Determine direction or unitization according to actual needs.
There's already three points, and you can construct two vectors.
The normal vector is perpendicular to these two vectors, which can be expressed by cross multiplication of the two vectors.
And n=AB×AC. or n=AC×AB
Determine direction or unitization according to actual needs.
Given that the coordinates of three points A, B and C are (1,-1,-1)(0,1,2)(0,6,6), respectively, the projection of vector OC in the normal vector direction of plane OAB Given that the coordinates of three points A, B and C are (1,-1,1-1)(0,1,2)(0,6,6), respectively, the projection of vector OC in the normal vector direction of the plane OAB
OA=(1,-1,1-1)
OB=(0,1,2)
Let OP=(m, n, p) and be perpendicular to the plane OAB.
Theorem: Perpendicular to the plane is any line perpendicular to the plane. That is, OP perpendicular to OA, OB
Therefore: OP.OA=0, OP.OB=0
I.e. m-n-p=0(1)
N+2p=0(2)
Take p as parameter, get
N=-2p
M=n+p=-2p+p=-p
I.e. OP=(-p,-2p, p)
P may be any value other than zero. For convenience, p=1
OP =(-1,-2,1) is the normal vector of the plane OAB.
OC=(0,6,6)
Its projection on the OP is:
L=|OC|* cos (OP, angle of OC)
=|OC OP|* cos (OP, OC angle)/|OP|(constant deformation)
That is: L=|OC|* cos (OP, angle of OC)
=|OC OP|* cos (OP, OC angle)/|OP|
=OC.OP/|OP|
=(-12+6)/Root 6
=-6/ Root 6=- Root 6
If the OP takes the opposite direction, the projection is negative,
I.e. L = root 6.
The question does not specify which direction to take, so both
Why is the normal vector of a plane equal to the product of two non-parallel vectors? Can you explain it more clearly?
1. The normal vector of the plane is perpendicular to the plane 2. The vector product of parallel vectors is equal to zero 3. The vector product of two non-parallel vectors in the plane is the normal vector perpendicular to the plane (right hand rule)
The vectors a, b, c are known to be three vectors in the same plane, where a =(1,2), if the module of vector b =2nd root number 5, and vectors a+2 vector b and 2 vector a One of the answers is this (a+2b)(2a-b)=2|a|2-2|b|2-ab I can't figure it out. I just figure it out. Finally, my value cosa=10/10=1... something's wrong. The answer is equal to -1 with an angle of 180° Why?
Topic incomplete
Given plane vector A =(root 3,-1), vector B =(1/2, root 3/2) If there is a real number KT which is different from 0, so that vector X=A+(T-3) B, vector Y=-KA+TB, and vector X⊥ vector Y, find the function relation K=F (T)
A=(√3,-1), b=(1/2.√3/2), x=a+(t^2-3) b, y=-ka+tb, x⊥y, then the vector x•y=0,(a+bt^2-3b)•(-ka+tb)=0,-ka^2-kabt^2+3abk+tab+t^3b^2-3tb^2=0, where a^2=3+1=4, b^2=1, a•b=-√3/2+√3/2=0,(a+bt^2-3b)•...
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