How to find the normal vector of a plane made of triangles?

If the triangle is ABC, AB vector and BC vector should be expressed in coordinates. Suppose AB is (x1, y1, z1) and BC is (x2, y2, z2). Then try to find the vector is (x, y1). Note that there are countless normal vectors. We can not determine the length, so let (x, y1). If you want to set (x, y, z), the final solution...

If the triangle is ABC, the AB vector and the BC vector should all be expressed in coordinates. Suppose AB is (x1, y1, z1) and BC is (x2, y2, z2). Then try to find the vector is (x, y1). Note that there are countless normal vectors and the length can not be determined. Let (x, y1). If you want to set (x, y, z), the final solution...

If this triangle is ABC, AB vector and BC vector should be expressed in coordinates. Suppose AB is (x1, y1, z1)1) and BC is (x2, y2, z2). Then try to find the vector is (x, y1). Note that there are countless normal vectors, and we can not determine the length. So let (x, y,1). If you want to set (x, y, z), the final solution...

Using Vector Method to Prove the Intersection of Three Centerlines of a Triangle

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Using Vector Method to Prove that Three Centerlines of a Triangle Intersect a Point RT Please use vector knowledge to solve Using Vector Method to Prove that Three Centerlines of a Triangle Intersect a Point RT uses vector knowledge to solve

Let the intersection point of two center lines be O, and let the vector of three sides of a triangle be the vector a, b, c, and the midpoint of three sides be D, E, F. If the two center lines are AD and BE, then the vector CO and OF can be expressed by a, b, c, and the vector CO and OF can be found to be parallel.

Let the intersection point of the two center lines be O, and let the vector of the three sides of the triangle be the vector a, b, c, and the midpoint of the three sides be D, E, F. If the two center lines are AD and BE, then the vector CO and OF can be expressed by a, b, c, and the vector CO and OF can be found to be parallel.

Let the intersection point of two center lines be O, and let the vector of three sides of a triangle be vector a, b, c, and the midpoint of three sides be D, E, F. If the two center lines are AD and BE, then the vector CO and OF can be expressed by a, b, c, and the vector CO and OF can be found to be parallel.

How to prove that the three center lines of three sides of a triangle intersect at vector method? How to prove that the three center lines of the three sides of a triangle intersect at a point by the vector method?

Let two neutral lines AD, BE intersect at one point G, and connect CG by triangle rule
CG=CA+AG=CA+2/3AD=.=1/3(CA+CB)
Take point F in AB, AF=1/2(CA+CB), so CG is parallel to AF (the above letters shall be added with arrow)

Let two neutral lines AD, BE intersect at one point G, and connect CG by triangle rule
CG = CA + AG = CA +2/3 AD =.=1/3(CA + CB)
Take point F in AB, AF=1/2(CA+CB), so CG is parallel to AF (the above letters shall be added with arrow)

The vectors a, b are known as two non-zero vectors and satisfy the vector absolute value a = vector absolute value b = vector absolute value (a-b) Find the angle between vector a and vector (a+b) The vectors a, b are known as two non-zero vectors and satisfy the vector absolute value a=vector absolute value b=vector absolute value (a-b) Find the angle between vector a and vector (a+b)

A*(a+b)=|a||a+b|cosθ
Let a=(acosα, asinα), b=(bcosβ, bsinβ)
Then: a-b=(acosα-bcosβ, asinα-bsinβ)
(|A|^2)=(a^2)=(|b|^2)=(b^2),(|a-b|^2)=(a^2)+(b^2)-2abcos (α-β)=β)=2(a^2)-2(a^2) cos (α-β)=(a^2) The solution is: cos (α-β)=(1/2). The angle between a and b is 60°,|a|=|b|, therefore: a+b, is the angle bisector of a and b. Then: the calculated angle is 30°.

How to find the normal vector of a plane made of triangles?