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The side length of equilateral triangle ABC is 1, vector AB=a, vector BC=b, vector CA=c, then a*b+b*c+c*a is equal to
0
Given that the plane vector abc is not collinear and the angle between the two is equal, if a modulus=2, b modulus=2, c modulus=1, then the angle between vector a+vector b+vector c and vector a is
The angle between a, b and c is 2π/3
|A|=|b|=2, so:|a+b|=2, and a+b and c are opposite
|C|=1, so a+b=-2c, so a+b+c=-c
Therefore: the included angle between a+b+c and a is π/3
The angle between a, b and c is 2π/3
|A|=|b|=2, so:|a+b|=2, and a+b and c are opposite
|C|=1, so a+b=-2c, so a+b+c=-c
Therefore: the angle between a+b+c and a is π/3
Given the points A (a,0,0), B (0, b,0), C (0,0, c), the unit normal vector of the plane ABC is __________. Ask each road master, find the normal vector have plus or minus two, or only one~ Given the points A (a,0,0), B (0, b,0), C (0,0, c), the unit normal vector of the plane ABC is __________. Would the experts from all walks of life be able to determine whether there were two positive or negative normal vectors or only one?
AB=(-a, b,0) AC=(-a,0, c) BC=(0,-b, c) try to vector u=(x, y, z) from AB*u=0 AC*u=0 BC*u=0 to get ax=by=cz1 and unit vector to get x^2+y^2+z^2=1 2 simultaneous 1 2 to get x=±bc/√(a^2c^2+a^2b^2+b^2c^2) with back 1 to get y, z There are two...
Given points A (a,0,0), B (0, b,0) C (0,0, c), find the unit normal vector?
That is, the plane is
(Bc) x+(ac) y+(ab) z=abc
Because you put x=a, y=0, z=0
X=0, y=b, z=0
X=0, y=0, z=c all hold
So a plane normal vector is the coefficient before x, y, z
N=(bc, ac, ab)
Given points A (3,0,0), B (0,4,0), C (0,0,5), find a unit normal vector of plane ABC
Find the vectors AB, AC, BC respectively. Then according to the normal vector and desperately perpendicular, and mutually perpendicular vector product is 1. Try to vector =(a, b, c) and the vectors AB, AC, BC, respectively multiplied by equal to 1, list three equations, stand together as a ternary first-order equations, can be solved a, b, c.
Find the vectors AB, AC, BC respectively. Then according to the normal vector and desperately perpendicular, and mutually perpendicular vector product is 1. Try to vector =(a, b, c) and vector AB, AC, BC, respectively multiplied by equal to 1, list three equations, together as a ternary first-order equations, can be solved a, b, c.
Given A (3,0,0), B (0,4,0), C (0,0,5), find the unit normal vector of the plane ABC
Try vector n=(x, y, z)
AB=(-3,4,0), BC=(0,-4,5)
The equation group -3x+4y=0;-4y+5z=0; x2+y2+z2=1
Solve to n=(
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