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In the concept of the product of must the vector be non-zero? In the concept of the product of the vector must be non-zero?
In the definition of vector quantity product, the two vectors of quantity product are not certain to be nonzero. The quantity product of two vectors is equal to the product of the module of one vector and the projection of the other vector in the direction of this vector. The quantity product of two vectors a and b: a·b=|a b|cosθ; where |a|,|b| are the product of two vectors.
The Problem of Solving Normal Vector in Mathematical Space Vector First establish the spatial rectangular coordinate system xyz, o is the origin Then take a point m (0,0,1) on the z-axis Take point n (1,0,0) on x-axis So what is the normal vector of the mno plane? I set the normal vector to (a, b, c) Can be brought into the equation C=0 A =0 There's no point. What's b=? Every time I ask for a normal vector, if I encounter one related to the origin, I can't find it. Please guide us.
First of all, your face is an XOZ plane. The normal vector must be in the format of Y-axis direction vector (0, y,0). The value of y can be taken arbitrarily. The normal vector is generally set as unit vector for easy calculation. Generally,(0,1,0)
30 Points. In the known parallelogram ABCD, A (4,1,3),(2,-5,1), C (5,7,5) find the coordinates of vertex D and the coordinates of diagonal intersection point O and IABI. 30 Points. A (4,1,3),(2,-5,1), C (5,7,-5) in a parallelogram ABCD of vertex D and the coordinates of diagonal intersection point O and IABI.
Vector AB=(-2,-6,-2), vector AC=(1,6,-7), vector BC=(3,12,-6), D=(x, y, z) When vector AD=vector BC,(x-4, y-1, z-3)=(3,12,-6), D (7,13,-3), midpoint O (9/2,4,-1) When vector CD=vector AB,(x-5, y-7, z+5)=(-2,-6,-2), D (3,1,-7), midpoint O (7/2,...
Mathematical Problems of Space Vector In the spatial rectangular coordinate system O-xyz, i, j, k are the direction vectors of the x-axis, y-axis, z-axis, respectively. Let a be a non-zero vector and =45°,=60°, then equal to There are steps Mathematical Problems of Space Vector In the spatial rectangular coordinate system O-xyz, i, j, k are the direction vectors of the x-axis, y-axis, z-axis, respectively, and if a is a non-zero vector and =45°,=60°, then equals to Step by step Mathematical Problems of Space Vector In the spatial rectangular coordinate system O-xyz, i, j, k are the direction vectors of the x-axis, y-axis, z-axis, respectively. Let a be a non-zero vector and =45°,=60°, then equal to There must be steps
By cos2+ cos2+ cos2=1
Cos2=1/4
Cos=1/2
=60.
Of space vectors Vectors OA=a, OB=b, points A', B' are known to be orthographic projections of A, B on OA, OB, respectively. Of space vectors It is known that the vectors OA=a, OB=b, points A', B' are orth orthographic projections of A, B on OA, OB respectively. Of space vectors The vectors OA=a, OB=b, points A', B' are the orthographic affine of A, B on OA, OB respectively.
In fact, the two vectors must be coplanar. Draw the two vectors in the plane by a·b=|a|b|cosθ.
OA · OB'= OB · OA'=| a | b | cos θ, so equal
Let two different unit vectors a=(x, y,0) b=(d, f,0) and vector c=(1,1,1) be π/4 Ask the size of a, b The answer is π/3
It can be found that x+y = two-part root number six xy =1/4, so can d and f.
According to the included angle formula of space vector,
Cos angle =1/2
RELATED INFORMATIONS
- 1. Given the rectangle ABCD-A' B' C' D', simplify the following vector expressions and mark out the vector of the simplified result. {1} Small arrow above AB + small arrow above BB'; {2} AB plus small arrow + BC plus small arrow + BB' plus small arrow. Given the rectangle ABCD-A' B' C' D', the following vector expression is simplified and the vector of the simplified result is marked. {1} Small arrow above AB + small arrow above BB'; {2} AB plus small arrow + BC plus small arrow + BB' plus small arrow.
- 2. Excuse me. A (5,1). B (7,10). Vector AB (how to calculate? , Answer is (2,9)
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