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Given that a, b is a nonzero vector and satisfies (a-2b) perpendicular to a and (b-2a) perpendicular to b, then what is the angle between a and b? Given that a, b is a nonzero vector and satisfies (a-2b) perpendicular to a and (b-2a) perpendicular to b, then is the angle between a and b?
A, b are known to be non-zero vectors and satisfy (a-2b) perpendicular to a,(b-2a) perpendicular to b
Then (a-2b) a=0(b-2a) b=0
So a^2=2ab b^2=2ab
So |a|=|b|
Let the angle between a and b be θ
Then cosθ=ab/|a||b||=ab/|a|^2=ab/a^2=ab/2ab=1/2
So θ=60°
The Direction Problem of Zero Vector and Zero Vector If the sum of the module length of vector a and the module length of vector b is equal to the module length of the sum of vector a and vector b, then vector a and vector b are in the same direction and collinear to judge the right and wrong
This judgment is not correct. The two vectors must be parallel, the reason for the error is "same direction."
This judgment is not correct. The two vectors must be parallel, the reason for the error is "the same direction."
On the problem of high - medium vector theorem. The formula in the book is: vector OP=vector OM+x vector MA+y vector MB. Vector OP=x vector OA+y vector OB+z vector OM. Now we encounter a problem: Given that the three points A, B and M are not collinear, for any point O outside the plane ABM, determine whether the point P is coplanar with A, B and M under the following conditions? (1) Vector OP=vector OM+vector PA+vector PB, then P, A, B, M are coplanar... Please elaborate why? (2) Vector OP=1/3 vector OA+1/3 vector BA+1/3 vector MA, then P, A, B, M are collinear... Please elaborate why? This doesn't match the previous formula. Can those two formulas be extended? If yes, please deduce it. The first two formulas are coplanar vector formulas. That is, the necessary and sufficient conditions for vector OP to be coplanar with vector MA and vector MB.
In fact, you don't understand the formula in the book: vector OP=vector OM+x vector MA+y vector MB.
Vector OP=x vector OA+y vector OB+z vector OM.
In the second formula, x, y, z must satisfy x+y+z=1 in order to judge the coplanar of P, A, B, M.
That is, if vector OP=x vector OA+y vector OB+z vector OM., and x+y+z=1, then P, A, B, M are coplanar.(1)
(1) Vector OP=vector OM+vector PA+vector PB
= Vector OM+ vector OA-vector OP+ vector OB-vector OP
Therefore,3-vector OP=vector OA+vector OB+vector OP
From OP=1/3 vector OA+1/3 vector OB+1/3 vector OP
Meet the book's conclusions, so P, A, B, M coplanar.
(2) Vector OP=1/3 vector OA+1/3 vector BA+1/3 vector MA
=1/3 Vector OA +1/3(vector OA - vector OB)+1/3(vector OA - vector OM)
= Vector OA-1/3 Vector OB-1/3 Vector OM
The preceding coefficients add up to 1, so P, A, B, M are coplanar.
Collinear vector theorem In the plane there is a vector OA=(1,7), OB=(5,1), OP=(2,1), point X is a moving point on the straight line OP. (1) When vector XA*XB gets the minimum value, find the coordinate of vector OX (2) When point X satisfies the condition and conclusion of (1), find the cosine value of angle AXB Why XA*XB =(1-2m)(5-2m)=(7-m)(1-m)
For point X on OP, let's assume that the coordinate of X is (2m, m), then XA=(1-2m,7-m), XB=(5-2m,1-m), XA*XB=(1-2m)(5-2m)+(7-m)(1-m)=(5-12m+4m2)+(7-8m+m2)=5m2-20m+12=5(m-2)2-8, when m=2, XA*XB gets the minimum value, then the coordinate of X is (4,2), OX=(4,...
VECTOR FORMULA FOR CALCULATING LINEAR AND SURFACE ANGLE If the problem is to use the vector to solve the geometric problem, the vector formula can be described in words
Plane geometry?
Line surface angle: the line L intersects with the plane S at point A. Take any point P on the line L to make a vertical line perpendicular to the plane. Set the vertical foot as B and connect AB, then the angle PAB is the line surface angle.
Face angle: If planes A and B intersect the straight line L, then you can make two straight lines L1 and L2 on planes A and B so that L1 is perpendicular to L and L2 is perpendicular to L. The angle between L1 and L2 is the face angle.
As for how to calculate these two angles, you can use the vector. For example, the two lines that make up this angle, the two-point equation of the straight line, remember?
Straight through A (X1, Y1) B (X2, Y2)
So its slope is k =(Y2-Y1)/(X2-X1)
The direction vector of the line is (X2-X1, Y2-Y1)
Now that the direction vector of the two lines is known, the angle formula of the two vectors can be used to calculate.
For example, the angle is a, and the direction vector of two lines is (X2-X1, Y2-Y1)
(X4-X3, Y4-Y3)
The angle a can be obtained by a formula
The formula is: product of two vector modules * COSa = dot product of two vectors
The module of vector:(X2-X1, Y2-Y1), its module is:[ square of (X2-X1)+ square of (Y2-Y1)] under root sign
Dot product of vector:(X2-X1, Y2-Y1),(X4-X3, Y4-Y3):
Their dot product is:(X2-X1)*(X4-X3)+(Y2-Y1)*(Y4-Y3)
The entire formula is:
Square of [(X2-X1)+ Square of (Y2-Y1)]* Square of [(X4-X3)+ Square of (Y4-Y3)]* COSa =(X2-X1)*(X4-X3)+(Y2-Y1)*(Y4-Y3)
To find the value of a
How to prove that the line and plane are parallel and perpendicular by the space vector method? What does this line have to do with normal vectors?
If the line direction vector and the plane normal vector are parallel, the line is perpendicular to the plane.
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