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Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Let a · b +2> m ((2/(a · b))+1). How did x get? Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Let a · b +2> m ((2/(a · b))+1). How was x found? Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Given vector a=(1, x), vector b=(x^2+x,-x), m is constant and m≤-2 Let a·b+2> m ((2/(a·b))+1). How did you get x?
By x! =2,! = Means not equal to
A·b+2> m ((2/(a·b))+1) is equivalent to
X+2> mx/x-2
1. When x-2>0, i.e. x >2, there is x^2-mx-4>0, and m^2+16<0 is obtained from Weida's theorem. If this formula is not valid, then x <2
2.X <2, x^2-mx-4<0,由韦达定理得m^2+16>0, there are two roots
X1=(m+sqrt (m^2+16))/2
X2=(m-sqrt (m^2+16))/2
Where sqrt is the root number
X1>0, x2<0
M >0 of x1>2 does not match the question, so x gets x2 from this
Given points A (6,-4), B (1,2), C (x, y), O are coordinate origin. OC = OA OB (R), then the trajectory equation of point C is () A.2x-y+16=0 B.2x-y-16=0 C. x-y+10=0 D. x-y-10=0
∵
OC =
OA
OB (R),
(X, y)=(6,-4)+λ(1,2),
X=6+λ, y=-4+2λ,
λ Is eliminated to obtain y=2x-16,
The trajectory equation of point C is:2x-y-16=0.
Therefore, B.
In the triangular ABC, M, N, P are the trisection points near A, B, C on the edges of AB, BC, CA, respectively, and O is any point on the triangular ABC plane. In the triangular ABC, M, N, P are the trisection points near A, B, C on AB, BC, CA, respectively, and O is any point on the triangular ABC plane.
, M, N, P are AB, BC, CA sides respectively
Close to the trisection point of A, B, C
OM=OA+AM=OA+1/3AB
ON=OB+BN=OB+1/3BC
OP=OC+CN=OC+1/3CA
OM+ON+OP
=OA+OB+OC+1/3(AB+BC+CA)
=E1/3-e2/2+0(vector)
=1/3*E1-1/2*e2
In a plane rectangular coordinate system, O is the origin of the coordinates, the vector a=(-1,2), point A (1,0), B (cosθ, t) (1) If the vector a⊥AB (the arrow above is omitted, the same below), and |AB|=√5|OA|, find the vector OB (2) If vector a is collinear with vector AB, find the minimum value of OB·OA ----------------------------------------------------- Note: Symbol copied from:θ·||⊥‖√
(1) A⊥AB (-1,2)·(cosθ-1, t)=0. cosθ=1+2t
| AB |=√5| OA |(2t)2+ t2=5, t2=1, t=-1[ t=1, cos θ=3, delete]
Vector OB=(-1,-1)
(2) Vector a is collinear with vector AB:(cosθ-1)/(-1)=t/2, cosθ=1-t/2
OB·OA=cosθ=-1min,[ t=4, B (-1,4]
If the planar quadrilateral ABCD satisfies AB+CD=0,(AB-AD)*AC=0, then the quadrilateral is A square B right-angle trapezoid C rectangular diamond Diagonal lines are perpendicular to each other.
AB + CD =0 means that the AB vector is opposite to the CD vector,(AB-AD)*AC =0 means that the AB-AD is DB, DB *AC =0 means that the two vectors are perpendicular and are rhombus (the square is a special rectangle, rhombus, parallelogram, quadrangle, and the rhombus is equal in four sides, the diagonal is perpendicular, and the four corners of the square are 90 degrees,...
AB + CD =0 means AB vector is opposite to CD vector,(AB-AD)*AC =0 means AB-AD, this means DB, DB *AC =0 means two vectors are perpendicular, diamond (square is special rectangle, diamond, parallelogram, quadrangle, diamond is four sides equal, diagonal is perpendicular, square has four angles of 90 degrees,...
AB + CD =0 means AB vector is opposite to CD vector,(AB-AD)*AC =0 means AB-AD, which means DB, DB *AC =0 means the two vectors are perpendicular, and are rhombus (square is special rectangle, rhombus, parallelogram, quadrangle, rhombus is four sides equal, diagonal is perpendicular, and square has four angles of 90 degrees,...
Given the vector OA=(3,-4), OB=(6,-3), OC=(5-X,-3-Y),(1) points A, B, C can form a triangle, find the condition that X, Y satisfies Given vectors OA=(3,-4), OB=(6,-3), OC=(5-X,-3-Y),(1) If points A, B, C can form a triangle, find the condition that X, Y satisfies
AB=(3,1).|AB|=√10. BC=(-1-x,-y),|BC|=√[(1+x)2+y2] CA=(-2+x,-1+y),|CA|=√[(2-x)2+(1-y)2] If A, B, C can form a triangle, x, y should satisfy the condition of 1231√10.[(1+x)2+y2...
RELATED INFORMATIONS
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