Three vertices A (0,2), B (-1,-2), C (3,1) of a quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3) Three vertices A (0,2), B (-1,-2), C (3,1) of the quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3)

Three vertices A (0,2), B (-1,-2), C (3,1) of a quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3) Three vertices A (0,2), B (-1,-2), C (3,1) of the quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3)

Let the coordinates of vertex D be (x, y)
∵

BC =(4,3),

AD =(x, y−2),
And

BC =2

AD,
∴
2X=4
2Y−4=3⇒
X =2
Y =7
2
Guxuan A

Given the point a1,2 b3,4 c -2,0d-3,3, the projection of vector AB on vector CD is Given the point a1,2 b3,4 c -2,0 d-3,3, the projection of vector AB on vector CD is

Given points A (1,2); B (3,4); C (-2,0); D (-3,3); then the projection of vector AB on vector CD is?
AB=(2,2); CD=(-1,3); AB =√8; CD =√10;
If the included angle of ABCD is α, then cosα=AB·CD/ AB CD =(-2+6)/√80=4/(4√5)=1/√5.
Therefore, the projection of AB on CD = AB cosα=(√8)×(1/√5)=√(8/5)=(2/5)√10.

Given vector AB=(3,-2,-6), vector CD=(6,2,3), find the projection of AB on CD?

From AB.CD=|AB CD|cosθ
Prj (|CD|)=|AB|cosθ=AB.CD/|CD|=-4/7

Given points A (-1,1), B (1,2), C (4,3), D (1,-1), the projection of vector AB in the CD direction is

Vector AB=(2,1), vector CD=(-3,-4).
|Vector AB|=√5,|Vector CD|=5.
Vector AB. Vector CD=2*(-3)+1*(-4)=-10.
Cos=AB.CD/|AB||CD|.
=-10/5*√5.
=-(2√5)/5.
The projection of vector AB in vector CD direction =| AB | cos =√5*(-2√5/5)=-2.

Given vector A, B and vector AB=vector a+vector 2b, vector BC=vector -5a+vector 6b, vector CD=vector 7a-vector 2b, the three points of a certain collinear are

Assuming that abc is collinear, ab=kbc1=-5k2=6k is not valid.
Assuming abd is collinear, ab=kbd1=2k2=4k holds, k=0.5
So abd is collinear
You can also try acd and bcd.

Assuming that abc is collinear, ab=kbc1=-5k2=6k is not valid.
Assuming abd is collinear, ab=kbd1=2k2=4k holds, k=0.5
So abd is collinear
You can also try acd and bcd

It is known that the vector a=(sinα,-2) and b=(1, cosα) are perpendicular to each other, where α belongs to (0,2/π),(1) find the values of sinα and cosα(2) if 5cos (α-β)=3,5cosβ,0<β<π/2, find the values of cosβ. I'm in a hurry.

Vector a=(sinθ,-2) and b=(1, cosθ) are perpendicular to each other, then sinθ-2cosθ=0, i.e. tanθ=2sinθ=2/√5, cosθ=1/√5, cos (α-β)=cos cos sin sinβ, then 5cosθcosb+5sinθsinb=√5cosb+2√5sinb=3√5cosb, then sinb=cosbtanb=1, because 0 is small.