In triangle ABC, absolute value of AC vector =10, absolute value of AD vector =5, AD vector =5/11DB vector, CD vector * AB vector =0(1) find (AB vector - AC Vector)

In triangle ABC, absolute value of AC vector =10, absolute value of AD vector =5, AD vector =5/11DB vector, CD vector * AB vector =0(1) find (AB vector - AC Vector)

14
AD vector =5/11 DB vector D AD =5 BD =11 on AB
CD vector * AB vector =0 CD perpendicular to AB
Absolute value of (AB vector-AC vector)= Absolute value of CB vector
Pythagorean DC^2=75 CB=14

Let vector OA=(-2, m), vector OB=(n,1), if A.B.C is collinear and vector OA⊥ vector OB, then the value of m+n is

Let A, B, C in the plane be in a straight line, vector 0A=(-2, M), OB=(N,1), OC=(5,-1), and vector OA is perpendicular to vector OB.

Given A (-1,-3), B (0,-1) and vector a =(2, m), and vector AB//a, then the value of m is

Find AB=(1,2), then m=4

Given a point M (1,-1,2) in the plane α, one of the normal vectors of the plane α is N=(6,-3,6), then () A.P (2,3,3) B.P (-2,0,1) C.P (-4,4,0) D.P (3,-3,4) Given a point M (1,-1,2) in the plane α, a normal vector of the plane α is N=(6,-3,6), then () A.P (2,3,3) B.P (-2,0,1) C.P (-4,4,0) D.P (3,-3,4)

Let P (x, y, z) be a point in the plane α, then:


MP=(x-1, y+1, z-2)
∵

N=(6,-3,6) is the normal vector of the plane α,
∴

N⊥

MP,

N•

MP=6(x-1)-3(y+1)+6(z-2)=6x-3y+6z-21,
Yanyou

N•

6X-3y+6z-21 for MP=0=0
2X-y+2z=7
Substitute the coordinate data of each option into the above formula to verify that A is suitable.
Therefore: A

Given vector a=(2,3), b=(-4,7), then the projection of vector a in vector b is? The question is why not ab=(2,3)(-4,7)=-8+21=13 Given vector a=(2,3), b=(-4,7), the projection of vector a in vector b is? The question is why not ab=(2,3)(-4,7)=-8+21=13

The projection of vector a in vector b is a*b/|b|=13/√4 2+7 2=13/√65

As shown in the figure, in the triangle ABC, the absolute value of the vector of AB =3, the absolute value of the AC vector =1, L is the vertical bisector of BC and intersects BC Point D, E is any point different from D on L, F is any point on line segment AD, find: AD vector multiplied by vector of AB minus vector of AC, judge whether the value of AE vector multiplied by vector of AB minus vector of AC is constant, explain the reason. If AC is vertical BC, find the maximum value of AF vector multiplied by vector FB + vector FC.

The following is used to represent that the vector ABD is BC midpoint=(+)/2(-)=(+)(-)/2=(^2-^2)/2=(3^2-1^2)/2=4==0(-)=(+)(-)=+(-)=4 Set = k=-=-(+)=(+)=(--2)=2(-)=2 k (1-k)^2 Easy to know |BC |=2 Root 2 So |AD |= Root 3(+)=6 k (1-k)=-6(k-1/2)^2+3/2 The maximum value is 3...