A spatial analytic geometry problem, kneel down and ask for a solution

1.Vector ac=(-2,-2,1), vector bc=(-1,1,0), the product of vector ac and vector bc = (-2)*(1)+(2)*1+1*0=0, so ac bc

2.Vector ab=(-1,-3,1), let the vector ah=tab, we get h(1 (1+t),(2-t) (1+t),t (1+t)), vector ch=((2+t) (1+t), (2-t) (1+t),-1 (1+t)), ch 丄ab gets ch*ab=0 to get t=9 2, substitution can get h

3.m is obtained by using the midpoint formula, and then t is obtained by using the line segment fixed score point formula, and the coordinates of t are the average of the coordinates of a, b, and c.

1.The square of the line segment ab = z ab + (|xa|+|xb|)²ya|+|yb|Here the lowercase letter is the x, y, z, and distinction sign that distinguishes the two points of ab.

Substituting the two-point coefficient of ab into the solution yields: ab = 11Again, the above formula is used to solve: ac=3, bc=2

Derived from the Pythagorean theorem: abc is a right triangle, and c=90° is used for reasoning, and it can be obtained by reverse deduction.

Solution: Let form a straight line x=1, from 2x-y=0: y=2, from 3x-2y+z=1 to get :z=2;

Let x=0, get: y=0, z=1; Straight lines pass through points (1,2,2) and points (0,0,1).

The vector composed of the crossing point (-2,3,1) and the crossing point (1,2,2) is v1=; The vector formed with the crossing point (0,0,1) is v2=, and the normal vector n=v1xv2=x=

The plane equation between the crossing point (-2, 3, 2) and the straight line l is: 3(x+2)+2(y-3)-6(z-1)=0, that is, 3x+2y-6z+6=0.

Answer: 3x+2y-6z+6=0.

Let a and b be the two moving points on the plane oxz and oyz, respectively, and satisfy the vectors oa*ob=1. Find the trajectory equation for the midpoint of AB.

Solution: Let the midpoint of ab be m(x,y,z), then a(2x,0,z1),b(0,2y,2z-z1), and the vector oa*ob=1 yields z1(2z-z1)=1

Please check the question.

The plane 2x+y+kz-1=0 is the general equation of the plane with the normal vector n=(2, 1, k).

The straight line (x-4) 3=(y+3) 4=z (-2) is a point-oriented equation with the direction vector s=(3, 4, -2).

If the line is parallel to the plane, the normal vector of the plane is perpendicular to the direction vector of the line.

That is, the product of the quantity of the two vectors is 0. That is, 2*3+1*4-2k=0 gives k=5

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