How to write the equation of the perpendicular from the point on the line

The question relates to analytic geometry. In this case there are two possible situations. The first of these simple relating to direct on the plane. The second task refers to the direct and planes in space. The reader should have an understanding of the simplest methods of vector algebra.

Instruction

1. The first time. Video given y=kx+b in the plane. You want to find the equation of the perpendicular line passing through the point M(M, n). The equation of the line search in the form y=cx+d. Use the geometric meaning of k-factor. This is the tangent of the angle of inclination α of a straight line to the x-axis k=tgα. Then s=tg(α+π/2)=-ctgα=-1/tgα=-1/k. At the moment, found the equation of the perpendicular line in the form y=-(1/k)x+d, where d remains to clarify. To do this, use the coordinates of the specified point M(m, n). Write down the equation n=-(1/k)m+d, from which d=n-(1/k)m. You can now give the answer y=-(1/k)x+n(1/k)m. There are other types of equations the flat straight. So there are other ways of solutions. However, they are easily converted into each other.
2. Spatial case. Let f video known set of canonical equations (if not, bring them to canonical form). f: (x-x0)/m=(y-y0)/n=(z-z0)/p, where M0(x0, y0, z0) is an arbitrary point of this straight line, and s={m,n,p} is its direction. Target point M(a,b,c). First, find the plane α, perpendicular line f that contains M. To do this, use one of the forms of the General equation of the straight line A(x-a)+B(y-b)+C(z-c)=0. The guiding vector n={A,B,C} coincides with the vector s (see Fig. 1). Therefore, n={m,n,p} and the equation α: m(x-a)+n(y-b)+p(z-c)=0.
3. Now find the point M1(x1,y1,z1) of the intersection of plane α and the straight f by solving the system of equations (x-x0)/m=(y-y0)/n=(z-z0)/p, and m(x-a)+n(y-b)+p(z-c)=0. In the decision process will be the same for all the searched coordinate value u= [m(x0-a)+n(y0-b)+p(z0-c)]/(m^2+n^2+p^2). Then the solution x1=x0-mu, y1=y0-nu, z1=z0-pu.
4. In this step, search the perpendicular line l, find the directional vector g=M1M={x1-a,y1-b z1-c}={x0-mu-a,y0-nu-b,z0-pu-c}. Put the coordinates of the vector m1=x0-mu-a, n1=y0-nu-b, p1=z0-pu-c and record response l: (x-a)/(x0-mu-a)=(y-b)/(y0-nu-b)=(z-c)/(z0-pu-c).