find the equation of the line perpendicular to the plane 3x+2y-5z=2 containing the point (1,2,-1)?

Question

how would you go about solving this type of problem? can someone solve and explain so i have some reference to look onto while i solve problems similar to this one.thanks.

umm im not sure. thats all the problem says =/

Answer 1

♠ cutie, so you want a rod, not fish? fine! here you are!
Classic equation of a plain looks:
a*x +b*y +c*z =d, where (a,b,c) is a unit vector, normal to the plain, i.e. a^2+b^2+c^2=1, while |d| is the shift of the plain from (0,0,0) point;
Classic equation of a 3D line looks:
♦ (x-x1)/A =(y-y1)/B =(z-z1)/C, where (A,B,C) is a vector parallel to the line;
♣ now your task: A=k*a=3, B=k*b=2, C=k*c=-5; x1=1, y1=2, z1=-1;
put them into (♦) and see the result! ♥ is your Dad tough?

Answer 2

The normal vector of the plane is also the directional vector of the line perpendicular to it. The equation of the line is:

r = <1, 2, -1> + t<3, 2, -5>
where t is a scalar ranging over the real numbers

Answer 3

to get the equation for a line, you may wish a factor and the direction of the line represented by making use of the vector. a vector perpendicular to the airplane 3x+2y-5z=2 may well be its regular vector, purely <3,2,-5>. so now you have your factor (a million,2,-a million) and your direction vector. the vector equation for the line may well be = (a million,2,-a million) + t<3,2,-5>. resolving it into aspects, the parametric equations for the line may well be x=a million+3t y=2+2t z= -a million-5t removing the parameter, the symmetric equations for the line may well be (x-a million)/3 = (y-2)/2 = (z+a million)/-5 any a form of equations may well be equivalent and fabulous.

Answer 4

Is your vertex closed or open?
It's very important.