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

Answer 1

put your equation in slope intercept form first then find the slope of the line and that way you can get the equation...remmember that the equations in order to be parallel you must have the same slope in both of them...

Answer 2

to get the equation for a line, you would need a point and the direction of the line represented by the vector.

a vector perpendicular to the plane 3x+2y-5z=2 would be its normal vector, simply <3,2,-5>. so now you have your point (1,2,-1) and your direction vector.

the vector equation for the line would be
= (1,2,-1) + t<3,2,-5>.

resolving it into components, the parametric equations for the line would be

x=1+3t
y=2+2t
z= -1-5t

removing the parameter, the symmetric equations for the line would be

(x-1)/3 = (y-2)/2 = (z+1)/-5

any one of these equations would be equivalent and correct.

Answer 3

the traditional vector of the airplane is likewise the directional vector of the line perpendicular to it. The equation of the line is: r = + t<3, 2, -5> the place t is a scalar ranging over the real numbers

Answer 4

Hint:

(x-1)/u1=(y-2)/u2=(z+1)/u3

determine (u1,u2,u3)