Find an equation for the plane that contains the line v=(−4,−3,−1)+t(4,−3,1) and is perpendicular to the plane 2x+5y+4z+2=0. (use symbolic notation and fractions where needed. note that the solution expects you to solve for z. you may need to scale your answer suitably. )

The desired plane will have a direction vector that is perpendicular to that of the given line (4, -3, 1) and that of the given plane (2, 5, 4). Computing the cross product of these gives (-17, -14, 26).

Since the plane contains the point (-4, -3, -1), we can find the constant (d) in ax+by+cz=d using the dot-product of the point and the direction vector just found.

  -17·(-4) -14·(-3) +26·(-1) = 68 +42 -26 = 84

Your plane can be written as

  -17x -14y +26z = 84

Solving for z, you get

  z = (17x +14y +84)/26

or, if you prefer, ...

  z = (17/26)x +(7/13)y +42/13

tip $2.90

total $17.40