the plane p has equation 2x-3y+6z=16. The plane q is parallel to p and contains the point with position vector i+4j+2k. i) Find the equation of q. ii) Calcualte the perpendicular distance between p and q. iii) The line is paraller to the plane p and also parallel to the plane with equation x-2y+2z-5. Given that l passes through the origin, find a vector eqauation for l

Question

turingtest · Answer

if the two planes are parallel they must have the parallel normal vectors
what is the nomal vector of the given plane?

turingtest · Answer

normal*

anonymous · Answer

i am confused....

turingtest · Answer

what is the normal vector to the plane 2x-3y+6z=16 ???

anonymous · Answer

2,-3,6...?

turingtest · Answer

|dw:1341331018259:dw|yes$$\vec n=23\hat i-3\hat j+6\hat k$$

turingtest · Answer

oops$$\vec n=2\hat i-3\hat j+6\hat k$$

anonymous · Answer

lol :P

turingtest · Answer

so if we draw our normal vector from the given point in the plane $\vec P_0=\langle1,4,2\rangle$, then we can draw a vector between that point and any random point $\vec P=\langle x,y,z\rangle$ and get a vector in the plane p$$\vec v=\vec P-\vec P_0$$|dw:1341331381439:dw|