Question

the plane that contains the line x=-2+3t , y=4+2t , z=3-t and is perpandicular to the plane x-2y+z=5.

Answer 1

say, \(a(x-x_1) + b(y-y_1) + c(z-z_1)\) is the required plane

Answer 2

if this plane contains the given line, the point (-2, 4, 3) must satisfy the plane :

\(a(x+2) + b(y-4)+c(z-3) = 0\)

Answer 3

Also, the direction vector of line and normal to plane must be perpendicular :

\(3a + 2b - c = 0\)

Since the plane is also perpendicular to another plane \(x-2y+z=5\), the normal vectors must be perpendicular :
\(a - 2b + c = 0\)

solving these two equations gives,
\(<a, b, c> = <0, 1, 2>\)

Answer 4

that gives,

\(0(x+2) + 1(y-4)+2(z-3) = 0\)

Answer 5

simplify

Answer 6

thanks alot :)

Answer 7

u wlc :)