Question

the plane that contains the line x=3t , y=1+t , z=2t and is parallel to the intersection of the planes y+z=-1 and 2x -y+z=0.

Answer 1

first find the point of intersection of planes
then assume plane ax+by+cz+d=0
this point will satisfy the equation.
and the three lines willl be perpendicular to normal of the plane
so u will get 4 equation and four variables
now you can solve

Answer 2

in detail please

Answer 3

say, \(a(x-x_1) + b(y-y_1 ) + c(z-z_1) = 0 \) is the equation of required plane

Answer 4

since it contains the line, x=3t , y=1+t , z=2t
and since normal is perpendicular to every line on the plane, we have :
\(3a + b + 2c = 0\)

Answer 5

Also, since the plane is parallel to line of intersection of plane y+z=-1 and 2x -y+z=0 :
direction vector of this line of intersection = \(\left<0, 1, 1\right> \times \left<2, -1, 1\right> = \left<1, 1, -1\right>\)

the normal of required plane is perpendicular to this direction vector also :
\(\large a +b -c = 0\)

Answer 6

solving above two equations for a, b, c gives :

(a, b, c) = (-3, 5, 2)

Answer 7

so the equation of required plane is :
\(-3(x-x_1) + 5(y-y_1 ) + 2(z-z_1) = 0\)

Answer 8

plugin the point (0, 1, 0) :

\(-3(x-0) + 5(y-1 ) + 2(z-0) = 0\)

Answer 9

simplify

Answer 10

let me knw if smthng doesnt make sense...

Answer 11

okay thanks again :)

Answer 12

np :)