Find parametric equations for the line through the point (0, 3, 2) that is parallel to the plane x + y + z = 5 and perpendicular to the line x = 1 + t, y = 3 − t, z = 2t. (use the parameter t.)
'inconus'= 'unknowns'
No, you are misinterpreting what Plato said. Any line in three dimension can be written as $\displaystyle x= x_0+ at$, $\displaystyle y= y_0+ bt$, $\displaystyle z= z_0+ ct$. In order that the line contain the point (0, 3, 2) we must have x= 0+ at= at, y= 3+ bt, and z= 2+ ct. In order that the line be parallel to the plane x+ y+ z= 5, its direction vector <a, b, c> must be perpendicular to the normal vector of the plane, <1, 1, 1>. We must have a+ b+ c= 0. Finally, in order that it be perpendicular to the line x= 1+ t, y= 3- t, z= 2t, it must intersect that line and a- b+ 2c= 0. We have the two equations a+ b+ c= 0 and a- b+ 2c= 0 and the requirement that the two lines must intersect.
The point P(0,3,2) is in the plane x+y+z=5 and therefore a line (L) through P and parallel to the plane must be in the plane.
(L) goes through the point Q(3/2, 5/2, 1) where x = 1 + t, y = 3 − t, z = 2t meets the plane.
hello : What do you think of this solution?let A(0,3,2) and (Δ) this line , v vector parallel to (Δ).
M∈ (Δ) : vector (AM) = t v..... t ∈ R1 )
(Δ) parallel to the plane x + y + z = 5 : let : n an vector perpendicular to the plane :
n ⊥ v .... n(1,1,1) so : n.v =0
means : n.vector (AM) = 0(1)(x)+(1)(y-3)+(1)(z -2) =0 ( vector (AM) = ( x, y -3 , z-2 )x+y+z - 5=0 ...(1)
2)
(Δ) perpendicular to the line (Δ') : x = 1+t , y = 3 - t , z = 2t :vector (u) ⊥ v .... vector(u) parallel to (Δ') and vector(u) = (1 , -1 ,1)vector (u) ⊥ vector (AM) means : (1)(x)+(-1)(y-3)+(2)(z -2) =0x - y+2z - 1 = 0 ...(2)so the system : x+y+z - 5=0 ...(1)x - y+2z - 1 = 0 ...(2) (1)+(2) :
2x+3z - 6 =0x = 3 - (3/2)z subsct in (1) : 3 - (3/2)z +y +z - 5 =0y = 1/2z +2let : z=t an parametric equations for the line (Δ) is : x = 3 - (3/2)t y = (1/2)t +2 z=t
verifiy : 1) (Δ) parallel to the plane x + y + z = 5 : (-3/2 , 1/2 ,1) perpendicular to (1,1,1)because : (1)(-3/2)+(1)(1/2)+(1)(1) = -1 +1 = 02) (Δ) perpendicular to the line (Δ') : (-3/2 , 1/2 ,1) perpendicular to (1,-1,2)because : (1)(-3/2)+(-1)(1/2)+(1)(2) = -2 +2 = 0A(0, 3, 2)∈(Δ) : 0 = 3-(3/2)t3 = (1/2)t+22 =t same : t = 2