Hey!
We could rewrite the problem as:
Since we have a negative number, we could rewrite the fraction like this.
Let's simplify it...
Our answer would be:
Thanks!
-TetraFish
Hello :
let A(0,3,2) and (Δ) this line , v vector parallel to (<span>Δ).
M</span>∈ (Δ) : vector (AM) = t v..... t ∈ R
1 ) (Δ) parallel to the plane x + y + z = 5 : let : n an vector <span>perpendicular
to the plane : n </span>⊥ 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) =0
x - 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 =0
x = 3 - (3/2)z
subsct in (1) : 3 - (3/2)z +y +z - 5 =0
y = 1/2z +2
let : 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) <span>perpendicular to (1,1,1)
</span>because : (1)(-3/2)+(1)(1/2)+(1)(1) = -1 +1 = 0
2) (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 = 0
A(0, 3, 2)∈(Δ) :
0 = 3-(3/2)t
3 = (1/2)t+2
2 =t
same : t = 2
I think this is how you solve this
