Find a parametric form for the intersection of the planes P

Find a parametric form for the intersection of the planes P and Q in R 3 whose parametric forms are given below:

P : [4 ;3 ;7]+ t[1; 2; 3 ]+ s [1 ;8 ;9]

Q : [0; 2 ;3]+ t[ 1 ;2 ;1]+ s[1; 6 ;7]

Solution

We will first convert the parametric equation of the two planes into Cartesian form as under:

The parametric equation of the first plane is

x = 4 + t + s …(i) , y = 3 + 2t + 8s…(ii), z = 7 + 3t + 9s …(iii)

On multiplying the first equation by 8, we get 8x = 32 + 8t + 8s …(iv). Now, on subtracting the 2^nd equation from the 4^th equation, we get 8x – y = 32 + 8t + 8s – (3 + 2t + 8s) or, 8x –y = 29 + 6t …(v)

On multiplying the 2^nd and the 3^rd equations by 9 and 8 respectively, we get ,

9y = 27 + 18t + 72s …(vi) and 8z = 56 + 24t + 72s …(vii) .On subtracting the 6^th equation from the 7^th equation, we get 8z – 9y = (56 + 24t + 72s) – (27 + 18t + 72s) = 29 + 6t …(viii). Now on comparing the equation numbers (v) and (viii), we get 8x – y = 8z - 9y or, 8x + 8y - 8z = 0 or x + y – z = 0…( 1^st Plane)

The parametric equation of the second plane is

x = 0- t + s …(i), y = 2 +2t + 6s…(ii), z = 3 + t + 7s…(iii)

On multiplying the 1^st equation by 6, we get, 6x = - 6t + 6s…(iv). On subtracting the 4^th equation from the 2^nd equation, we get, y – 6x = (2 +2t + 6s) – (- 6t + 6s) or, y - 6x = 2 + 8t or, 8t = y – 6x -2 …(v)

On multiplying the 2^nd equation by 7, we get 7y = 14 + 14t + 42s…(vi). On multiplying the 3^rd equation by 6, we get 6z = 18+ 6t + 42s…(vi). On subtracting the 5^th equation from the 6^th equation, we get,

6z – 7y =(18+ 6t + 42s) – (14 + 14t + 42s) = 4 - 8t …(vii). Now, on substituting the value of 8t from the 5^th equation in the 7^th equation, we get, 6z – 7y = 4 – (y – 6x -2 ) = 4 – y + 6x + 2 or, 6z – 7y = 6x – y + 6 or,

or, 6z -7y + y -6x = 6 or, 6z – 6y -6x = 6 or, z – y –x = 1 or, x + y –z = - 1….(2^nd Plane).

Since the equation of the 1^st plane is x + y – z = 0, the two planes are parallel and do not intersect.