Answer this Question of iit jee . Chapter → Three dimensional Geometry.
Answers
Answer:
Option (c) x+3y+6z = 7
Step-by-step explanation:
Let equation of plane containing the lines
2x - 5y + z = 3 and x + y + 4z = 5 be
(2x - 5y + z - 3) + λ ( x+y+4z -5) = 0
(2+λ ) x + (λ + 5) y + (4λ + 1) z - 3 - 5λ = 0 ____(1)
This plane is parallel to the plane x+3y + 6z = 1
∴ 2 + λ / 1 = λ - 5/3 = 4λ + 1 / 6
On taking first two equalities ,we get
- → 6 + 3λ = λ - 5
- → 2λ = - 11
- → λ = -11/2
On taking last two equalities , We get
- → 6λ - 30 = 3 + 12λ
- → -6λ = 33
- → λ = 33/6
So ,the equation of required plane is
(2- 11/2) x + (-11/2-5)y + (-44/2 + 1) z -3 + 5 × 11/2 =0
- → -7/2x - 21/2y - 42/2z + 49/2 = 0
- → x + 3y + 6z - 7 = 0
- or, x+ 3y + 6z = 7 Answer
Answer:
The Plane is containing the lines 2x-5y+z=3 and x+y+4z=5
So, we can write the eqn of plane as :-
(2x-5y+z) + λ(x+y+4z) = 0
→ (2+λ)x + (-5+λ)y + (1+4λ)z = 0. -----(1)
The plane is parallel to another plane x+3y+6z=1
→ 2+λ/1 = -5+λ/3 = 1+4λ/6
→ 3(λ+2) = -5+λ
→ 3λ+6 = λ-5
→ 2λ = -5-6
→ λ = -11/2
Now, put the value of λ in eqn (1) and we will get the eqn of plane as
-7x/2 - 21y/2 -21z + 49/2 = 0
→ x + 3y +6z -7 = 0