JEE 2023 Mathematics PYQ — Let the plane contain the line and be parallel to the line . Then… | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let the plane P contain the line 2x+y−z−3=0=5x−3y+4z+9 and be parallel to the line 2x+2=−43−y=5z−7. Then, the distance of the point A(8,−1,−19) from the plane P measured parallel to the line −3x=4y−5=−122−z is equal to _____.
Choose the correct answer:
A.
26
(Correct Answer)
B.
27
C.
28
D.
29
Correct Answer:
26
Explanation
Step 1: Plane P ki Equation nikalna
Plane P un do planes ke intersection se guzar raha hai, isliye uski equation Family of Planes se hogi:
(2x+y−z−3)+λ(5x−3y+4z+9)=0
(2+5λ)x+(1−3λ)y+(−1+4λ)z+(9λ−3)=0— (i)
Plane ka normal vector: n=(2+5λ,1−3λ,−1+4λ).
Di gayi parallel line hai: 2x+2=4y−3=5z−7. Iska direction vector hai b=(2,4,5).
Kyunki plane line ke parallel hai, toh n⋅b=0:
2(2+5λ)+4(1−3λ)+5(−1+4λ)=0
4+10λ+4−12λ−5+20λ=0⟹18λ+3=0⟹λ=−61
λ ki value equation (i) mein rakhne par Plane P ki equation milti hai:
Plane P:7x+9y−10z−27=0
Step 2: Point A se guzarti hui Line ki equation
Hamein point A(8,−1,−19) se distance nikalna hai jo ek dusri line ke parallel ho.
Di gayi line: −3x=4y−5=12z−2 (yahan −122−z ko 12z−2 likha gaya hai).
Is line ke direction ratios hain: (−3,4,12).
Point A se guzarne wali aur is line ke parallel line ki equation hogi:
−3x−8=4y+1=12z+19=r
Is line par kisi bhi general point Q ke coordinates honge:
Q=(−3r+8,4r−1,12r−19)
Step 3: Point Q aur Plane P ka Intersection
Kyunki distance "measured parallel to the line" hai, isliye point Q plane P par hona chahiye:
7(−3r+8)+9(4r−1)−10(12r−19)−27=0
−21r+56+36r−9−120r+190−27=0
−105r+210=0
r=105210=2
Step 4: Distance AQ nikalna
Distance d ka formula r aur direction ratios (a,b,c) ke terms mein hota hai:
d=∣r∣a2+b2+c2
Yahan r=2 aur direction ratios (−3,4,12) hain:
d=2(−3)2+42+122
d=29+16+144
d=2169=2×13=26
Final Answer:
Point A ka plane P se distance 26 hai.
Explanation
Step 1: Plane P ki Equation nikalna
Plane P un do planes ke intersection se guzar raha hai, isliye uski equation Family of Planes se hogi:
(2x+y−z−3)+λ(5x−3y+4z+9)=0
(2+5λ)x+(1−3λ)y+(−1+4λ)z+(9λ−3)=0— (i)
Plane ka normal vector: n=(2+5λ,1−3λ,−1+4λ).
Di gayi parallel line hai: 2x+2=4y−3=5z−7. Iska direction vector hai b=(2,4,5).
Kyunki plane line ke parallel hai, toh n⋅b=0:
2(2+5λ)+4(1−3λ)+5(−1+4λ)=0
4+10λ+4−12λ−5+20λ=0⟹18λ+3=0⟹λ=−61
λ ki value equation (i) mein rakhne par Plane P ki equation milti hai:
Plane P:7x+9y−10z−27=0
Step 2: Point A se guzarti hui Line ki equation
Hamein point A(8,−1,−19) se distance nikalna hai jo ek dusri line ke parallel ho.
Di gayi line: −3x=4y−5=12z−2 (yahan −122−z ko 12z−2 likha gaya hai).
Is line ke direction ratios hain: (−3,4,12).
Point A se guzarne wali aur is line ke parallel line ki equation hogi:
−3x−8=4y+1=12z+19=r
Is line par kisi bhi general point Q ke coordinates honge:
Q=(−3r+8,4r−1,12r−19)
Step 3: Point Q aur Plane P ka Intersection
Kyunki distance "measured parallel to the line" hai, isliye point Q plane P par hona chahiye:
7(−3r+8)+9(4r−1)−10(12r−19)−27=0
−21r+56+36r−9−120r+190−27=0
−105r+210=0
r=105210=2
Step 4: Distance AQ nikalna
Distance d ka formula r aur direction ratios (a,b,c) ke terms mein hota hai:
