JEE 2023 Mathematics PYQ — Let the plane be rotated by an angle about its line of intersecti… | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let the plane P:4x−y+z=10 be rotated by an angle 2π about its line of intersection with the plane x+y−z=4. If α is the distance of the point (2,3,−4) from the new position of the plane P, then 35α is equal to
Choose the correct answer:
A.
90
B.
105
C.
85
D.
126
(Correct Answer)
Correct Answer:
126
Explanation
Step 1: Family of Planes ka use karna
Naya plane (P′) un do planes ki line of intersection se guzarta hai, isliye uski equation hogi:
P′=P1+λP2=0
(4x−y+z−10)+λ(x+y−z−4)=0
Isse rearrange karne par:
(4+λ)x+(λ−1)y+(1−λ)z−(10+4λ)=0
Step 2: Perpendicularity Condition (2π rotation)
Sawal mein diya gaya hai ki plane P ko 2π (90°) rotate kiya gaya hai. Iska matlab purana plane (P) aur naya plane (P′) aapas mein perpendicular hain.
Perpendicular planes ke liye unke normal vectors ka dot product zero hota hai (a1a2+b1b2+c1c2=0):
4(4+λ)+(−1)(λ−1)+1(1−λ)=0
16+4λ−λ+1+1−λ=0
2λ+18=0⟹λ=−9
Step 3: Naye Plane ki Equation
λ=−9 ko P′ ki equation mein rakhein:
(4−9)x+(−9−1)y+(1−(−9))z−(10+4(−9))=0
−5x−10y+10z+26=0
Ya phir: 5x+10y−10z−26=0
Step 4: Distance α Nikalna
Point (2,3,−4) se plane 5x+10y−10z−26=0 ki distance:
α=52+102+(−10)2∣5(2)+10(3)−10(−4)−26∣
α=25+100+100∣10+30+40−26∣
α=225∣54∣=1554
Isse simplify karne par: α=518
Step 5: Final Value
Hame 35α ki value nikalni hai:
35α=35×518
35α=7×18=126
Sahi Answer: (4) 126
Explanation
Step 1: Family of Planes ka use karna
Naya plane (P′) un do planes ki line of intersection se guzarta hai, isliye uski equation hogi:
P′=P1+λP2=0
(4x−y+z−10)+λ(x+y−z−4)=0
Isse rearrange karne par:
(4+λ)x+(λ−1)y+(1−λ)z−(10+4λ)=0
Step 2: Perpendicularity Condition (2π rotation)
Sawal mein diya gaya hai ki plane P ko 2π (90°) rotate kiya gaya hai. Iska matlab purana plane (P) aur naya plane (P′) aapas mein perpendicular hain.
Perpendicular planes ke liye unke normal vectors ka dot product zero hota hai (a1a2+b1b2+c1c2=0):
4(4+λ)+(−1)(λ−1)+1(1−λ)=0
16+4λ−λ+1+1−λ=0
2λ+18=0⟹λ=−9
Step 3: Naye Plane ki Equation
λ=−9 ko P′ ki equation mein rakhein:
(4−9)x+(−9−1)y+(1−(−9))z−(10+4(−9))=0
−5x−10y+10z+26=0
Ya phir: 5x+10y−10z−26=0
Step 4: Distance α Nikalna
Point (2,3,−4) se plane 5x+10y−10z−26=0 ki distance:
