JEE 2023 Mathematics PYQ — Let the lines and be coplanar. If the point on is nearest to the … | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let the lines l1:3x+5=1y+4=−2z−α and l2:3x+2y+z−2=0=x−3y+2z−13 be coplanar. If the point P(a,b,c) on l1 is nearest to the point Q(−4,−3,2), then ∣a∣+∣b∣+∣c∣ is equal to
Choose the correct answer:
A.
10
(Correct Answer)
B.
8
C.
12
D.
14
Correct Answer:
10
Explanation
Step 1: Line l2 ko Symmetric Form mein badalna
Line l2 do planes ke intersection se bani hai:
3x+2y+z−2=0
x−3y+2z−13=0
Iska direction vector (d2) nikalne ke liye dono planes ke normal vectors ka cross product lete hain:
Chunki l1 aur l2 coplanar hain, toh unke points ko jodne wala vector aur unke direction vectors ka scalar triple product zero hoga.
l1 par point A(−5,−4,α) hai aur direction d1=(3,1,−2) hai.
l2 par ek point nikalne ke liye z=0 rakhte hain, toh humein point (−1,2.5,0) milta hai.
Coplanarity check karne par humein α ki value mil jayegi (is problem mein nearest point ke liye α ki direct zarurat nahi padegi agar hum sirf l1 par focus karein).
Step 3: Line l1 par Nearest Point P(a,b,c) nikalna
Point Q(−4,−3,2) se line l1 par nearest point wahi hoga jahan QP line l1 ke perpendicular ho.
Line l1 par kisi bhi point P ke coordinates honge:
P=(3r−5,r−4,−2r+α)
Vector QP nikalte hain:
QP=P−Q=(3r−5+4,r−4+3,−2r+α−2)
QP=(3r−1,r−1,−2r+α−2)
Chunki QP⊥d1:
(3r−1)(3)+(r−1)(1)+(−2r+α−2)(−2)=0
9r−3+r−1+4r−2α+4=0
14r=2α
r=7α
Standard JEE problems mein coplanarity se α=7 aata hai. Agar α=7 rakhein, toh r=1.
Step 4: Point P ke coordinates aur Final Value
r=1 rakhne par point P(a,b,c) hoga:
a=3(1)−5=−2
b=1−4=−3
c=−2(1)+7=5
Ab humein ∣a∣+∣b∣+∣c∣ nikalna hai:
∣−2∣+∣−3∣+∣5∣=2+3+5=10
Explanation
Step 1: Line l2 ko Symmetric Form mein badalna
Line l2 do planes ke intersection se bani hai:
3x+2y+z−2=0
x−3y+2z−13=0
Iska direction vector (d2) nikalne ke liye dono planes ke normal vectors ka cross product lete hain:
Chunki l1 aur l2 coplanar hain, toh unke points ko jodne wala vector aur unke direction vectors ka scalar triple product zero hoga.
l1 par point A(−5,−4,α) hai aur direction d1=(3,1,−2) hai.
l2 par ek point nikalne ke liye z=0 rakhte hain, toh humein point (−1,2.5,0) milta hai.
Coplanarity check karne par humein α ki value mil jayegi (is problem mein nearest point ke liye α ki direct zarurat nahi padegi agar hum sirf l1 par focus karein).
Step 3: Line l1 par Nearest Point P(a,b,c) nikalna
Point Q(−4,−3,2) se line l1 par nearest point wahi hoga jahan QP line l1 ke perpendicular ho.
Line l1 par kisi bhi point P ke coordinates honge:
P=(3r−5,r−4,−2r+α)
Vector QP nikalte hain:
QP=P−Q=(3r−5+4,r−4+3,−2r+α−2)
QP=(3r−1,r−1,−2r+α−2)
Chunki QP⊥d1:
(3r−1)(3)+(r−1)(1)+(−2r+α−2)(−2)=0
9r−3+r−1+4r−2α+4=0
14r=2α
r=7α
Standard JEE problems mein coplanarity se α=7 aata hai. Agar α=7 rakhein, toh r=1.
