Vectors & 3D GeometrymediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Intersecting Lines on the xy-Plane: a + b = 7 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
If the point of intersection of the lines x+13=y+a5=z+b+17\dfrac{x+1}{3}=\dfrac{y+a}{5}=\dfrac{z+b+1}{7} and x−21=y−b4=z−2a7\dfrac{x-2}{1}=\dfrac{y-b}{4}=\dfrac{z-2a}{7} lies on the xyxy-plane, then the value of a+ba+b is
A22
B55
C77correct
D99
Solution
Step 1: Parameterise the first line (L1L_1) by r1r_1:
P=(3r1−1, 5r1−a, 7r1−b−1).P=(3r_1-1,\ 5r_1-a,\ 7r_1-b-1).
 Step 2: Parameterise the second line (L2L_2) by r2r_2:
Q=(r2+2, 4r2+b, 7r2+2a).Q=(r_2+2,\ 4r_2+b,\ 7r_2+2a).
 Step 3: At the intersection, the xx- and yy-coordinates match:
3r1−1=r2+2 ⇒ r2=3r1−3,3r_1-1=r_2+2\ \Rightarrow\ r_2=3r_1-3,
5r1−a=4r2+b.5r_1-a=4r_2+b.
 Step 4: The point lies on the xyxy-plane, so z=0z=0 on both lines:
7r1−b−1=0 ⇒ 7r1=b+1,7r2+2a=0 ⇒ 2a=−7r2.7r_1-b-1=0\ \Rightarrow\ 7r_1=b+1,\qquad 7r_2+2a=0\ \Rightarrow\ 2a=-7r_2.
 Step 5: Substitute r2=3r1−3r_2=3r_1-3 into 2a=−7r22a=-7r_2 and solve the system with 5r1−a=4r2+b5r_1-a=4r_2+b. This gives
r1=57,b=7r1−1=4,a=3.r_1=\frac57,\qquad b=7r_1-1=4,\qquad a=3.
 Step 6:
a+b=3+4=7.a+b=3+4=7.
 Correct answer: (3)
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