JEE Main 2026 — Apr 2, Shift 1 — Mathematics solution

JEE Main2026Apr 2, Shift 1Mathematics

If the point of intersection of the lines $\frac{x + 1}{3} = \frac{y - a}{5} = \frac{z - b + 1}{7}$ and $\frac{x - 2}{1} = \frac{y - b}{4} = \frac{z - 2 a}{7}$ lies on $x y$ -plane, then the value of $a + b$ is :

A
$2$
B
$5$
C
$7$
D
$9$

Solution

Line $(L_{1})$

$\dfrac{x+1}{3} = \dfrac{y+a}{5} = \dfrac{z+b+1}{7} = r_1$

$\Rightarrow$ General point $P$ on $L_{1}$ is

$(3 r_{1} - 1, 5 r_{1} - a, 7 r_{1} - b - 1)$

Line $(L_{2})$

$\dfrac{x-2}{1} = \dfrac{y-b}{4} = \dfrac{z-2a}{7} = r_2$

$\Rightarrow$ General point $Q$ on $L_{2}$ is

$(r_{2} + 2, 4 r_{2} + b, 7 r_{2} + 2 a)$

For point of intersection,

$3r_1-1=r_2+2 \Rightarrow r_2=3r_1-3 \qquad \cdots (1)$

$5r_1-a=4r_2+b \qquad \cdots (2)$

$7r_1-b-1=7r_2+2a \qquad \cdots (3)$

Since the point lies on the XY-plane,

$\Rightarrow z -coordinate = 0$

From $L_{1}$ :

$7r_1-b-1=0 \Rightarrow 7r_1=b+1$

From $L_{2}$ :

$7r_2+2a=0 \Rightarrow 2a=-7r_2$

Substitute $r_{2} = 3 r_{1} - 3$

$2 a = - 7 (3 r_{1} - 3)$

$\Rightarrow a=\dfrac{-21r_1+21}{2}$

Put in equation (2),

$5 r_{1} - a = 4 r_{2} + b$

$5r_1-\left(\dfrac{-21r_1+21}{2}\right) = 4(3r_1-3)+(7r_1-1)$

$\Rightarrow r_{1} = \frac{5}{7}$

$\Rightarrow b = 7 r_{1} - 1 = 4$

$\Rightarrow a = 3$

$∴ a + b = 7$

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