JEE Main 2026 · 2 Apr · Shift 1 · Mathematics Q13

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MathematicsSingle correct+4 / -1Question 13 of 75

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

Answer and worked solution

Official answer: C

General points. Line 1: $(-1+3s,\,-a+5s,\,-b-1+7s)$ . Line 2: $(2+t,\,b+4t,\,2a+7t)$ .

On the xy-plane ( $z=0$ ): $s=\frac{b+1}7$ and $t=-\frac{2a}7$ .

Equate x and y.

$x$ : $-1+\frac{3(b+1)}7=2-\frac{2a}7$ , so $2a+3b=18$ .
$y$ : $-a+\frac{5(b+1)}7=b-\frac{8a}7$ , so $a-2b=-5$ .

Solve. $b=4$ , $a=3$ , so $a+b=7$ .

(C)

Topics:Equation of a line in space

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JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 13