JEE Main 2014 · 12 Apr · Shift 1 · Mathematics Q83

MathematicsSingle correct+4 / -1Question 83 of 90

A symmetrical form of the line of intersection of the planes $x=ay+b$ and $z=cy+d$ is :

\dfrac{x-b}a=\dfrac{y-1}1=\dfrac{z-d}c

\dfrac{x-b-a}a=\dfrac{y-1}1=\dfrac{z-d-c}c

\dfrac{x-a}b=\dfrac{y-0}1=\dfrac{z-c}d

\dfrac{x-b-a}b=\dfrac{y-1}0=\dfrac{z-d-c}d

Answer and worked solution

Official answer: B

Direction: $x=ay+b$ , $z=cy+d$ give direction $(a,1,c)$ . At $y=1$ the point is $(a+b,1,c+d)$ :
$\frac{x-b-a}a=\frac{y-1}1=\frac{z-d-c}c.$

JEE Main 2014 · 12 Apr · Shift 1, Mathematics question 83