Question

The value of '$a$' for which the lines $\frac{x-2}{1}=\frac{y-9}{2}=\frac{z-13}{3}$ and $\frac{x-a}{-1}=\frac{y-7}{2}=\frac{z+2}{-3}$ intersect, is

• A. $-5$
• B. $-2$
• C. $5$
• D. $-3$

Answer 1

The correct option is D $-3$

Let the intersection point of two given lines be $(x,y,z)$ therefore;
$\frac{z-13}{3}=\frac{z+2}{-3}$
on solving, we get
$z=\frac{11}{2}$
Now,
$\frac{x-a}{-1}=\frac{y-7}{2}=\frac{z+2}{-3}=r$

$\therefore$ $\frac{x-2}{1}=\frac{z-13}{3}$

$\frac{x-2}{1}=\frac{\frac{11}{2}-13}{3}$
On solving, we get
$x=\frac{-1}{2}$

Now putting the value of $x$ in other equation of line
$\frac{x-a}{-1}=\frac{z+2}{-3}$

$\frac{\frac{-1}{2}-a}{-1}=\frac{\frac{11}{2}+2}{-3}$
On solving, we get
$a=-3$