If the point of intersection of the lines (x+1)/(3) = (y+a)/(5) = (z+b+1)/(7) and (x-2)/(1…

Question

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-2a}{7} $  lies on xy-plane, then the value of $a+b$ is:

Accepted Answer

Let the common point of intersection of the two lines be $P$.

Since the point lies on the $xy$-plane, its $z$-coordinate must be $0$.

We write both lines in parametric form.

For the first line,

$\frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}=\lambda $

So,

$x=3\lambda -1,\, y=5\lambda -a,\, z=7\lambda -b-1$

For the second line,

$\frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}=\mu $

So,

$x=\mu +2,\, y=4\mu +b,\, z=7\mu +2a$

At the point of intersection, coordinates are equal.

So we get:

$3\lambda -1=\mu +2$

$5\lambda -a=4\mu +b$

$7\lambda -b-1=7\mu +2a$

Also, since the point lies on the $xy$-plane,

$z=0$

Hence from both lines:

$7\lambda -b-1=0$

and

$7\mu +2a=0$

Now solve step by step.

From

$3\lambda -1=\mu +2$

we get

$3\lambda -\mu =3\, \Rightarrow \, \mu =3\lambda -3$

From $z=0$ in the first line,

$b=7\lambda -1$

From $z=0$ in the second line,

$2a=-7\mu \, \Rightarrow \, a=-\frac{7\mu }{2}$

Now use

$5\lambda -a=4\mu +b$

Substitute $\mu =3\lambda -3$, $b=7\lambda -1$:

$5\lambda -a=4(3\lambda -3)+(7\lambda -1)$

$5\lambda -a=12\lambda -12+7\lambda -1$

$5\lambda -a=19\lambda -13$

$-a=14\lambda -13$

$a=13-14\lambda $

Also,

$a=-\frac{7\mu }{2}=-\frac{7(3\lambda -3)}{2}$

$a=\frac{-21\lambda +21}{2}$

Equate the two values of $a$:

$13-14\lambda =\frac{-21\lambda +21}{2}$

Multiply by $2$:

$26-28\lambda =-21\lambda +21$

$5=7\lambda $

$\lambda =\frac{5}{7}$

Then

$\mu =3\lambda -3=3\cdot \frac{5}{7}-3=\frac{15}{7}-\frac{21}{7}=-\frac{6}{7}$

Now find $a$ and $b$.

$b=7\lambda -1=5-1=4$

$a=-\frac{7\mu }{2}=-\frac{7}{2}(-\frac{6}{7})=3$

Therefore,

$a+b=3+4=7$

So the correct answer is:

$7$

Option C

Solutions · 1