Question

The point of intersection of the lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}and\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$ is

• A. $(\frac{1}{2},\frac{1}{2},-\frac{3}{2})$
• B. $(-\frac{1}{2},-\frac{1}{2},\frac{3}{2})$
• C. $(\frac{1}{2},-\frac{1}{2},-\frac{3}{2})$
• D. $(-\frac{1}{2},\frac{1}{2},\frac{3}{2})$

Answer 1

The correct option is C $(\frac{1}{2},-\frac{1}{2},-\frac{3}{2})$

Let $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda ....\left(i\right)$
Then $x=3\lambda -1,y=5\lambda -3,z=7\lambda -5$
General point on this line is ($3\lambda -1,5\lambda -3,7\lambda -5)$
Again $let\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu$ ...(ii)
Then $x=\mu +2,y=3\mu +4,z=5\mu +6$
A general point on this line is $(\mu +2,3\mu +4,5\mu +6)$
For intersection, they have a common point, for some values of $\lambda$ and $\mu$, we must have $(3\lambda -1)=(\mu +2),(5\lambda -3)=(3\mu +4),(7\lambda -5)=(5\mu +6)$
From first two we have, $\mu =3\lambda -3$ ....(iii)
and $3\mu =5\lambda -7$ ...(iv)
From (iii), put the values of $\mu$ in (iv), we have $3(3\lambda -3)=5\lambda -7$
$\Rightarrow 9\lambda -9=5\lambda -7or4\lambda =2or\lambda =\frac{1}{2}$
Put $\lambda =\frac{1}{2}$ in (iii), we get $\mu =-\frac{3}{2}(Putting\lambda =\frac{1}{2})$
The required point of intersection is
$[\frac{3}{2}-1],[\frac{5}{2}-3],[\frac{7}{2}-5]=[\frac{1}{2},\frac{1}{2},-\frac{3}{2}].$