Question

The equation of the line which passes through the point $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ is

• A. $\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}$
• B. $\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{-5}$
• C. $\frac{x-1}{-2}=\frac{y-1}{1}=\frac{z-1}{-4}$
• D. $\frac{x-1}{8}=\frac{y-1}{-2}=\frac{z-1}{3}$

Answer 1

Answer: (A)

Any line passing through the point $(1,1,1)$ is
$\frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}\dots$(i)
This line intersects the line
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$.
If $a:b:c\ne 2:3:4$ and
$\left|\begin{array}{ccc}1-1& 2-1& 3-1\\ a& b& c\\ 2& 3& 4\end{array}\right|=0$
$\Rightarrow a-2b+c=0\dots$(ii)
Again, line (i) intersects line
$\frac{x-(-2)}{1}=\frac{y-3}{2}=\frac{z-(-1)}{4}$
If $a:b:c\ne 2:3:4$ and
$\left|\begin{array}{ccc}-2-1& 3-1& -1-1\\ a& b& c\\ 1& 2& 4\end{array}\right|=0$
$6a+5b-4c=0\dots$(iii)
From (ii) and (iii) by cross multiplication, we have
$\frac{a}{8-5}=\frac{b}{6+4}=\frac{c}{5+12}$ or
$\frac{a}{3}=\frac{b}{10}=\frac{c}{17}$
So, the required line is
$\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}$.