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 $a: b: c \neq 2: 3: 4$ and
$\begin{vmatrix}1-1 & 2-1 & 3-1 \\ a & b & c \\ 2 & 3 & 4\end{vmatrix}=0$
$\Rightarrow a-2 b+c=0 \dots$(ii)
Again, line (i) intersects line
$\frac{x-(-2)}{1}=\frac{y-3}{2}=\frac{z-(-1)}{4}$
If $a: b: c \neq 2: 3: 4$ and
$\begin{vmatrix}-2-1 & 3-1 & -1-1 \\ a & b & c \\ 1 & 2 & 4\end{vmatrix}=0$
$6 a+5 b-4 c=0\dots$(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
$\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}$.