Question

The symmetric equation of line by the planes $3x+2y+z-5=0$ and $x+y-2z-3=0$, is

• A. $\frac{x-1}{5}=\frac{y-4}{7}=\frac{z-0}{1}$
• B. $\frac{x+1}{5}=\frac{y+4}{7}=\frac{z-0}{1}$
• C. $\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$
• D. $\frac{x-1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$

Answer 1

The correct option is C $\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$

Let $a,b,c$ be the direction ratios of required line,
$\therefore 3a+2b+c=0$ and $a+b-2c=0$
$\Rightarrow \frac{a}{-4-1}=\frac{b}{1+6}=\frac{c}{3-2}$
$\Rightarrow \frac{a}{-5}=\frac{b}{7}=\frac{c}{1}$
In order to find a point on the required line we put $z=0$ in the two given equations to obtain, $3x+2y=5$ and $x+y=3$. Solving these two equations, we get $x=-1,y=4$.
$\therefore$ Coordinates of point on required line are $(-1,4,0)$.
Hence, required line is $\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$