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The symmetric equation of lines $3 x+2 y+z-5=0$ and $x+y-2 z-3=0$, is
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2868 Upvotes
Verified Answer
The correct answer is:
$\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$
Let $a, b, c$ be the direction ratios of required line. $\therefore 3 a+2 b+c=0$ and $a+b-2 c=0$
$\Rightarrow \quad \frac{a}{-4-1}=\frac{b}{1+6}=\frac{c}{3-2}$
$\Rightarrow$
$\frac{a}{-5}=\frac{b}{7}=\frac{c}{1}$
In order to find a point on the required line we put $z=0$ in the two given equations to obtain, $3 x+2 y=5$ 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 $\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$
$\Rightarrow \quad \frac{a}{-4-1}=\frac{b}{1+6}=\frac{c}{3-2}$
$\Rightarrow$
$\frac{a}{-5}=\frac{b}{7}=\frac{c}{1}$
In order to find a point on the required line we put $z=0$ in the two given equations to obtain, $3 x+2 y=5$ 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 $\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}$
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