Question

The symmetric form of the equation of the line x + y – z = 1, 2x – 3y + z = 2 is

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

Answer 1

The correct option is C $\frac{x-1}{2}=\frac{y}{3}=\frac{z}{5}$

To find one point on the line, y = z = 0, then x = 1
suppose direction cosines of line of intersection are l, m, n
Then,
l + m – n = 0, because line and the normal of first plane are perpendicular
2l – 3m + n = 0, because line and the normal of second plane are perpendicular
Then $\frac{l}{2}=\frac{m}{3}=\frac{n}{5}$
$\therefore$ Equation of line in symmetric form is $\frac{x-1}{2}=\frac{y}{3}=\frac{z}{5}$