Question

The equation of the line $x+y+z-1=0,4x+y-2z+2=0$ written in the symmetrical form is:

• A. $\frac{x+1}{1}=\frac{y-2}{-2}=\frac{z-0}{1}$
• B. $\frac{x}{1}=\frac{y}{-2}=\frac{z-1}{1}$
• C. $\frac{x+1/2}{1}=\frac{y-1}{-2}=\frac{z-1/2}{1}$
• D. $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z-2}{2}$

Answer 1

Answer: (A), (B), (C)

$\frac{x+1}{1}=\frac{y-2}{-2}=\frac{z-0}{1}$

$\frac{x+1/2}{1}=\frac{y-1}{-2}=\frac{z-1/2}{1}$

$\frac{x}{1}=\frac{y}{-2}=\frac{z-1}{1}$

The equation of a line passing through $(x_0,y_0,z_0)$ and direction ratio $(a,b,c)$ is given by $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$. The line of intersection of plane perpendicular to the normal vectors of the plane.$(i+j+k)\times (4i+j-2k)=-3(i-2j+k).$

Hence the direction ratio of the line is given by $(1,-2,1)$.

Put $z=0$ on the equation of the planes we get $x+y=1$ and $4x+y=-2$ solving we get $x=-1,y=2$.

Hence any point on the line given by $(-1,2,0)+t(1,-2,1)$.

Plugging $t=1$ get the point $(0,0,1)$ and plugging $t=1/2$ get the point $(-\frac{1}{2},1,\frac{1}{2})$.

Hence the options A,B,C are correct.

Clearly option D is wrong since the given the direction ratios different.