Question

Find in a symmetrical form, the equations of the line formed by the planes $x+y+z+1=0,4x+y-2z+2=0$ and find its direction-cosines.

• A. $\frac{x-\frac{1}{3}}{1}=\frac{y+\frac{2}{3}}{-2}=\frac{z-0}{1};-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}}$
• B. $\frac{x-\frac{1}{3}}{-1}=\frac{y-\frac{2}{3}}{2}=\frac{z-0}{1};\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}$
• C. $\frac{x+\frac{1}{3}}{-1}=\frac{y+\frac{2}{3}}{2}=\frac{z+0}{-1};-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}$
• D. $\frac{x+\frac{1}{3}}{1}=\frac{y-\frac{2}{3}}{2}=\frac{z+0}{1};\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}}$

Answer 1

Answer: (C)

$\frac{x+\frac{1}{3}}{-1}=\frac{y+\frac{2}{3}}{2}=\frac{z+0}{-1};-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}$

The lines should be $x+y+z=0$

and, $4x+y-2z=(-2)$

Now, equation should lie on both planes.

So, the equation of line should be perpendicular to both planes.

Let the DRs of the line $(a,b,c)$

So, $(a,b,c)\perp (1,1,1)$

and, $(a,b,c)\perp (4,,1,-2)$

$\therefore a+b+c=0$

$4a+b-2c=0$

On solving,

$\frac{a}{-2-1}=\frac{-b}{-2-4}=\frac{c}{1-4}$

$\Rightarrow a=-3,b=6,c=-3$

$\Rightarrow a=-1,b=2,c=-1$

Let the coordinates of the point lying in $xy$ plane $(x,y,0)$

on subtracting $x+y=-1-\left(1\right)$

and $4x+y=-2$

$\Rightarrow -3x=1$

$\Rightarrow x=\frac{-1}{3}$

Putting vbalue of $x$ in $\left(1\right)$

$\frac{-1}{3}+y=-1$

$y=-1+\frac{-1}{3}$

$=\frac{-2}{3}$

Coordinate $(\frac{-1}{3},\frac{-2}{3},0)$

$\therefore Equation=\frac{x+\frac{1}{3}}{-1}=\frac{y+\frac{2}{3}}{2}=\frac{z-0}{-1}$

DC = $\frac{-1}{\sqrt{(-1{)}^2+(2{)}^2+(1{)}^2}},\frac{2}{\sqrt{(-1{)}^2+(2{)}^2+(1{)}^2}},\frac{-1}{\sqrt{(-1{)}^2+(2{)}^2+(1{)}^2}}$

$=(\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}})$