Question

Assertion :The lines $\frac{x-1}{1}=\frac{y}{-1}=\frac{z+1}{1}$ and $\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z}{3}$ are coplanar and the equation of the plane containing them is $5x+2y-3z-8=0$ Reason: The line $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z}{3}$ is perpendicular to the plane $3x+6y+9z-8=0$ and parallel to the plane $x+y-z=0$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect and reason is correct

Answer 1

The correct option is D Assertion is incorrect and reason is correct

$\frac{x-1}{1}=\frac{y-0}{-1}=\frac{z+1}{1}=L_1$

$\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z}{3}=L_2$

For checking co-planarity

$\left|\begin{array}{ccc}(1-2)& (0-(-1\left)\right)& (-1-0)\\ 1& -1& 1\\ 2& -2& 3\end{array}\right|$

$=\left|\begin{array}{ccc}-1& 1& -1\\ 1& -1& 1\\ 2& -2& 3\end{array}\right|=0$

So it is co-planer

So, normal of plane is parallel

$\overrightarrow{L_1}\times \overrightarrow{L_2}$ i.e. $(i-j+k)\times (2i-2j+3k)$

i.e. $-i-j$

so equation of plane is $-x-y+0.z=c$

as point on line is on the plane

i.e. $(1,0,-1)$ is on the plane,so we get c=-1

So $-x-y=-1$

$x+y=1$ is the plane

Reason:

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

drs of line are $(1,2,3)$

$x+y-z=0$, drs of normal are$(1,1,-1)$

Dot produst of drs $=1.1+2.1+3.-1=0$

So line and normal are perpendicular.

$\Rightarrow$line$\parallel$plane $x+y-z=0$

Drs of normal to plane$3x+6y+9z-8=0$ are $(1,2,3)$

$\Rightarrow$ line $\parallel$normal

$\Rightarrow$ line$\perp$ plane $3x+6y+9z-8=0$

So Assertion is false, Reason is true.