Question

If $xi+j+k$ , $i+yj+k$ and $i+j+zk$ are co-planar and $x\ne 1$,$y\ne 1$, and $z\ne 1$, then the value of $\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}$ is:

• A. $-1$
• B. $-3$
• C. $2$
• D. $-2$

Answer 1

Answer: (D)

$-2$

Given that $xi+j+k$ , $i+yj+k$ and $i+j+zk$ are coplanar

So we get $\left|\begin{array}{ccc}x& 1& 1\\ 1& y& 1\\ 1& 1& z\end{array}\right|=0$

$\Rightarrow x(yz-1)-1(z-1)+1(1-y)=0$

$\Rightarrow xyz-x-y-z+2=0$

$\Rightarrow x+y+z=xyz+2$

We have to find the value of $\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}$

$\Rightarrow \frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}=\frac{x+y+z+3xyz-2(xy+yz+zx)}{(1-x)(1-y)(1-z)}=\frac{-2(xy+yz+zx-2xyz)}{(1-x)(1-y)(1-z)}$

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

Therefore option $D$ is correct