Question

If the vectors $a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+c\hat{k},$ $(a\ne b\ne c\ne 1)$ are coplanar, then find the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$.

• A. 0
• B. 1
• C. Not defined
• D. None of the above

Answer 1

Answer: (B)

1

Since $ai+j+k,i+bj+k$ and $i+j+ck=0$ are coplanar, there exist scalar x,y,z (not all zero) such that
$x(ai+j+k)+y(i+bj+k)+z(i+j+ck)=0\Rightarrow (ax+y+z)i+(x+by+z)j+(x+y+cz)k=0\Rightarrow ax+y+z=0,x+by+z=0,x+y+cz=0$
Since at least one of x,y,z is different from zero, the above system must have a non-zero solution.
$\therefore \left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& b\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}a& 1& 1\\ 1-2& b-1& 0\\ 1-a& 0& c-1\end{array}\right|=0\Rightarrow a(b-1)(c-1)-(1-a)(b-1)-(c-1)(1-a)=0$
Dividing by $(1-a)(1-b)(1-c)$, we get
$\frac{a}{1-a}+\frac{1}{1-c}+\frac{1}{1-b}=0$
$\Rightarrow \frac{1-(1-a)}{1-a}+\frac{1}{1-c}+\frac{1}{1-b}=0$
$\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$