Question

Let \[a=i-k,b=xi+j+(1-x)k,c=yi+xj+(1+x-y)k\]. Then $[abc]$ depends on
[IIT Screening 2001; AIEEE 2005]
A. Only $x$
B. Only $y$
C. Neither $x$ nor $y$
D. Both $x$ and $y$

Answer 1

Hint: Solve for $[abc]$ with the given values of a, b and c. Then check for the dependency. If you get any integer value then that means $[abc]$ depends neither $x$ nor $y$ .



Formula Used:\[(a\times b).c=[abc]=\left| \begin{matrix}
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right|\]



Complete step by step solution:Given that, \[a=i-k,b=xi+j+(1-x)k,c=yi+xj+(1+x-y)k\]
Therefore,
$[abc]=\left| \begin{matrix}
   1 & 0 & -1 \\
   x & 1 & 1-x \\
   y & x & 1+x-y \\
\end{matrix} \right|$
By applying ${{C}_{3}}\to {{C}_{3}}+{{C}_{1}}$ we get,
$[abc]=\left| \begin{matrix}
   1 & 0 & 0 \\
   x & 1 & 1 \\
   y & x & 1+x \\
\end{matrix} \right|$
$[abc]=1(1+x-x)-0(x(1+x)-y)+0({{x}^{2}}-y)$
$[abc]=1+x-x=1$
Hence, $[abc]$ depends neither $x$ nor $y$.



Option ‘C’ is correct

Note: While simplifying the determinants other transitions can also be applied but in this question, this is the simplest and easiest one. And while solving the determinant after simplification try not to do silly mistakes.