Question
Let $$\vec a = \hat i - \hat k,\,\vec b = x\hat i + \hat j + \left( {1 - x} \right)\hat k$$ and $$\vec c = y\hat i + x\hat j + \left( {1 + x - y} \right)\,\hat k.$$ Then $$\left[ {\vec a,\,\vec b,\,\vec c} \right]$$ depends on :
A.
only $$y$$
B.
only $$x$$
C.
both $$x$$ and $$y$$
D.
neither $$x$$ nor $$y$$
Answer :
neither $$x$$ nor $$y$$
Solution :
\[\begin{array}{l}
\vec a = \hat i - \hat k,\,\vec b = x\hat i + \hat j + \left( {1 - x} \right)\hat k\,\,{\rm{and}}\\
\vec c = y\hat i + x\hat j + \left( {1 + x - y} \right)\,\hat k\\
\left[ {\vec a,\,\vec b,\,\vec c} \right] = \vec a.\vec b \times \vec c = \left| \begin{array}{l}
1\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\, - 1\\
x\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,1 - x\\
y\,\,\,\,\,\,\,\,x\,\,\,\,1 + x - y
\end{array} \right|\\
= 1\left[ {1 + x - y - x + {x^2}} \right] - \left[ { - {x^2} - y} \right]\\
= 1 - y + {x^2} - {x^2} + y = 1
\end{array}\]
Hence $$\left[ {\vec a,\,\vec b,\,\vec c} \right]$$ is independent of $$x$$ and $$y$$ both.