Question

If a, b, c are any three vectors and their inverse are \[{{\bf{a}}^{ - 1}},{{\bf{b}}^{ - 1}},{{\bf{c}}^{ - 1}}\] and \[[{\bf{abc}}] \ne 0\], then \[\left[ {{{\bf{a}}^{ - 1}}{{\bf{b}}^{ - 1}}{{\bf{c}}^{ - 1}}} \right]\]
A. Zero
B. One
C. Non-zero
D. \[\left[ {\begin{array}{*{20}{l}}{\rm{a}}&{{\rm{c}}]}\end{array}} \right.\]

Answer 1

Hint: Perhaps the following definition of the inverse vector will be useful: An inverse rectilinear vector' is a vector that is co-directed (in the same direction as) a vector but differs in magnitude from it as follows: . We could treat vectors like any other variable if vector multiplication was invertible. We could do basic algebra. The cancellations make sense in general. However, invertible vector multiplication exists and may be demonstrated.



Formula Used:Dot product of two vectors formula
\[\vec a \cdot \vec b = |a||b|\cos \theta \]
Here, \[|a|\]is the magnitude of \[\vec a\]
\[|b|\]is the magnitude of \[\vec b\]



Complete step by step solution:We have been given in the question that a, b, c are any three vectors and their inverse are \[{{\bf{a}}^{ - 1}},{{\bf{b}}^{ - 1}},{{\bf{c}}^{ - 1}}\] and \[[{\bf{abc}}] \ne 0\]
Now, we have to write the inverse of each vector as follows:
 \[{{\bf{a}}^{ - 1}} = \dfrac{{{\bf{b}} \times {\bf{c}}}}{{[{\bf{abc}}]}},\left| {{{\bf{c}}^{ - 1}} = \dfrac{{{\bf{a}} \times {\bf{b}}}}{{[{\bf{abc}}]}},} \right|{{\bf{b}}^{ - 1}} = \dfrac{{{\bf{c}} \times {\bf{a}}}}{{[{\bf{abc}}]}}\]
Now, on multiplying the all the three vectors, we have
\[ \Rightarrow \left[ {{{\bf{a}}^{ - 1}}{{\bf{b}}^{ - 1}}{{\bf{c}}^{ - 1}}} \right] = \dfrac{{({\bf{b}} \times {\bf{c}})}}{{[{\bf{abc}}]}} \cdot \left( {\dfrac{{({\bf{c}} \times {\bf{a}})}}{{[{\bf{abc}}]}} \times \dfrac{{({\bf{a}} \times {\bf{b}})}}{{[{\bf{abc}}]}}} \right)\]
According to the given data, the above expression can be restructured by the concept of dot product as below, we obtain
\[ = \dfrac{{{\bf{b}} \times {\bf{c}}}}{{[{\bf{abc}}]}} \cdot \left[ {\dfrac{{\bf{a}}}{{[{\bf{abc}}]}}} \right] = \dfrac{1}{{[{\bf{abc}}]}} \ne 0\]
Therefore, if a, b, c are any three vectors and their inverse are \[{{\bf{a}}^{ - 1}},{{\bf{b}}^{ - 1}},{{\bf{c}}^{ - 1}}\] and \[[{\bf{abc}}] \ne 0\], then \[\left[ {{{\bf{a}}^{ - 1}}{{\bf{b}}^{ - 1}}{{\bf{c}}^{ - 1}}} \right]\]will be \[\dfrac{1}{{[{\bf{abc}}]}} \ne 0\]
That is, it is Non Zero.



Option ‘C’ is correct

Note:The possibility of ambiguity in this question exists at the moment where we have utilized\[(\vec a \cdot \vec b) \cdot \vec c = 0 \Rightarrow \vec a \cdot \vec b = 0\]. Because \[\left| {\overrightarrow c } \right|\] is a unit vector, this is conceivable. ‘\[\left| {\overrightarrow c } \right|\]’ is the unit vector \[\left| {\overrightarrow c } \right| = 1\]. So the only remaining possibility is that \[\overrightarrow a \cdot \overrightarrow b = 0\]. Examining scalars and vectors yields the important understanding. We'll look at their magnitudes and concentrate on the units in which they're measured.