Question

Find the correct option
The vectors \[\overrightarrow a = \widehat i + \widehat j + m\widehat k\], \[\overrightarrow b = \widehat i + \widehat j + (m + 1)\widehat k\] and \[\overrightarrow c = \widehat i - \widehat j + m\widehat k\] are coplanar, if \[m\] is equal to
A. \[1\]
B. \[4\]
C. \[3\]
D. no value of \[m\] for which vectors are coplanar

Answer 1

Hint: Using the determinant formula for coplanar vectors, we will establish an equation taking the components of the three given vectors to find the value of \[m\] and then we will choose the correct option.

Formulae Used: Let, \[\overrightarrow a ,\overrightarrow b ,\overrightarrow c \] are three vectors represented as
\[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\] , \[\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\] , \[\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k\] ,
Then, \[\overrightarrow a ,\overrightarrow b ,\overrightarrow c \] will be coplanar, if \[\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} \right| = 0\]

Complete step-by-step solution:
It has been given that \[\overrightarrow a ,\overrightarrow b ,\overrightarrow c \] are three vectors represented as
\[\overrightarrow a = \widehat i + \widehat j + m\widehat k\]
\[\overrightarrow b = \widehat i + \widehat j + (m + 1)\widehat k\]
\[\overrightarrow c = \widehat i - \widehat j + m\widehat k\]
The above three vectors are coplanar i.e. the vectors lie in the same plane or parallel to the same plane.
Now, we have to form a determinant with the components of the given vectors and equate it to zero
to satisfy the above-mentioned formulae for coplanar vectors.
Thus, assigning the numerical values to the variables of the determinant, we have
\[\begin{array}{l}{a_1} = 1\\{a_2} = 1\\{a_3} = m\end{array}\] , \[\begin{array}{l}{b_1} = 1\\{b_2} = 1\\{b_3} = m + 1\end{array}\] and \[\begin{array}{l}{c_1} = 1\\{c_2} = - 1\\{c_3} = m\end{array}\]
Substituting the corresponding values of the variables in the formula, we have
\[\left| {\begin{array}{*{20}{c}}1&1&1\\1&1&{ - 1}\\m&{m + 1}&m\end{array}} \right| = 0\]
\[ \Rightarrow 1\left| {\begin{array}{*{20}{c}}1&{ - 1}\\{m + 1}&m\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}1&{ - 1}\\m&m\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}1&1\\m&{m + 1}\end{array}} \right| = 0\]
Further solving the above, we have
\[1(m + m + 1) - 1(m + m) - 1(m + 1 - m) = 0\]
\[ \Rightarrow 2m + 1 - 2m - 1 = 0\]
\[ \Rightarrow 0 = 0\]
It is observed that the value of \[m\] can not be found from the above equation.
So, there is no value of \[m\], for which, the vectors are coplanar.

Hence, option D. is the correct answer.

Note: Any three vectors will be coplanar, if their scalar triple product is equal to zero as the parallelepiped defined by them would be flat and have no volume. The scalar product of three vectors is given by the determinant of the components of the three vectors.