Answer
Verified
396.9k+ views
Hint: To solve this question, we will use the concept of coplanarity of three vectors. Three vectors $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $are said to be coplanar if their scalar triple product is zero. i.e. \[\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = 0\].
Complete step by step answer:
Those vectors which lie on the same plane in a three-dimensional space, are called coplanar vectors in a three-dimensional space. These vectors are parallel to the same plane.
Given that,
\[\overrightarrow a = \hat i + \hat j + m\hat k,\overrightarrow b = \hat i + \hat j + \left( {m + 1} \right)\hat k\] and $\overrightarrow c = \hat i - \hat j + m\hat k$ are coplanar.
We have to find the value of m.
We know that, the scalar triple product in terms of components is,
If \[\overrightarrow a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k,\] \[\overrightarrow b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\] and \[\overrightarrow c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\] be three vectors. Then,
$\left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}} \\
{{b_1}}&{{b_2}}&{{b_3}} \\
{{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right|$.
According to the question, vectors \[\overrightarrow a = \hat i + \hat j + m\hat k,\overrightarrow b = \hat i + \hat j + \left( {m + 1} \right)\hat k\] and $\overrightarrow c = \hat i - \hat j + m\hat k$ are coplanar.
Therefore,
$ \Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}} \\
{{b_1}}&{{b_2}}&{{b_3}} \\
{{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right| = 0$
$ \Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}
1&1&m \\
1&1&{(m + 1)} \\
1&{ - 1}&m
\end{array}} \right| = 0$
Solving this, we get
$
\Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = 1(m - ( - 1)(m + 1)) - 1(m - (m + 1)) + m( - 1 - 1) = 0 \\
\Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = 1(m + m + 1) - 1(m - m - 1) + m( - 2) = 0 \\
\Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = 2m + 1 - 2m = 0 \\
$
Here we can see that we cannot find out the value of m from this.
So, the correct answer is “Option D”.
Note: Whenever we ask such types of questions, we have to remember the concept of coplanarity. We will use the formula of the scalar triple product of three vectors. Then we will equate that scalar triple product to zero because three vectors are coplanar if their scalar triple product is equal to zero. Through this we will get the required unknown value. Then we will get the answer.
Complete step by step answer:
Those vectors which lie on the same plane in a three-dimensional space, are called coplanar vectors in a three-dimensional space. These vectors are parallel to the same plane.
Given that,
\[\overrightarrow a = \hat i + \hat j + m\hat k,\overrightarrow b = \hat i + \hat j + \left( {m + 1} \right)\hat k\] and $\overrightarrow c = \hat i - \hat j + m\hat k$ are coplanar.
We have to find the value of m.
We know that, the scalar triple product in terms of components is,
If \[\overrightarrow a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k,\] \[\overrightarrow b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\] and \[\overrightarrow c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\] be three vectors. Then,
$\left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}} \\
{{b_1}}&{{b_2}}&{{b_3}} \\
{{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right|$.
According to the question, vectors \[\overrightarrow a = \hat i + \hat j + m\hat k,\overrightarrow b = \hat i + \hat j + \left( {m + 1} \right)\hat k\] and $\overrightarrow c = \hat i - \hat j + m\hat k$ are coplanar.
Therefore,
$ \Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}} \\
{{b_1}}&{{b_2}}&{{b_3}} \\
{{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right| = 0$
$ \Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}}
1&1&m \\
1&1&{(m + 1)} \\
1&{ - 1}&m
\end{array}} \right| = 0$
Solving this, we get
$
\Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = 1(m - ( - 1)(m + 1)) - 1(m - (m + 1)) + m( - 1 - 1) = 0 \\
\Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = 1(m + m + 1) - 1(m - m - 1) + m( - 2) = 0 \\
\Rightarrow \left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = 2m + 1 - 2m = 0 \\
$
Here we can see that we cannot find out the value of m from this.
So, the correct answer is “Option D”.
Note: Whenever we ask such types of questions, we have to remember the concept of coplanarity. We will use the formula of the scalar triple product of three vectors. Then we will equate that scalar triple product to zero because three vectors are coplanar if their scalar triple product is equal to zero. Through this we will get the required unknown value. Then we will get the answer.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE