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Question: The vectors \[\overrightarrow a = \hat i + \hat j + m\hat k,\overrightarrow b = \hat i + \hat j + \l...

The vectors a→=i^+j^+mk^,b→=i^+j^+(m+1)k^\overrightarrow a = \hat i + \hat j + m\hat k,\overrightarrow b = \hat i + \hat j + \left( {m + 1} \right)\hat k and c→=i^−j^+mk^\overrightarrow c = \hat i - \hat j + m\hat k are coplanar if m is equals to
A. 1.
B. 4.
C. 3.
D. None of these.

Explanation

Solution

To solve this question, we will use the concept of coplanarity of three vectors. Three vectors a→,b→,c→\overrightarrow a ,\overrightarrow b ,\overrightarrow c are said to be coplanar if their scalar triple product is zero. i.e. [a→b→c→]=0\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,
a→=i^+j^+mk^,b→=i^+j^+(m+1)k^\overrightarrow a = \hat i + \hat j + m\hat k,\overrightarrow b = \hat i + \hat j + \left( {m + 1} \right)\hat k and c→=i^−j^+mk^\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 a→=a1i^+a2j^+a3k^,\overrightarrow a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k, b→=b1i^+b2j^+b3k^\overrightarrow b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k and c→=c1i^+c2j^+c3k^\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 a→=i^+j^+mk^,b→=i^+j^+(m+1)k^\overrightarrow a = \hat i + \hat j + m\hat k,\overrightarrow b = \hat i + \hat j + \left( {m + 1} \right)\hat k and c→=i^−j^+mk^\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
⇒[a→ b→ c→]=1(m−(−1)(m+1))−1(m−(m+1))+m(−1−1)=0 ⇒[a→ b→ c→]=1(m+m+1)−1(m−m−1)+m(−2)=0 ⇒[a→ b→ c→]=2m+1−2m=0  \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.