Question

If the given vectors \[(-bc,{{b}^{2}}+bc,{{c}^{2}}+bc)\], \[({{a}^{2}}+ac,-ac,{{c}^{2}}+ac)\], and \[({{a}^{2}}+ab,{{b}^{2}}+ab,-ab)\] are coplanar, where none of $a$, $b$, and $c$ is zero, then
A. \[({{a}^{2}}+{{b}^{2}}+{{c}^{2}})=1\]
B. \[(bc+ca+ab)=0\]
C. \[(a+b+c)=0\]
D. \[({{a}^{2}}+{{b}^{2}}+{{c}^{2}})=(bc+ca+ab)\]

Answer 1

Hint: In the above question, we need to find the condition that holds true if we solve the three given coplanar vectors. In order to find which condition holds true on solving the given vectors we should know the concept of coplanarity, and the matrix method of determination of determinants of vectors.

Formula used: Scalar triple product of three vectors:
We have the vectors \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] as
\[\begin{align}
  & \overrightarrow{a}={{a}_{1}}\overrightarrow{i}+{{a}_{2}}\overrightarrow{j}+{{a}_{3}}\overrightarrow{k} \\
 & \overrightarrow{b}={{b}_{1}}\overrightarrow{i}+{{b}_{2}}\overrightarrow{j}+{{b}_{3}}\overrightarrow{k} \\
 & \overrightarrow{c}={{c}_{1}}\overrightarrow{i}+{{c}_{2}}\overrightarrow{j}+{{c}_{3}}\overrightarrow{k} \\
\end{align}\]
Then, the triple product is calculated by,
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\]
Thus, the volume of a 3D structure is calculated by $V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]$

Complete step by step solution: Here, we are given the three vectors which are coplanar
\[\begin{align}
  & (-bc,{{b}^{2}}+bc,{{c}^{2}}+bc) \\
 & ({{a}^{2}}+ac,-ac,{{c}^{2}}+ac) \\
 & ({{a}^{2}}+ab,{{b}^{2}}+ab,-ab) \\
\end{align}\]
Then, by the scalar triple product of these vectors, we get
\[\begin{align}
  & [\overrightarrow{A}\text{ }\overrightarrow{B}\text{ }\overrightarrow{C}]=\left| \begin{matrix}
   -bc & {{b}^{2}}+bc & {{c}^{2}}+bc \\
   {{a}^{2}}+ac & -ac & {{c}^{2}}+ac \\
   {{a}^{2}}+ab & {{b}^{2}}+ab & -ab \\
\end{matrix} \right| \\
 & =-bc\left[ (-ac\times -ab)-(({{c}^{2}}+ac)({{b}^{2}}+ab)) \right]- \\
 & ({{b}^{2}}+bc)\left[ ({{a}^{2}}+ac)(-ab)-({{a}^{2}}+ab)({{c}^{2}}+ac) \right]+ \\
 & ({{c}^{2}}+bc)\left[ ({{a}^{2}}+ac)({{b}^{2}}+ab)-({{a}^{2}}+ab)(-ac) \right] \\
 & =-bc\left[ {{a}^{2}}bc-({{c}^{2}}{{b}^{2}}+ab{{c}^{2}}+a{{b}^{2}}c+{{a}^{2}}bc) \right]- \\
 & ({{b}^{2}}+bc)\left[ (-{{a}^{3}}b-{{a}^{2}}bc)-({{a}^{2}}{{c}^{2}}+{{a}^{3}}c+ab{{c}^{2}}+{{a}^{2}}bc) \right]+ \\
 & ({{c}^{2}}+bc)\left[ ({{a}^{2}}{{b}^{2}}+{{a}^{3}}b+a{{b}^{2}}c+{{a}^{2}}bc)-(-{{a}^{3}}c-{{a}^{2}}bc) \right] \\
 & =-bc({{a}^{2}}bc-{{c}^{2}}{{b}^{2}}-ab{{c}^{2}}-a{{b}^{2}}c-{{a}^{2}}bc)- \\
 & ({{b}^{2}}+bc)(-{{a}^{3}}b-{{a}^{2}}bc-{{a}^{2}}{{c}^{2}}-{{a}^{3}}c-ab{{c}^{2}}-{{a}^{2}}bc)+ \\
 & ({{c}^{2}}+bc)({{a}^{2}}{{b}^{2}}+{{a}^{3}}b+a{{b}^{2}}c+{{a}^{2}}bc+{{a}^{3}}c+{{a}^{2}}bc) \\
\end{align}\]
\[\begin{align}
  & =-bc(-{{c}^{2}}{{b}^{2}}-ab{{c}^{2}}-a{{b}^{2}}c)- \\
 & ({{b}^{2}}+bc)(-{{a}^{3}}b-2{{a}^{2}}bc-{{a}^{2}}{{c}^{2}}-{{a}^{3}}c-ab{{c}^{2}})+ \\
 & ({{c}^{2}}+bc)({{a}^{2}}{{b}^{2}}+{{a}^{3}}b+a{{b}^{2}}c+2{{a}^{2}}bc+{{a}^{3}}c) \\
 & =bc({{c}^{2}}{{b}^{2}}+ab{{c}^{2}}+a{{b}^{2}}c)+ \\
 & ({{b}^{2}}+bc)({{a}^{3}}b+2{{a}^{2}}bc+{{a}^{2}}{{c}^{2}}+{{a}^{3}}c+ab{{c}^{2}})+ \\
 & ({{c}^{2}}+bc)({{a}^{2}}{{b}^{2}}+{{a}^{3}}b+a{{b}^{2}}c+2{{a}^{2}}bc+{{a}^{3}}c) \\
 & ={{(ab+bc+ca)}^{3}} \\
\end{align}\]
Since the determinant of the coplanar matrix is equal to zero, we can write
$\begin{align}
  & {{(ab+bc+ca)}^{3}}=0 \\
 & (ab+bc+ca)=0 \\
\end{align}$

Thus, Option (B) is correct.

Additional Information: Important vector identities for solving vector equations are:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
\[\begin{align}
  & \overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1 \\
 & \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k} \\
 & \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\
 & \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j} \\
\end{align}\]

Note: We can see from the above concept that the value of the determinant of the coefficient matrix is 0, which is calculated using the concept of coplanarity and the matrix method of determination of determinants of vectors.