Question

If the vectors \[2\widehat{i}+\widehat{j}-\widehat{k},-\widehat{i}+2\widehat{j}+\lambda \widehat{k}\] and \[-5\widehat{i}+2\widehat{j}-\widehat{k}\] are coplanar, then the value of \[\lambda \] is equal to
A. \[-13\]
B. \[\dfrac{13}{9}\]
C. \[-\dfrac{13}{9}\]
D. \[-\dfrac{9}{13}\]

Answer 1

Hint: In this question, we have to find the value of \[\lambda \]. In a three-dimensional space when vectors lie on the same plane, then they are called Coplanar vectors. These vectors are parallelly aligned to the same plane. When the scalar triple product is zero for the given three vectors in 3d-space, then the vectors are coplanar

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,
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\]

Complete step by step solution: It is given that,
\[\begin{align}
  & \overrightarrow{a}=2\widehat{i}+\widehat{j}-\widehat{k} \\
 & \overrightarrow{b}=-\widehat{i}+2\widehat{j}+\lambda \widehat{k} \\
 & \overrightarrow{c}=-5\widehat{i}+2\widehat{j}-\widehat{k} \\
\end{align}\]
These three vectors are coplanar. Then, \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=0\]
On substituting,
\[\begin{align}
  & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right| \\
 & \text{ }=\left| \begin{matrix}
   2 & 1 & -1 \\
   -1 & 2 & \lambda \\
   -5 & 2 & -1 \\
\end{matrix} \right| \\
 & \text{ }=2(-2-2\lambda )-1(1+5\lambda )-1(-2+10) \\
 & \text{ }=-4-4\lambda -1-5\lambda +2-10 \\
 & \text{ }=-9\lambda -13 \\
\end{align}\]
Thus,
\[\begin{align}
  & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-9\lambda -13=0 \\
 & \Rightarrow 9\lambda =-13 \\
 & \Rightarrow \lambda =\dfrac{-13}{9} \\
\end{align}\]

Thus, Option (C) is correct.

Note: To solve this problem conditions for coplanarity must be remembered i.e., if three vectors are coplanar, then their scalar triple product is zero. One should know the concept of dot product and cross product before tackling such questions.