Question

Let \[\vec{A}=\hat{i}+\hat{j}+\hat{k},\vec{B}=\hat{i},\vec{C}={{C}_{1}}\hat{i}+{{C}_{2}}\hat{j}+{{C}_{3}}\hat{k}\]. If ${{C}_{2}}=-1$ and ${{C}_{3}}=1$, then to make three vectors coplanar [AMU 2000]
A. ${{C}_{1}}=0$
B. ${{C}_{1}}=1$
C. ${{C}_{1}}=2$
D. No value of ${{C}_{1}}$ can be found

Answer 1

Hint: 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.
Conditions for the Coplanar Vectors
· When the scalar triple product is zero for the given three vectors in 3d-space, then the vectors are coplanar.
· When in 3d-space vectors are linearly independent, then the vectors are coplanar.



Formula Used:For three vectors to be coplanar, their triple product should be equals to zero i.e.,
$\vec{A}.(\vec{B}\times \vec{C})=0$



Complete step by step solution:Given that \[\vec{A}=\hat{i}+\hat{j}+\hat{k},\vec{B}=\hat{i},\vec{C}={{C}_{1}}\hat{i}+{{C}_{2}}\hat{j}+{{C}_{3}}\hat{k}\]
For three vectors to be coplanar, their triple product should be equals to zero i.e.,
$\vec{A}.(\vec{B}\times \vec{C})=0$
If ${{C}_{2}}=-1$ and ${{C}_{3}}=1$, then we have
\[\vec{C}={{C}_{1}}\hat{i}-\hat{j}+\hat{k}\]
Therefore,
\[(\hat{i}+\hat{j}+\hat{k}).[\hat{i}\times ({{C}_{1}}\hat{i}-\hat{j}+\hat{k})]=0\]
\[\Rightarrow (\hat{i}+\hat{j}+\hat{k}).\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   1 & 0 & 0 \\
   {{C}_{1}} & -1 & 1 \\
\end{matrix} \right|=0\]
\[\Rightarrow (\hat{i}+\hat{j}+\hat{k}).[\hat{i}(0-0)-\hat{j}(1-0)+\hat{k}(-1-0)]=0\]
$\Rightarrow (\hat{i}+\hat{j}+\hat{k}).(-\hat{j}-\hat{k})=0$
$\Rightarrow -2\ne 0$
Therefore, we cannot find value for ${{C}_{1}}$.



Option ‘D’ is correct

Note: Dot product of UNIT VECTORS: -
$\begin{align}
  & \hat{i}.\hat{i}=1 \\
 & \hat{j}.\hat{j}=1 \\
 & \hat{k}.\hat{k}=1 \\
 & \hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{k}.\hat{i}=0 \\
\end{align}$
Cross product of unit vectors: -
\[\begin{align}
  & \hat{i}\times \hat{j}=\hat{k} \\
 & \hat{j}\times \hat{k}=\hat{i} \\
 & \hat{k}\times \hat{i}=\hat{j} \\
 & \hat{j}\times \hat{i}=-\hat{k} \\
 & \hat{k}\times \hat{j}=-\hat{i} \\
 & \hat{i}\times \hat{k}=-\hat{j} \\
 & \hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0 \\
\end{align}\]