Question

If \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are three non-coplanar vectors, then \[\dfrac{\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}}{\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}}+\dfrac{\overrightarrow{b}\cdot \overrightarrow{a}\times \overrightarrow{c}}{\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}}=\]
A. $0$
B. $2$
C. $3$
D. None of these

Answer 1

Hint: In this question, the dot and cross products of vectors are applied to find the required vector expression. The dot product is said to be a scalar product and the cross product is said to be a skew product or vector product. By using appropriate formulae, the required vector equation is calculated.



Formula Used:The dot product of two vectors is
$\overrightarrow{a}\cdot \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos (\overrightarrow{a},\overrightarrow{b})$
The cross-product of two vectors is
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\overrightarrow{a},\overrightarrow{b})\overrightarrow{n}$
Scalar triple product of three vectors:
\[\begin{align}
  & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b} \\
 & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}]=[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}] \\
 & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]=-[\overrightarrow{c}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=-[\overrightarrow{a}\text{ }\overrightarrow{c}\text{ }\overrightarrow{b}] \\
\end{align}\]
Important vector identities for solving vector equations are:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]



Complete step by step solution:It is given that,
\[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are three non-coplanar vectors, then \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\ne 0\].
So,
\[\dfrac{\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}}{\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}}+\dfrac{\overrightarrow{b}\cdot \overrightarrow{a}\times \overrightarrow{c}}{\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}}=\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}+\dfrac{[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}\text{ }...(1)\]
Since we have a scalar triple product as
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}\]
We have the vector identity
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]\]
On substituting this identity in equation (1), we get
\[\begin{align}
  & \dfrac{\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}}{\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}}+\dfrac{\overrightarrow{b}\cdot \overrightarrow{a}\times \overrightarrow{c}}{\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}}=\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}+\dfrac{[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]} \\
 & \text{ }=\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}-\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]} \\
 & \text{ }=0 \\
\end{align}\]
Therefore, the value of the given vector equation is 0.



Option ‘A’ is correct

Note:Here we may go wrong with the vector identities and scalar triple product. Here are the simple formulae used for solving the given vector equation. I.e., \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]\] and \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}\]. By applying appropriate vector products, the given vector equation is evaluated.