Question

If $ b $ , $ c $ are any two non-collinear unit vectors and $ a $ is any vector, then $ \left( a.b \right)b+\left( a.c \right)c+\dfrac{a.\left( b+c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)= $
A. $ a $
B. $ b $
C. $ c $
D. None of these

Answer 1

Hint: In the problem we have given that $ b $ , $ c $ are any two non-collinear unit vectors and $ a $ is any vector, so we will assume the values of all the vectors we have given. Now we need to calculate the value of $ \left( a.b \right)b+\left( a.c \right)c+\dfrac{a.\left( b+c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)= $ , for this we need to calculate the individual values of $ a.b $ , $ a.c $ , $ a.\dfrac{b\times c}{\left| b\times c \right|} $ . Now we will substitute the calculated values in the given equation and simplify them to get the result.

Complete step by step answer:
Given that,
 $ b $ , $ c $ are any two non-collinear unit vectors and $ a $ is any vector. We know that unit vectors are represented by $ \hat{i} $ , $ \hat{j} $ , $ \hat{k} $ .
Let $ b=\hat{i} $ , $ c=\hat{j} $ . Now the values of $ \left| b \right| $ , $ \left| c \right| $ is given by
 $ \begin{align}
  & \left| b \right|=\left| {\hat{i}} \right| \\
 & \Rightarrow \left| b \right|=\sqrt{{{1}^{2}}} \\
 & \Rightarrow \left| b \right|=1 \\
\end{align} $
Similarly, the value of $ \left| c \right|=1 $ .
Now the value of $ b\times c $ is given by
 $ b\times c=\left| b \right|\left| c \right|\sin \alpha .k $
Substituting the values of $ \left| b \right| $ , $ \left| c \right| $ , then we will get
 $ \begin{align}
  & \Rightarrow b\times c=1.1\sin \alpha .k \\
 & \Rightarrow b\times c=\sin \alpha .k \\
\end{align} $
Now the value of $ \left| b\times c \right| $ value is given by
 $ \begin{align}
  & \Rightarrow \left| b\times c \right|=\sqrt{{{\left( \sin \alpha \right)}^{2}}} \\
 & \Rightarrow \left| b\times c \right|=\sin \alpha \\
\end{align} $
Now the value of $ \dfrac{b\times c}{\left| b\times c \right|} $ can be written as
 $ \begin{align}
  & \dfrac{b\times c}{\left| b\times c \right|}=\dfrac{\sin \alpha k}{\sin \alpha } \\
 & \Rightarrow \dfrac{b\times c}{\left| b\times c \right|}=k \\
\end{align} $
Let $ a $ is any vector and that can be written as
 $ a={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} $
Now,
 $ \begin{align}
  & a.b=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right).\hat{i} \\
 & \Rightarrow a.b={{a}_{1}}\left( \hat{i}.\hat{i} \right) \\
\end{align} $
We know that $ \hat{i}.\hat{i}=1 $ , then we will get
 $ a.b={{a}_{1}} $
Similarly, the value of $ a.c={{a}_{2}} $ .
Now the value of $ a.\dfrac{b\times c}{\left| b\times c \right|} $ can be written as
 $ \begin{align}
  & a.\dfrac{b\times c}{\left| b\times c \right|}=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right).\hat{k} \\
 & \Rightarrow a.\dfrac{b\times c}{\left| b\times c \right|}={{a}_{3}}\left( \hat{k}.\hat{k} \right) \\
 & \Rightarrow a.\dfrac{b\times c}{\left| b\times c \right|}={{a}_{3}} \\
\end{align} $
Finally, the value of given equation is $ \left( a.b \right)b+\left( a.c \right)c+\dfrac{a.\left( b+c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right) $
Substituting the all the values we have calculated in the above equation, then we will get
 $ \left( a.b \right)b+\left( a.c \right)c+\dfrac{a.\left( b+c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)={{a}_{1}}.b+{{a}_{2}}.c+{{a}_{3}}\dfrac{b\times c}{\left| b\times c \right|} $
We have the $ b=\hat{i} $ , $ c=\hat{j} $ , $ \dfrac{b\times c}{\left| b\times c \right|}=k $ , then
 $ \Rightarrow \left( a.b \right)b+\left( a.c \right)c+\dfrac{a.\left( b+c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)={{a}_{1}}.\hat{i}+{{a}_{2}}.\hat{j}+{{a}_{3}}\hat{k} $
Substituting the values of $ a={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} $ in the above equation, then we will get
 $ \begin{align}
  & \Rightarrow \left( a.b \right)b+\left( a.c \right)c+\dfrac{a.\left( b+c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)={{a}_{1}}.\hat{i}+{{a}_{2}}.\hat{j}+{{a}_{3}}\hat{k} \\
 & \Rightarrow \left( a.b \right)b+\left( a.c \right)c+\dfrac{a.\left( b+c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)=a \\
\end{align} $
$ \therefore $ Option - A is correct answer.

Note:
In the problem we have used dot product and vector product seamlessly. The dot product and vector products of the two vectors say $ a={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} $ , $ b={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} $ can be written as
 $ \begin{align}
  & a.b=\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right).\left( {{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} \right) \\
 & \Rightarrow a.b={{a}_{1}}{{b}_{1}}\left( \hat{i}.\hat{i} \right)+{{a}_{2}}{{b}_{2}}\left( \hat{j}.\hat{j} \right)+{{a}_{3}}{{b}_{3}}\left( \hat{k}.\hat{k} \right) \\
\end{align} $
We know that $ \hat{i}.\hat{i}=1 $ , $ \hat{j}.\hat{j}=1 $ , $ \hat{k}.\hat{k}=1 $ hence
 $ a.b={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} $
Similarly, we will calculate the dot product but we will use the formulas $ \hat{i}\times \hat{j}=\hat{k} $ , $ \hat{j}\times \hat{k}=\hat{i} $ , $ \hat{k}\times \hat{i}=\hat{j} $ , then we will get a vector quantity as the cross product of the two vectors.