Question

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

Answer 1

Hint: We need to simplify the problem by using the given hints of vectors b and c being any two non-collinear vectors. We treat them as unit vectors. We find the general form of the given equation which is also similar to its specific form.

Complete step-by-step solution:
We know that $\overrightarrow{b}\times \overrightarrow{c}\bot \overrightarrow{b},\overrightarrow{b}\times \overrightarrow{c}\bot \overrightarrow{c}$.
The three parts $\left( a.b \right),\left( a.c \right),a.\left( b\times c \right)$ are all scalar values.
We are trying to take the projection of a vector “a” on vector b, vector a on vector c, vector “a” on vector $\left( b\times c \right)$.
As the vectors, b and c are any two non-collinear vectors, and a is any vector, to simplify the problem we take vectors b and c as unit vectors.
So, $\overrightarrow{b}=\widehat{i},\overrightarrow{c}=\widehat{j},\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}$.
Now we find the value of \[\left( b\times c \right)\].
So, \[\left( b\times c \right)=\left( \widehat{i}\times \widehat{j} \right)=\widehat{k}\]. And the modulus value will be unit. So, \[\left| \left( b\times c \right) \right|=\left| \widehat{k} \right|=1\].
We solve $\left( a.b \right)b+\left( a.c \right)c+\dfrac{a.\left( b\times c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)$.
Value of $\left( a.b \right)b=\left( \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\widehat{i} \right)\widehat{i}=x\widehat{i}$
Value of $\left( a.c \right)c=\left( \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\widehat{j} \right)\widehat{j}=y\widehat{j}$
Value of $\dfrac{a.\left( b\times c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)=\dfrac{\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\widehat{k}}{{{1}^{2}}}\widehat{k}=\dfrac{z}{1}\widehat{k}=z\widehat{k}$
So, $\left( a.b \right)b+\left( a.c \right)c+\dfrac{a.\left( b\times c \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)=x\widehat{i}+y\widehat{j}+z\widehat{k}=\overrightarrow{a}$.
The correct option is A.
Note: In the specific cases we can consider a vector $\overrightarrow{p}={{p}_{x}}\widehat{i}+{{p}_{y}}\widehat{j}+{{p}_{z}}\widehat{k}$. Here the axes are normal X, Y, Z axes but we can change it to any form possible. We change them in $\widehat{\alpha },\widehat{\beta },\widehat{\lambda }$. Then we can treat $\widehat{\alpha },\widehat{\beta },\widehat{\lambda }$ as \[\dfrac{\overrightarrow{b}}{\left| \overrightarrow{b} \right|},\dfrac{\overrightarrow{c}}{\left| \overrightarrow{c} \right|},\dfrac{\overrightarrow{\left( b\times c \right)}}{\left| \overrightarrow{\left( b\times c \right)} \right|}\] respectively. The projection will be also similar to vector a.