Question

# For any vector $\overrightarrow r$, prove that $\overrightarrow r = \left( {\overrightarrow r .\mathop i\limits^ \wedge } \right)\mathop i\limits^ \wedge + \left( {\overrightarrow r .\mathop j\limits^ \wedge } \right)\mathop j\limits^ \wedge + \left( {\overrightarrow {r.} \mathop k\limits^ \wedge } \right)\mathop k\limits^ \wedge$.

Hint: To prove the given problem we have to take the standard equation of vector $\overrightarrow r$i.e., $\overrightarrow r = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge$. So, use this concept to reach the solution of the given problem.

Given $\overrightarrow r = \left( {\overrightarrow r .\mathop i\limits^ \wedge } \right)\mathop i\limits^ \wedge + \left( {\overrightarrow r .\mathop j\limits^ \wedge } \right)\mathop j\limits^ \wedge + \left( {\overrightarrow {r.} \mathop k\limits^ \wedge } \right)\mathop k\limits^ \wedge ..................................................\left( 1 \right)$
Let $\overrightarrow r = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge ............................................................\left( 2 \right)$
From equation (1) and (2) we have
$\overrightarrow r = \left( {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\mathop i\limits^ \wedge } \right)\mathop i\limits^ \wedge + \left( {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\mathop j\limits^ \wedge } \right)\mathop j\limits^ \wedge + \left( {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\mathop k\limits^ \wedge } \right)\mathop k\limits^ \wedge$
Now first consider $\overrightarrow r = \left( {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\mathop i\limits^ \wedge } \right)\mathop i\limits^ \wedge$
$\overrightarrow r = \left( {x\mathop i\limits^ \wedge .\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge .\mathop i\limits^ \wedge + z\mathop k\limits^ \wedge .\mathop i\limits^ \wedge } \right)\mathop i\limits^ \wedge$
By using the formulae $\mathop i\limits^ \wedge .\mathop i\limits^ \wedge = 1{\text{ , }}\mathop j\limits^ \wedge .\mathop i\limits^ \wedge = 0{\text{ and }}\mathop k\limits^ \wedge .\mathop i\limits^ \wedge = 0$ we have
$\overrightarrow r = \left( {x\left( 1 \right) + y\left( 0 \right) + z\left( 0 \right)} \right)\mathop i\limits^ \wedge = x\mathop i\limits^ \wedge$
Then consider $\overrightarrow r = \left( {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\mathop j\limits^ \wedge } \right)\mathop j\limits^ \wedge$
$\overrightarrow r = \left( {x\mathop i\limits^ \wedge .\mathop j\limits^ \wedge + y\mathop j\limits^ \wedge .\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge .\mathop j\limits^ \wedge } \right)\mathop j\limits^ \wedge$
By using the formulae $\mathop i\limits^ \wedge .\mathop j\limits^ \wedge = 0{\text{ , }}\mathop j\limits^ \wedge .\mathop j\limits^ \wedge = 1{\text{ and }}\mathop k\limits^ \wedge .\mathop j\limits^ \wedge = 0$ we have
$\overrightarrow r = \left( {x\left( 0 \right) + y\left( 1 \right) + z\left( 0 \right)} \right)\mathop j\limits^ \wedge = y\mathop j\limits^ \wedge$
Next consider $\overrightarrow r = \left( {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\mathop k\limits^ \wedge } \right)\mathop k\limits^ \wedge$
$\overrightarrow r = \left( {x\mathop i\limits^ \wedge .\mathop k\limits^ \wedge + y\mathop j\limits^ \wedge .\mathop k\limits^ \wedge + z\mathop k\limits^ \wedge .\mathop k\limits^ \wedge } \right)\mathop k\limits^ \wedge$
By using the formulae $\mathop i\limits^ \wedge .\mathop k\limits^ \wedge = 0{\text{ , }}\mathop j\limits^ \wedge .\mathop k\limits^ \wedge = 0{\text{ and }}\mathop k\limits^ \wedge .\mathop k\limits^ \wedge = 1$
$\overrightarrow r = \left( {x\left( 0 \right) + y\left( 1 \right) + z\left( 1 \right)} \right)\mathop k\limits^ \wedge = z\mathop k\limits^ \wedge$
Using the above information, we have
$\overrightarrow r = \left( {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\mathop i\limits^ \wedge } \right)\mathop i\limits^ \wedge + \left( {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\mathop j\limits^ \wedge } \right)\mathop j\limits^ \wedge + \left( {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\mathop k\limits^ \wedge } \right)\mathop k\limits^ \wedge$ equals to
$\overrightarrow r = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge ...........................................\left( 3 \right)$
From equations (2) and (3) we can conclude that
$\overrightarrow r = \left( {\overrightarrow r .\mathop i\limits^ \wedge } \right)\mathop i\limits^ \wedge + \left( {\overrightarrow r .\mathop j\limits^ \wedge } \right)\mathop j\limits^ \wedge + \left( {\overrightarrow {r.} \mathop k\limits^ \wedge } \right)\mathop k\limits^ \wedge$
Hence proved.

Note: Here we have used dot products of vectors. The formulae which are used in the solution are
$\mathop i\limits^ \wedge .\mathop i\limits^ \wedge = 1{\text{ , }}\mathop i\limits^ \wedge . \mathop j\limits^ \wedge = 0{\text{ and }}\mathop i\limits^ \wedge .\mathop k\limits^ \wedge = 0 \\ \mathop j\limits^ \wedge .\mathop i\limits^ \wedge = 0{\text{ , }}\mathop j\limits^ \wedge .\mathop j\limits^ \wedge = 1{\text{ and }}\mathop j\limits^ \wedge .\mathop k\limits^ \wedge = 0 \\ \mathop i\limits^ \wedge .\mathop k\limits^ \wedge = 0{\text{ , }}\mathop j\limits^ \wedge .\mathop k\limits^ \wedge = 0{\text{ and }}\mathop k\limits^ \wedge .\mathop k\limits^ \wedge = 1 \\$