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

Answer Verified Verified
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.

Complete step-by-step answer:
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
 \[\mathop i\limits^ \wedge .\mathop i\limits^ \wedge = 1{\text{ , }}\mathop i\limits^ \wedge . \mathop j\limits^ \wedge = 0{\text{ and }}\mathop i\limits^ \wedge .\mathop k\limits^ \wedge = 0 \\
  \mathop j\limits^ \wedge .\mathop i\limits^ \wedge = 0{\text{ , }}\mathop j\limits^ \wedge .\mathop j\limits^ \wedge = 1{\text{ and }}\mathop j\limits^ \wedge .\mathop k\limits^ \wedge = 0 \\
  \mathop i\limits^ \wedge .\mathop k\limits^ \wedge = 0{\text{ , }}\mathop j\limits^ \wedge .\mathop k\limits^ \wedge = 0{\text{ and }}\mathop k\limits^ \wedge .\mathop k\limits^ \wedge = 1 \\
\]
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