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

Question

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

Vedantu Content Team · Accepted Answer

Hint: To prove the given problem we have to take the standard equation of vector

\vec{r}

i.e.,

\vec{r} = x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}

. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given

\vec{r} = (\vec{r} . \overset{\land}{i}) \overset{\land}{i} + (\vec{r} . \overset{\land}{j}) \overset{\land}{j} + (\vec{r .} \overset{\land}{k}) \overset{\land}{k} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)

Let

\vec{r} = x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)

From equation (1) and (2) we have

\vec{r} = ((x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}) . \overset{\land}{i}) \overset{\land}{i} + ((x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}) . \overset{\land}{j}) \overset{\land}{j} + ((x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}) . \overset{\land}{k}) \overset{\land}{k}

Now first consider

\vec{r} = ((x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}) . \overset{\land}{i}) \overset{\land}{i}

\vec{r} = (x \overset{\land}{i} . \overset{\land}{i} + y \overset{\land}{j} . \overset{\land}{i} + z \overset{\land}{k} . \overset{\land}{i}) \overset{\land}{i}

By using the formulae

\overset{\land}{i} . \overset{\land}{i} = 1, \overset{\land}{j} . \overset{\land}{i} = 0 and \overset{\land}{k} . \overset{\land}{i} = 0

we have

\vec{r} = (x (1) + y (0) + z (0)) \overset{\land}{i} = x \overset{\land}{i}

Then consider

\vec{r} = ((x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}) . \overset{\land}{j}) \overset{\land}{j}

\vec{r} = (x \overset{\land}{i} . \overset{\land}{j} + y \overset{\land}{j} . \overset{\land}{j} + z \overset{\land}{k} . \overset{\land}{j}) \overset{\land}{j}

By using the formulae

\overset{\land}{i} . \overset{\land}{j} = 0, \overset{\land}{j} . \overset{\land}{j} = 1 and \overset{\land}{k} . \overset{\land}{j} = 0

we have

\vec{r} = (x (0) + y (1) + z (0)) \overset{\land}{j} = y \overset{\land}{j}

Next consider

\vec{r} = ((x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}) . \overset{\land}{k}) \overset{\land}{k}

\vec{r} = (x \overset{\land}{i} . \overset{\land}{k} + y \overset{\land}{j} . \overset{\land}{k} + z \overset{\land}{k} . \overset{\land}{k}) \overset{\land}{k}

By using the formulae

\overset{\land}{i} . \overset{\land}{k} = 0, \overset{\land}{j} . \overset{\land}{k} = 0 and \overset{\land}{k} . \overset{\land}{k} = 1

\vec{r} = (x (0) + y (1) + z (1)) \overset{\land}{k} = z \overset{\land}{k}

Using the above information, we have

\vec{r} = ((x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}) . \overset{\land}{i}) \overset{\land}{i} + ((x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}) . \overset{\land}{j}) \overset{\land}{j} + ((x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k}) . \overset{\land}{k}) \overset{\land}{k}

equals to

\vec{r} = x \overset{\land}{i} + y \overset{\land}{j} + z \overset{\land}{k} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)

From equations (2) and (3) we can conclude that

\vec{r} = (\vec{r} . \overset{\land}{i}) \overset{\land}{i} + (\vec{r} . \overset{\land}{j}) \overset{\land}{j} + (\vec{r .} \overset{\land}{k}) \overset{\land}{k}

Hence proved.

Note: Here we have used dot products of vectors. The formulae which are used in the solution are

\overset{\land}{i} . \overset{\land}{i} = 1, \overset{\land}{i} . \overset{\land}{j} = 0 and \overset{\land}{i} . \overset{\land}{k} = 0 \overset{\land}{j} . \overset{\land}{i} = 0, \overset{\land}{j} . \overset{\land}{j} = 1 and \overset{\land}{j} . \overset{\land}{k} = 0 \overset{\land}{i} . \overset{\land}{k} = 0, \overset{\land}{j} . \overset{\land}{k} = 0 and \overset{\land}{k} . \overset{\land}{k} = 1

For any vector r→, prove that r→=(r→.i∧)i∧+(r→.j∧)j∧+(r.→k∧)k∧.

For any vector $\vec{r}$ , prove that $\vec{r} = (\vec{r} . \overset{\land}{i}) \overset{\land}{i} + (\vec{r} . \overset{\land}{j}) \overset{\land}{j} + (\vec{r .} \overset{\land}{k}) \overset{\land}{k}$ .