Question

Let $\overset{\to }{\mathop{a}}\,$ be any vector, then $\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{i}}\, \right)\overset{\to }{\mathop{i}}\,+\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{j}}\, \right)\overset{\to }{\mathop{j}}\,+\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{k}}\, \right)\overset{\to }{\mathop{k}}\,$?

Answer 1

In this problem, we are given that $\overset{\to }{\mathop{a}}\,$ is any vector, where we have to find the value of $\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{i}}\, \right)\overset{\to }{\mathop{i}}\,+\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{j}}\, \right)\overset{\to }{\mathop{j}}\,+\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{k}}\, \right)\overset{\to }{\mathop{k}}\,$. We can now assume $\overset{\to }{\mathop{a}}\,$ as $x\overset{\to }{\mathop{i}}\,+y\overset{\to }{\mathop{j}}\,+z\overset{\to }{\mathop{k}}\,$. We can then substitute the assumed value of the vector $\overset{\to }{\mathop{a}}\,$ in the given expression. We can then simplify using the dot product operations and simplify them. We will get the required answer.

Complete step by step answer:
Here we are given that $\overset{\to }{\mathop{a}}\,$ is any vector, where we have to find the value of,
$\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{i}}\, \right)\overset{\to }{\mathop{i}}\,+\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{j}}\, \right)\overset{\to }{\mathop{j}}\,+\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{k}}\, \right)\overset{\to }{\mathop{k}}\,$ ………. (1)
We can now assume $\overset{\to }{\mathop{a}}\,$ as $x\overset{\to }{\mathop{i}}\,+y\overset{\to }{\mathop{j}}\,+z\overset{\to }{\mathop{k}}\,$
$\Rightarrow \overset{\to }{\mathop{a}}\,=x\overset{\to }{\mathop{i}}\,+y\overset{\to }{\mathop{j}}\,+z\overset{\to }{\mathop{k}}\,$…….. (2)
We can now substitute the assume vector (2) in the given expression (1), we get
$\Rightarrow \left( \left( x\overset{\to }{\mathop{i}}\,+y\overset{\to }{\mathop{j}}\,+z\overset{\to }{\mathop{k}}\, \right)\cdot \overset{\to }{\mathop{i}}\, \right)\overset{\to }{\mathop{i}}\,+\left( \left( x\overset{\to }{\mathop{i}}\,+y\overset{\to }{\mathop{j}}\,+z\overset{\to }{\mathop{k}}\, \right)\cdot \overset{\to }{\mathop{j}}\, \right)\overset{\to }{\mathop{j}}\,+\left( \left( x\overset{\to }{\mathop{i}}\,+y\overset{\to }{\mathop{j}}\,+z\overset{\to }{\mathop{k}}\, \right)\cdot \overset{\to }{\mathop{k}}\, \right)\overset{\to }{\mathop{k}}\,$
We can now simplify the above step, we get
$\Rightarrow \left( x\left[ \overset{\to }{\mathop{i}}\,.\overset{\to }{\mathop{i}}\, \right]+y\left[ \overset{\to }{\mathop{j}}\,\cdot \overset{\to }{\mathop{i}}\, \right]+z\left[ \overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{i}}\, \right] \right)\overset{\to }{\mathop{i}}\,+\left( x\left[ \overset{\to }{\mathop{i\cdot }}\,\overset{\to }{\mathop{j}}\, \right]+y\left[ \overset{\to }{\mathop{j}}\,\cdot \overset{\to }{\mathop{j}}\, \right]+z\left[ \overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{j}}\, \right] \right)\overset{\to }{\mathop{j}}\,+\left( x\left[ \overset{\to }{\mathop{i\cdot }}\,\overset{\to }{\mathop{k}}\, \right]+y\left[ \overset{\to }{\mathop{j}}\,\cdot \overset{\to }{\mathop{k}}\, \right]+z\left[ \overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{k}}\, \right] \right)\overset{\to }{\mathop{k}}\,$
We can now simplify the above step using the operations in dot product.
We know that in dot product,
$\overset{\to }{\mathop{i}}\,\cdot \overset{\to }{\mathop{i}}\,=\overset{\to }{\mathop{j}}\,\cdot \overset{\to }{\mathop{j}}\,=\overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{k}}\,=1$ and $\overset{\to }{\mathop{i}}\,\cdot \overset{\to }{\mathop{j}}\,=\overset{\to }{\mathop{j}}\,\cdot \overset{\to }{\mathop{k}}\,=\overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{i}}\,=0$
We can now substitute the above formulas of dot product in the above step, we get
$\Rightarrow \left( x\left[ 1 \right]+y\left[ 0 \right]+z\left[ 0 \right] \right)\overset{\to }{\mathop{i}}\,+\left( x\left[ 0 \right]+y\left[ 1 \right]+z\left[ 0 \right] \right)\overset{\to }{\mathop{j}}\,+\left( x\left[ 0 \right]+y\left[ 0 \right]+z\left[ 1 \right] \right)\overset{\to }{\mathop{k}}\,$
We can further simplify the above step, by removing the terms which are multiplied with zero, we get
$\Rightarrow x\overset{\to }{\mathop{i}}\,+y\overset{\to }{\mathop{j}}\,+z\overset{\to }{\mathop{k}}\,=\overset{\to }{\mathop{a}}\,$
Therefore, the value of $\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{i}}\, \right)\overset{\to }{\mathop{i}}\,+\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{j}}\, \right)\overset{\to }{\mathop{j}}\,+\left( \overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{k}}\, \right)\overset{\to }{\mathop{k}}\,=\overset{\to }{\mathop{a}}\,$.

Note: We should always remember that there is a difference between dot product and cross product. The rules and formulas are different for both dot and cross products. We should always remember that the rule used in the dot product where the vectors are multiplied using the dot product$\overset{\to }{\mathop{i}}\,\cdot \overset{\to }{\mathop{i}}\,=\overset{\to }{\mathop{j}}\,\cdot \overset{\to }{\mathop{j}}\,=\overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{k}}\,=1$ and $\overset{\to }{\mathop{i}}\,\cdot \overset{\to }{\mathop{j}}\,=\overset{\to }{\mathop{j}}\,\cdot \overset{\to }{\mathop{k}}\,=\overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{i}}\,=0$. We should also simplify the terms step by step in order to avoid confusion.