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**Hint:**Formula for angle between two vectors is or

${{\theta = co}}{{{s}}^{ - 1}}{{ }}\left( {\dfrac{{{{\vec a}}{{.\vec b}}}}{{{{|a||b|}}}}} \right)$

Where ${{\vec a}}{{.\vec b}}$ = dot product of ${{\vec A}}$ and ${{\vec B}}$

${{|a||b|}}$= magnitude of vector ${{\vec A}}$ and ${{\vec B}}$

Now for the next part, the formula for the resultant vector is \[{{\vec R = \vec A + \vec B}}\].

Then after finding the resultant vector, find the projection of the vector on the x-axis by dot product of the resultant vector with ${{\hat i}}$.

**Complete step by step solution:**

Given: ${{\vec A = 2\hat i - 2\hat j - \hat k}}$ and ${{\vec B = \hat i + \hat j}}$

Let us consider that angle between both the vectors is ${{\theta }}$.

The angle between ${{\vec A}}$ and ${{\vec B}}$ is ${{\vec A}}{{.\vec B = |A||B| cos \theta }}$

Or ${{\theta = co}}{{{s}}^{ - 1}}{{ }}\left( {\dfrac{{{{\vec a}}{{.\vec b}}}}{{{{|a||b|}}}}} \right)$

Dot product is given by

${{\vec a}}{{.\vec b}}$= \[{{(2\hat i - 2\hat j - 2\hat k)(\hat i}} + {{\hat j)}}\]= 2 – 2 = 0

Magnitude of vectors is given by

${{|a| = }}\sqrt {{{{2}}^{{2}}}{{ + ( - 2}}{{{)}}^{{2}}}{{ + ( - 1}}{{{)}}^{{2}}}} {{ = }}\sqrt {{9}} $

${{|b| = }}\sqrt {{{{1}}^{{2}}}{{ + (1}}{{{)}}^{{2}}}} {{ = }}\sqrt {{2}} $

${{|a||b| = }}\sqrt 9 \sqrt 2 = 3.56$

Now substituting the values in above formula, we get

${{\theta = co}}{{{s}}^{ - 1}}{{ }}\left( {\dfrac{0}{{3.56}}} \right)$

\[

{{\theta = co}}{{{s}}^{{{ - 1}}}}{{(0)}} \\

or {{\theta }} = {90^0} \\

\therefore {{\theta = 9}}{{{0}}^{{0}}} \\

\]

Therefore, the angle between ${{\vec A}}$ and ${{\vec B}}$ is 90°.

The resultant of two or more than two vectors is a single vector which produces the same effect as that of the individual vectors together.

Resultant vector of ${{\vec A}}$ and ${{\vec B}}$ is given by

\[{{\vec R = \vec A + \vec B}}\]

On substituting the values, we get

$

{{\vec R = (2\hat i - 2\hat j - \hat k) + (\hat i + \hat j)}} \\

{{\vec R = 3\hat i - \hat j - \hat k}} \\

$

The projection of resultant vector of ${{\vec A}}$ and ${{\vec B}}$ on x-axis is given by

${{\vec R }}{{. \hat i = 3}}$

So, the angle between ${{\vec A}}$ and ${{\vec B}}$ is 90° and the projection of resultant vector of ${{\vec A}}$ and ${{\vec B}}$ on x-axis is 3.

**Note:**In the question projection of resultant vector of ${{\vec A}}$ and ${{\vec B}}$ on x-axis is asked then dot product of resultant vector with ${{\hat i}}$ is done. If projection of resultant vector of vectors on the y-axis is asked then dot product of resultant vector with ${{\hat j}}$. Similarly, when the resultant vector on the z-axis is asked then the dot product of the resultant vector with ${{\hat k}}$ is done. Scalars can be added algebraically but vectors do not as vectors possess both magnitude and direction.

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