Question

If ${{\vec A = 2\hat i - 2\hat j - \hat k}}$ and ${{\vec B = \hat i + \hat j}}$, then
A) Find angle between ${{\vec A}}$ and ${{\vec B}}$.
B) Find the projection of the resultant vector of ${{\vec A}}$ and ${{\vec B}}$ on x-axis.
${{cos\theta = }}\dfrac{{{{\vec a}}{{.\vec b}}}}{{{{|a||b|}}}}$

Answer 1

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.