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Question: If \({{\vec A = 2\hat i - 2\hat j - \hat k}}\) and \({{\vec B = \hat i + \hat j}}\), then A) Find ...

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

Explanation

Solution

Formula for angle between two vectors is or
θ=cos−1(a⃗.b⃗∣a∣∣b∣){{\theta = co}}{{{s}}^{ - 1}}{{ }}\left( {\dfrac{{{{\vec a}}{{.\vec b}}}}{{{{|a||b|}}}}} \right)
Where a⃗.b⃗{{\vec a}}{{.\vec b}} = dot product of A⃗{{\vec A}} and B⃗{{\vec B}}
∣a∣∣b∣{{|a||b|}}= magnitude of vector A⃗{{\vec A}} and B⃗{{\vec B}}
Now for the next part, the formula for the resultant vector is R⃗=A⃗+B⃗{{\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 i^{{\hat i}}.

Complete step by step solution:
Given: A⃗=2i^−2j^−k^{{\vec A = 2\hat i - 2\hat j - \hat k}} and B⃗=i^+j^{{\vec B = \hat i + \hat j}}
Let us consider that angle between both the vectors is θ{{\theta }}.
The angle between A⃗{{\vec A}} and B⃗{{\vec B}} is A⃗.B⃗=∣A∣∣B∣cosθ{{\vec A}}{{.\vec B = |A||B| cos \theta }}
Or θ=cos−1(a⃗.b⃗∣a∣∣b∣){{\theta = co}}{{{s}}^{ - 1}}{{ }}\left( {\dfrac{{{{\vec a}}{{.\vec b}}}}{{{{|a||b|}}}}} \right)
Dot product is given by
a⃗.b⃗{{\vec a}}{{.\vec b}}= (2i^−2j^−2k^)(i^+j^){{(2\hat i - 2\hat j - 2\hat k)(\hat i}} + {{\hat j)}}= 2 – 2 = 0
Magnitude of vectors is given by
∣a∣=22+(−2)2+(−1)2=9{{|a| = }}\sqrt {{{{2}}^{{2}}}{{ + ( - 2}}{{{)}}^{{2}}}{{ + ( - 1}}{{{)}}^{{2}}}} {{ = }}\sqrt {{9}}
∣b∣=12+(1)2=2{{|b| = }}\sqrt {{{{1}}^{{2}}}{{ + (1}}{{{)}}^{{2}}}} {{ = }}\sqrt {{2}}
∣a∣∣b∣=92=3.56{{|a||b| = }}\sqrt 9 \sqrt 2 = 3.56
Now substituting the values in above formula, we get
θ=cos−1(03.56){{\theta = co}}{{{s}}^{ - 1}}{{ }}\left( {\dfrac{0}{{3.56}}} \right)

θ=cos−1(0) orθ=900 ∴θ=900  {{\theta = co}}{{{s}}^{{{ - 1}}}}{{(0)}} \\\ or {{\theta }} = {90^0} \\\ \therefore {{\theta = 9}}{{{0}}^{{0}}} \\\

Therefore, the angle between A⃗{{\vec A}} and B⃗{{\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 A⃗{{\vec A}} and B⃗{{\vec B}} is given by
R⃗=A⃗+B⃗{{\vec R = \vec A + \vec B}}
On substituting the values, we get
R⃗=(2i^−2j^−k^)+(i^+j^) R⃗=3i^−j^−k^  {{\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 A⃗{{\vec A}} and B⃗{{\vec B}} on x-axis is given by
R⃗.i^=3{{\vec R }}{{. \hat i = 3}}
So, the angle between A⃗{{\vec A}} and B⃗{{\vec B}} is 90° and the projection of resultant vector of A⃗{{\vec A}} and B⃗{{\vec B}} on x-axis is 3.

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