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

If A⃗=2i^−2j^+2k^\vec{A}=2\hat{i}-2\hat{j}+2\hat{k} and B⃗=i^+j^\vec{B}=\hat{i}+\hat{j} then (a)Find the angle between $\vec{A}$ and $\vec{B}$ (b) Find the resultant vector of $\vec{A}$ and $\vec{B}$on $x-$axis.

Explanation

Solution

We use the formula for smaller angle between two vectors written in 3 components (a⃗=a1i^+a2j^+a3k^,b⃗=b1i^+b2j^+b3k^)\left( \vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} \right) that is cos⁡−1(a1b1+a2b2+a3b3){{\cos }^{-1}}\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right). We use the resultant of a⃗\vec{a} and b⃗\vec{b} which is given by vectorR⃗=a⃗+b⃗=a⃗=(a1+b1)i^+(a2+b2)j^+(a3+b3)k^\vec{R}=\vec{a}+\vec{b}=\vec{a}=\left( {{a}_{1}}+{{b}_{1}} \right)\hat{i}+\left( {{a}_{2}}+{{b}_{2}} \right)\hat{j}+\left( {{a}_{3}}+{{b}_{3}} \right)\hat{k} and then find the projection of R⃗\vec{R} along x−x-axis which same as along i^\hat{i} with formula R⃗⋅i^∣i^∣\vec{R}\cdot \dfrac{{\hat{i}}}{\left| {\hat{i}} \right|}.$$$$

Complete step-by-step answer:
We know that i^\hat{i},j^\hat{j} and k^\hat{k} are unit vectors(vectors with magnitude 1) along x,yx,y and zz axes respectively. So the magnitude of these vectors ∣i^∣=∣j^∣=∣k^∣=1 \left| {\hat{i}} \right|=\left| {\hat{j}} \right|=\left| {\hat{k}} \right|=1. The vectors just like their axes are perpendicular to each other which means any angle among i^\hat{i},j^\hat{j} and k^\hat{k}is 90∘.{{90}^{\circ }}. So i^⋅i^=j^⋅j^=k^⋅k^=1⋅1⋅cos⁡0∘=1\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\cdot 1\cdot \cos {{0}^{\circ }}=1 and i^⋅j^=j^⋅i^=j^⋅k^=k^⋅j^=i^⋅k^=k^⋅i^=1⋅1⋅cos⁡90∘=0\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{i}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{j}=\hat{i}\cdot \hat{k}=\hat{k}\cdot \hat{i}=1\cdot 1\cdot \cos {{90}^{\circ }}=0 We also know that the dot product of two vectors $\vec{a}$ and $\vec{b}$ is denoted as $\vec{a}\cdot \vec{b}$ and is given by $\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta $ where $\theta $ is the angle between the vectors $\vec{a}$ and $\vec{b}$.So if two vectors written in 3 components $\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k},$ their dot product is given by $\vec{a}\cdot \vec{b}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}$ and the angle between them is given by ${{\cos }^{-1}}\left( \vec{a}\cdot \vec{b} \right)={{\cos }^{-1}}\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right)$. The resultant of $\vec{a}$ and $\vec{b}$ is given by vector$\vec{R}=\vec{a}+\vec{b}=\vec{a}=\left( {{a}_{1}}+{{b}_{1}} \right)\hat{i}+\left( {{a}_{2}}+{{b}_{2}} \right)\hat{j}+\left( {{a}_{3}}+{{b}_{3}} \right)\hat{k}.$

We also know that projection of a⃗\vec{a} along b⃗\vec{b} is the component of a⃗\vec{a} along b⃗\vec{b} which is given by a⃗⋅b⃗∣b⃗∣\vec{a}\cdot \dfrac{{\vec{b}}}{\left| {\vec{b}} \right|} where ∣b⃗∣\left| {\vec{b}} \right| is the magnitude b⃗\vec{b}. (a) (a)
The given two vectors are A⃗=2i^−2j^+2k^\vec{A}=2\hat{i}-2\hat{j}+2\hat{k} and B⃗=i^+j^\vec{B}=\hat{i}+\hat{j} so we find the angle between them with inverse cosine of sum of product of respective components. The angle between them is
θ=cos⁡−1(A⃗⋅B⃗)=cos⁡−1(2(1)+(−2)1+(−1)0)=cos⁡−1(0)=90∘\theta ={{\cos }^{-1}}\left( \vec{A}\cdot \vec{B} \right)={{\cos }^{-1}}\left( 2\left( 1 \right)+\left( -2 \right)1+\left( -1 \right)0 \right)={{\cos }^{-1}}\left( 0 \right)={{90}^{\circ }}
We add the respective components to find the resultant of the vectors a⃗\vec{a} and b⃗\vec{b} . So we have,
R⃗=A⃗+B⃗=(2+1)i^+(−2+1)j^+(0+1)k=3i^−j^+k^\vec{R}=\vec{A}+\vec{B}=\left( 2+1 \right)\hat{i}+\left( -2+1 \right)\hat{j}+\left( 0+1 \right)k=3\hat{i}-\hat{j}+\hat{k}
(b)$$$$
The projection of the resultant vector along the x−x-axis is the same as projection along the unit orthogonal vector i^\hat{i} which is a vector in the direction of the x−x-axis . So the projection of R⃗\vec{R} along x−x-axis (denoted as P) is
P=R⃗⋅i^∣i^∣=R⃗⋅i^1=(3i^−j^+k^)⋅i^=3i^⋅i^=3P=\vec{R}\cdot \dfrac{{\hat{i}}}{\left| {\hat{i}} \right|}=\vec{R}\cdot \dfrac{{\hat{i}}}{1}=\left( 3\hat{i}-\hat{j}+\hat{k} \right)\cdot \hat{i}=3\hat{i}\cdot \hat{i}=3

Note: We need to be careful of the confusion between dot product and the resultant . The dot product involves multiplication of respective components while resultant involves addition. We can also find the angle from cross product as sin⁡−1(a⃗×b⃗){{\sin }^{-1}}\left( \vec{a}\times \vec{b} \right)