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Question: If \({\bar{A}}=3\hat{i}+2\hat{j}\) and \({\bar{B}}=2\hat{i}+3\hat{j}-\hat{k},\) then find a unit vec...

If Aˉ=3i^+2j^{\bar{A}}=3\hat{i}+2\hat{j} and Bˉ=2i^+3j^−k^,{\bar{B}}=2\hat{i}+3\hat{j}-\hat{k}, then find a unit vector along (Aˉ−Bˉ).\left( {\bar{A}}-{\bar{B}} \right).

Explanation

Solution

A unit vector is a vector of length one. We will find the magnitude of the given vector by taking the square root of the sum of the squares of the components of the vector. Then we will divide the vector with the magnitude to find the unit vector along the vector.

Complete step by step answer:
We are asked to find the unit vector along the vector Aˉ−Bˉ.{\bar{A}}-{\bar{B}}\text{.}
Let us consider the given vectors, vector A\text{A} and vector B.\text{B}\text{.}
So, Aˉ=3i^+2j^{\bar{A}}=3\hat{i}+2\hat{j} and Bˉ=2i^+3j^−k^.{\bar{B}}=2\hat{i}+3\hat{j}-\hat{k}.
Since Aˉ{\bar{A}} and Bˉ{\bar{B}} are vectors, their difference Aˉ−Bˉ{\bar{A}}-{\bar{B}} is also a vector.
To find this vector, we need to subtract Bˉ{\bar{B}} from Aˉ{\bar{A}} by the component wise subtraction.
So, we will get Aˉ−Bˉ=3i^+2j^−(2i^+3j^−k^)=3i^+2j^−2i^−3j^+k^=i^−j^+k^.{\bar{A}}-{\bar{B}}=3\hat{i}+2\hat{j}-\left( 2\hat{i}+3\hat{j}-\hat{k} \right)=3\hat{i}+2\hat{j}-2\hat{i}-3\hat{j}+\hat{k}=\hat{i}-\hat{j}+\hat{k}.
Now, since we got the difference vector by vector subtraction, we need to find the magnitude of the obtained vector so as to divide the vector by the magnitude to get the unit vector along this vector.
To find the magnitude of the vector, we need to find the square root of the sum of the squares of the three components.
So, the magnitude of the difference vector is given by ∣Aˉ−Bˉ∣=12+(−1)2+12=3.\left| {\bar{A}}-{\bar{B}} \right|=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{1}^{2}}}=\sqrt{3}.
Now, we divide the vector Aˉ−Bˉ{\bar{A}}-{\bar{B}} by 3\sqrt{3} to get the unit vector along Aˉ−Bˉ.{\bar{A}}-{\bar{B}}.
Now, we will get 13(Aˉ−Bˉ)=13(i−j+k).\dfrac{1}{\sqrt{3}}\left( {\bar{A}}-{\bar{B}} \right)=\dfrac{1}{\sqrt{3}}\left( i-j+k \right).
Therefore, the unit vector along the vector (Aˉ−Bˉ)=13(i−j+k)=13i−13j+13k.\left( {\bar{A}}-{\bar{B}} \right)=\dfrac{1}{\sqrt{3}}\left( i-j+k \right)=\dfrac{1}{\sqrt{3}}i-\dfrac{1}{\sqrt{3}}j+\dfrac{1}{\sqrt{3}}k.

Hence the unit vector n^=Aˉ−Bˉ∣Aˉ−Bˉ∣=13i−13j+13k.\hat{n}=\dfrac{{\bar{A}}-{\bar{B}}}{\left| {\bar{A}}-{\bar{B}} \right|}=\dfrac{1}{\sqrt{3}}i-\dfrac{1}{\sqrt{3}}j+\dfrac{1}{\sqrt{3}}k.

Note: Generally, the unit vector along a vector Pˉ{\bar{P}} is a vector that represents only the direction of the vector Pˉ{\bar{P}} and has a magnitude 1.1. It is obtained as n^=Pˉ∣Pˉ∣.\hat{n}=\dfrac{{{\bar{P}}}}{\left| {{\bar{P}}} \right|}. In the above problem, the magnitude e of the unit vector ∣n^∣=(13)2+(−13)2+(13)2=13+13+13=33=1=1.\left| {\hat{n}} \right|=\sqrt{{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}+{{\left( \dfrac{-1}{\sqrt{3}} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}=\sqrt{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}}=\sqrt{\dfrac{3}{3}}=\sqrt{1}=1.