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Question: Find the components of, \(\overrightarrow{a}=2\widehat{i}+3\widehat{j}\) along the direction of vect...

Find the components of, a→=2i^+3j^\overrightarrow{a}=2\widehat{i}+3\widehat{j} along the direction of vectors, i^+j^\widehat{i}+\widehat{j} and i^−j^\widehat{i}-\widehat{j} .

Explanation

Solution

To find the component of any vector along the direction of any other vector, we first calculate the unit vector of the directional vector (the vector along which we have to find components). Once we get this value, then the component of any vector along this direction is, the scalar product of the first vector with unit directional vector, multiplied by unit directional vector.

Complete answer:
Let the directional vectors given in the question be denoted with the following terms:
⇒m→=i^+j^\Rightarrow \overrightarrow{m}=\widehat{i}+\widehat{j}
⇒n→=i^−j^\Rightarrow \overrightarrow{n}=\widehat{i}-\widehat{j}
Now, we will calculate the unit vectors along the directions of these vectors. This can be easily done by dividing the given vectors by their magnitudes. Let us first calculate the unit vector along m→\overrightarrow{m}.
⇒m^=i^+j^(1)2+(1)2 ⇒m^=i^+j^2 \begin{aligned} & \Rightarrow \widehat{m}=\dfrac{\widehat{i}+\widehat{j}}{\sqrt{{{(1)}^{2}}+{{(1)}^{2}}}} \\\ & \Rightarrow \widehat{m}=\dfrac{\widehat{i}+\widehat{j}}{\sqrt{2}} \\\ \end{aligned}
Now, we will calculate the unit vector along n→\overrightarrow{n} . Therefore, we can write:
⇒n^=i^−j^(1)2+(1)2 ⇒n^=i^−j^2 \begin{aligned} & \Rightarrow \widehat{n}=\dfrac{\widehat{i}-\widehat{j}}{\sqrt{{{(1)}^{2}}+{{(1)}^{2}}}} \\\ & \Rightarrow \widehat{n}=\dfrac{\widehat{i}-\widehat{j}}{\sqrt{2}} \\\ \end{aligned}
Also, it has been given that we have to find the component of a→\overrightarrow{a} along these two given vectors.
⇒a→=2i^+3j^\Rightarrow \overrightarrow{a}=2\widehat{i}+3\widehat{j}
The component of a→\overrightarrow{a} along m→\overrightarrow{m} will be given as:
⇒am→=(a→.m^)m^\Rightarrow \overrightarrow{{{a}_{m}}}=(\overrightarrow{a}.\widehat{m})\widehat{m}
Putting all the respective values in the above equation, we get the required component as:
⇒am→=[(2i^+3j^).(i^+j^2)]i^+j^2 ∴am→=52i^+52j^ \begin{aligned} & \Rightarrow \overrightarrow{{{a}_{m}}}=\left[ \left( 2\widehat{i}+3\widehat{j} \right).\left( \dfrac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right) \right]\dfrac{\widehat{i}+\widehat{j}}{\sqrt{2}} \\\ & \therefore \overrightarrow{{{a}_{m}}}=\dfrac{5}{2}\widehat{i}+\dfrac{5}{2}\widehat{j} \\\ \end{aligned}
Now, finding the component of a→\overrightarrow{a} along n→\overrightarrow{n}. We will proceed as follows:
⇒an→=(a→.n^)n^\Rightarrow \overrightarrow{{{a}_{n}}}=(\overrightarrow{a}.\widehat{n})\widehat{n}
Putting all the respective values in the above equation, we can get the required component as:
⇒an→=[(2i^+3j^).(i^−j^2)]i^−j^2 ∴an→=−12i^+12j^ \begin{aligned} & \Rightarrow \overrightarrow{{{a}_{n}}}=\left[ \left( 2\widehat{i}+3\widehat{j} \right).\left( \dfrac{\widehat{i}-\widehat{j}}{\sqrt{2}} \right) \right]\dfrac{\widehat{i}-\widehat{j}}{\sqrt{2}} \\\ & \therefore \overrightarrow{{{a}_{n}}}=-\dfrac{1}{2}\widehat{i}+\dfrac{1}{2}\widehat{j} \\\ \end{aligned}
Hence, the component of a→\overrightarrow{a} along i^+j^\widehat{i}+\widehat{j}comes out to be 52i^+52j^\dfrac{5}{2}\widehat{i}+\dfrac{5}{2}\widehat{j}. And, the component of a→\overrightarrow{a}along i^−j^\widehat{i}-\widehat{j} comes out to be −12i^+12j^-\dfrac{1}{2}\widehat{i}+\dfrac{1}{2}\widehat{j}.

Note:
The component of one vector along the other is the projection of that vector on the directional vector. Here, we calculated the vector component along the direction of another vector but the question could be asked to find the vector component along the perpendicular to the given vector. In that case, we will first find the perpendicular directional vector and then proceed ahead.