What is the vector equally inclined to the vectors\[\hat{i}+3\hat{j}\] and\[3\hat{i}+\hat{j}\]?
A) \[\hat{i}+\hat{j}\]
B) \[2\hat{i}-\hat{j}\]
C) \[2\hat{i}+\hat{j}\]
D) None of theses
Correct Answer: A
Solution :
[a] Let the required vector be \[\hat{i}+\hat{j}\] Since the vector \[\hat{i}+\hat{j}\] is equally inclined to the vectors \[\hat{i}+3\hat{j}\] and \[3\hat{i}+\hat{j}\] therefore angle b/w \[\hat{i}+\hat{j}\] and \[\hat{i}+3\hat{j}={{\theta }_{1}}\] is equal to angle between \[\hat{i}+\hat{j}\] and \[\hat{i}+3\hat{j}\] \[={{\cos }^{-1}}\left[ \frac{(1)(1)+(1)(3)}{\sqrt{{{(1)}^{2}}+{{(1)}^{2}}}\sqrt{{{(1)}^{2}}+{{(3)}^{2}}}} \right]\] \[={{\cos }^{-1}}\left[ \frac{1+3}{\sqrt{2}\sqrt{10}} \right]={{\cos }^{-1}}\left[ \frac{4}{\sqrt{2}\sqrt{10}} \right]\] \[={{\cos }^{-1}}\left[ \frac{2}{\sqrt{5}} \right]\] and angle between \[\hat{i}+\hat{j}\] and \[(3\hat{i}+\hat{j})\] \[={{\cos }^{-1}}\left| \frac{1+3}{\sqrt{10}\sqrt{2}} \right|\] \[={{\cos }^{-1}}\left( \frac{4}{\sqrt{2}\sqrt{10}} \right)={{\cos }^{-1}}\left( \frac{2}{\sqrt{5}} \right)\] Hence, required vector is \[\hat{i}+\hat{j}\]
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