
If a and b are two unit vectors such that angle between them is \[60{}^\circ \] , then \[\left| a-b \right|\] =
1. \[\sqrt{5}\]
2. \[\sqrt{3}\]
3. \[0\]
4. \[1\]
Answer
507.6k+ views
Hint: To solve this question you must know the basic properties and formulas which we use in vectors now here since we know that the angle between the two vectors is \[60{}^\circ \] then we also know the cosine value between them which we can write as \[\theta =60{}^\circ \]. Now since it is also said that these are unit vectors we can say that \[\left| a \right|=\left| b \right|=1\]. We can solve this question by the known formula that
\[\left| a-b \right|=\sqrt{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}-2\left| a \right|\times \left| b \right|\times \cos \theta }\]
Complete step-by-step solution:
Now here since both of those vectors are unit vectors we can say that the magnitude of both of them will be equal to one. Now we let one vector to be a and other vector to be b. Therefore we can write the above expressed statement in a mathematical expression that
\[\left| a \right|=\left| b \right|=1\]
Now here it is also said that the angle between the two vectors is also \[60{}^\circ \] therefore we can write it in the form of cosine function so that we can substitute it in the formula we can write is as
\[\theta =60{}^\circ \]
Now substituting all of these values we know in the formula for vector subtraction which is \[\left| a-b \right|=\sqrt{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}-2\left| a \right|\times \left| b \right|\times \cos \theta }\] we get
\[\left| a-b \right|=\sqrt{{{1}^{2}}+{{1}^{2}}-2\times \cos 60}\]
Putting the value for cosine we get
\[\left| a-b \right|=\sqrt{{{1}^{2}}+{{1}^{2}}-2\times \dfrac{1}{2}}\]
Multiplying ;
\[\left| a-b \right|=\sqrt{{{1}^{2}}+{{1}^{2}}-1}\]
Taking the square and adding and subtracting
\[\left| a-b \right|=1\]
Hence this is how we got the answer we needed. The answer for this question is option 4.
Note: A vector is a type of line that will always have magnitude and direction. The length of the line shows its magnitude and the arrowhead points in the direction. We can add two vectors by joining those head-to-tail. We can also subtract two vectors by first taking a negative of a vector, that is take the vector completely opposite it and then add it with the other vector.
\[\left| a-b \right|=\sqrt{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}-2\left| a \right|\times \left| b \right|\times \cos \theta }\]
Complete step-by-step solution:
Now here since both of those vectors are unit vectors we can say that the magnitude of both of them will be equal to one. Now we let one vector to be a and other vector to be b. Therefore we can write the above expressed statement in a mathematical expression that
\[\left| a \right|=\left| b \right|=1\]
Now here it is also said that the angle between the two vectors is also \[60{}^\circ \] therefore we can write it in the form of cosine function so that we can substitute it in the formula we can write is as
\[\theta =60{}^\circ \]
Now substituting all of these values we know in the formula for vector subtraction which is \[\left| a-b \right|=\sqrt{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}-2\left| a \right|\times \left| b \right|\times \cos \theta }\] we get
\[\left| a-b \right|=\sqrt{{{1}^{2}}+{{1}^{2}}-2\times \cos 60}\]
Putting the value for cosine we get
\[\left| a-b \right|=\sqrt{{{1}^{2}}+{{1}^{2}}-2\times \dfrac{1}{2}}\]
Multiplying ;
\[\left| a-b \right|=\sqrt{{{1}^{2}}+{{1}^{2}}-1}\]
Taking the square and adding and subtracting
\[\left| a-b \right|=1\]
Hence this is how we got the answer we needed. The answer for this question is option 4.
Note: A vector is a type of line that will always have magnitude and direction. The length of the line shows its magnitude and the arrowhead points in the direction. We can add two vectors by joining those head-to-tail. We can also subtract two vectors by first taking a negative of a vector, that is take the vector completely opposite it and then add it with the other vector.
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