
Find the angle between the x-axis and the vector $ \hat i + \hat j + \hat k $ .
Answer
530.1k+ views
Hint: In the given question, we are required to find the angle between a vector given to us in the question and the x-axis. We can find the angle between two vectors by various methods such as cross product of the two vectors and dot product of the two vectors. X-axis can also be presented by the vector $ \hat i $ as $ \hat i $ is the vector in the direction of the positive x-axis and with magnitude 1 unit.
Complete step by step solution:
In the question given to us, we are asked to find the angle between the x-axis and the vector $ \hat i + \hat j + \hat k $ . We know that the x-axis is represented by the vector $ \hat i $ as $ \hat i $ is the unit vector in the direction of the positive x-axis. Now, we can find the angle between the two vectors by computing their dot product.
The dot product of two vectors $ \vec a $ and $ \vec b $ is computed as \[\vec a.\vec b = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta \] where \[\theta \] is the angle between the two vectors.
So, let the angle between the x-axis and the vector $ \hat i + \hat j + \hat k $ is $ \alpha $ .
Now, the dot product of the vectors $ \hat i $ and $ \hat i + \hat j + \hat k $ $ = 1\left( 1 \right) + 1\left( 0 \right) + 1\left( 0 \right) = 1 $
Magnitude of vector $ \hat i + \hat j + \hat k $ $ = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3 $
Magnitude of vector $ \hat i $ \[ = \sqrt {{1^2} + {0^2} + {0^2}} = \sqrt 1 = 1\]
Therefore, $ \left( {\sqrt 3 } \right)\left( 1 \right)\cos \alpha = 1 $
$ \Rightarrow \cos \alpha = \left( {\dfrac{1}{{\sqrt 3 }}} \right) $
\[ \Rightarrow \alpha = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)\]
The principal value branch of \[{\cos ^{ - 1}}\] function is $ \left[ {0,\pi } \right] $ .
So, the angle between the x-axis and the vector $ \hat i + \hat j + \hat k $ is \[{\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)\] lying between $ \left[ {0,\pi } \right] $ .
Note: Above question can also be solved by computing the cross product of the two vectors but the cross product also involves the sense of direction along with the magnitude which makes finding angle between two vectors an inaccurate and tedious process.
Complete step by step solution:
In the question given to us, we are asked to find the angle between the x-axis and the vector $ \hat i + \hat j + \hat k $ . We know that the x-axis is represented by the vector $ \hat i $ as $ \hat i $ is the unit vector in the direction of the positive x-axis. Now, we can find the angle between the two vectors by computing their dot product.
The dot product of two vectors $ \vec a $ and $ \vec b $ is computed as \[\vec a.\vec b = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta \] where \[\theta \] is the angle between the two vectors.
So, let the angle between the x-axis and the vector $ \hat i + \hat j + \hat k $ is $ \alpha $ .
Now, the dot product of the vectors $ \hat i $ and $ \hat i + \hat j + \hat k $ $ = 1\left( 1 \right) + 1\left( 0 \right) + 1\left( 0 \right) = 1 $
Magnitude of vector $ \hat i + \hat j + \hat k $ $ = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3 $
Magnitude of vector $ \hat i $ \[ = \sqrt {{1^2} + {0^2} + {0^2}} = \sqrt 1 = 1\]
Therefore, $ \left( {\sqrt 3 } \right)\left( 1 \right)\cos \alpha = 1 $
$ \Rightarrow \cos \alpha = \left( {\dfrac{1}{{\sqrt 3 }}} \right) $
\[ \Rightarrow \alpha = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)\]
The principal value branch of \[{\cos ^{ - 1}}\] function is $ \left[ {0,\pi } \right] $ .
So, the angle between the x-axis and the vector $ \hat i + \hat j + \hat k $ is \[{\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)\] lying between $ \left[ {0,\pi } \right] $ .
Note: Above question can also be solved by computing the cross product of the two vectors but the cross product also involves the sense of direction along with the magnitude which makes finding angle between two vectors an inaccurate and tedious process.
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