Find angle between unit vectors $ \hat i + \hat j + \hat k $ and $ \hat i + \hat j $ by cross product?

Hint: Here, we have to find the angle between unit vectors $ \hat i + \hat j + \hat k $ and $ \hat i + \hat j $ by cross product. Cross product of two vectors can be defined as a binary operation on two vectors in three dimensional spaces and denoted by $ \times $ . In order to find the angle between given vectors by using cross product we use the formula which is $ \left| {\vec c} \right| = \left| a \right|\left| b \right|\sin \theta $ where $ a $ and $ b $ are the magnitudes of the vector, $ c $ is the magnitude of the vector product and $ \theta $ is the angle between these two vectors.

Complete answer:
The cross- vector product, area product, or the vector product of two vectors can be defined as a binary operation on two vectors in three dimensional spaces and denoted by $ \times $ .
The magnitude of the vector product can be given as $ \left| {\vec c} \right| = \left| a \right|\left| b \right|\sin \theta $ , where $ a $ and $ b $ are the magnitudes of the vector, $ c $ is the magnitude of the vector product and $ \theta $ is the angle between these two vectors.
Here we have to find the angle between two unit vectors. So,
 $ \sin \theta = \dfrac{{\left| {\vec c} \right|}}{{\left| {\vec a} \right|\left| {\vec b} \right|}} $
Let $ \left| {\vec a} \right| = \hat i + \hat j + \hat k $ and $ \left| {\vec b} \right| = \hat i + \hat j $
Now, we will find the magnitude of the vectors. We have,
 $ \Rightarrow \left| {\vec a} \right| = \sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}} $
Solving the square root. We get,
 $ \Rightarrow \left| {\vec a} \right| = \sqrt 3 $
 $ \Rightarrow \left| {\vec b} \right| = \sqrt {{{(1)}^2} + {{(1)}^2}} $
Solving the square root. We get,
 $ \Rightarrow \left| {\vec b} \right| = \sqrt 2 $
Therefore, the magnitude of $ \left| {\vec a} \right| = \sqrt 3 $ and $ \left| {\vec b} \right| = \sqrt 2 $
Now, we will calculate the magnitude of the cross product for that we have to find the cross product of two vectors. So,
 $ \Rightarrow \vec a \times \vec b = (\vec i + \vec j + \vec k) \times (\vec i + \vec j) $
 $ \Rightarrow \vec a \times \vec b = (\hat i \times \hat i) + (\hat i \times \hat j) + (\hat j \times \hat i) + (\hat j \times \hat j) + (\hat k \times \hat i) + (\hat k \times \hat j) $ .
We know that in cross product $ \hat i \times \hat i = \hat j \times \hat j = \hat k \times \hat k = 0 $ and $ \hat i \times \hat j = \hat k,\,\,\,\hat j \times \hat i = - \hat k,\,\,\,\hat k \times \hat i = \hat j,\,\,\,\hat k \times \hat j = - \hat i $
Therefore,
 $ \Rightarrow \vec a \times \vec b = 0 + \hat k + ( - \hat k) + 0 + \hat j + ( - \hat i) $
Cancelling out the equal term with the opposite sign. We get,
 $ \Rightarrow \vec a \times \vec b = \hat j - \hat i $
Now, the magnitude of the cross product will be
 $ \Rightarrow \left| {\vec a \times \vec b} \right| = \sqrt {{{(1)}^2} + {{(1)}^2}} $
Solving the square root. We get,
 $ \Rightarrow \left| {\vec a \times \vec b} \right| = \sqrt 2 $
Now, angle between vector is given by $ \sin \theta = \dfrac{{\left| {\vec c} \right|}}{{\left| {\vec a} \right|\left| {\vec b} \right|}} $
Therefore,
 $ \Rightarrow \sin \theta = \left[ {\dfrac{{\sqrt 2 }}{{\sqrt 3 \cdot \sqrt 2 }}} \right] $
Cancelling out the equal terms. We get,
 $ \Rightarrow \sin \theta = \left[ {\dfrac{1}{{\sqrt 3 }}} \right] $
Taking the inverse of $ \sin $ function. we get,
 $ \Rightarrow \theta = {\sin ^{ - 1}}\left[ {\dfrac{1}{{\sqrt 3 }}} \right] $
 $ \Rightarrow \theta = 35.26^\circ $
Hence, the angle between unit vectors $ \hat i + \hat j + \hat k $ and $ \hat i + \hat j $ is $ 35.26^\circ $ .

Note:
The vector product of two vectors basically refers to a vector that is perpendicular to both of the vectors and can be obtained by multiplying their magnitudes by the $ \sin $ of the angle that exists between them. We can calculate the direction of the vector product with the help of the right- hand thumb rule in which we curl our fingers of the right hand around a line perpendicular to the plane of vectors $ a $ and $ b $ , then stretched thumb points in the direction of $ c $ .