Question

Find the angle between any diagonals of a cube.

Answer 1

Hint: The formula to be used for the for calculating the angle between the two vectors is
\[\cos \theta = \dfrac{{\vec a.\vec b}}{{\left| {\vec a} \right|\left| {\vec b} \right|}}\], where $ \vec a $ and $ \vec b $ are two vectors.

Complete step-by-step answer:
The figure below shows a cube of side length 1, in which OQ and OP are its diagonals. O is the origin of the cube.

From the above figure,
The coordinates of point $ P(1,1,1) $
The coordinates of point $ Q(1,1,0) $
Vector OP is given by,
 $ \mathop {OP}\limits^ \to = \hat i + \hat j + \hat k $
The modulus of the vector r is given by,
 $ \Rightarrow \vec r = a\hat i + b\hat j + c\hat k $ is $ \left| {\vec r} \right| = \sqrt {{a^2} + {b^2} + {c^2}} $
The modulus of $ \left| {\mathop {OP}\limits^ \to } \right| $ is given by,

 $
\Rightarrow \left| {\mathop {OP}\limits^ \to } \right| = \sqrt {{1^2} + {1^2} + {1^2}} \\
\Rightarrow \left| {\mathop {OP}\limits^ \to } \right| = \sqrt 3 \\
  $
Vector OQ is given by,
 \[\mathop {OQ}\limits^ \to = \hat i + \hat j\]
The modulus of vector OQ is given by,
 $
\Rightarrow \left| {\mathop {OQ}\limits^ \to } \right| = \sqrt {{1^2} + {1^2}} \\
\Rightarrow \left| {\mathop {OQ}\limits^ \to } \right| = \sqrt 2 \\
  $
The angle between $ \mathop {OP}\limits^ \to $ and $ \mathop {OQ}\limits^ \to $ is given by,
\[\Rightarrow \cos \theta = \dfrac{{\mathop {OP}\limits^ \to .\mathop {OQ}\limits^ \to }}{{\left| {\mathop {OP}\limits^ \to } \right|\left| {\mathop {OQ}\limits^ \to } \right|}} \cdots \left( 1 \right)\]
Substitute the value of $ \mathop {OP}\limits^ \to $ and $ \mathop {OQ}\limits^ \to $ in equation (1)
The dot product of same unit vector is $ \hat i.\hat i = 1 $ and that of different unit vector is $ \hat i.\hat j = 0 $
 $
\Rightarrow \cos \theta = \dfrac{{1 + 1 + 0}}{{\left( {\sqrt 3 } \right)\left( {\sqrt 2 } \right)}} \\
\Rightarrow \cos \theta = \dfrac{2}{{\left( {\sqrt 3 } \right)\left( {\sqrt 2 } \right)}} \\
\Rightarrow \cos \theta = \dfrac{{\sqrt 2 }}{{\sqrt 3 }} \\
  \theta = {\cos ^{ - 1}}\left( {0.8165} \right) \\
\Rightarrow \theta = {35.26^o} \\
  $
Hence, the angle between any two diagonals of a cube is $ {35.26^o} $

Note: The important point that is to be noted are,
The value of vector AB if vector A and vector B are given, is calculated as
 $ \left| {\mathop {AB}\limits^ \to } \right| = \mathop B\limits^ \to - \mathop A\limits^ \to $
The modulus of the vector $ \vec r = a\hat i + b\hat j + c\hat k $ is calculated by taking the square root of the sum of the square of the coefficients and is given by the formula
 $ \left| {\vec r} \right| = \sqrt {{a^2} + {b^2} + {c^2}} $
The modulus of the vectors tells about the magnitude of the vector.
The angle between the two vectors $ \vec m $ and $ \vec n $ is given by the formula
\[\cos \theta = \dfrac{{\vec m.\vec n}}{{\left| {\vec m} \right|\left| {\vec n} \right|}}\]
If the angle between the vectors is $ \dfrac{\pi }{2} $ , then $ \vec m.\vec n = 0 $
If the angle between the vector is $ 0 $ , then $ \vec m.\vec n = \left| {\vec m} \right|\left| {\vec n} \right| $ , and
If the angle between the vector is $ \pi $ , then $ \vec m.\vec n = - \left| {\vec m} \right|\left| {\vec n} \right| $