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 $\vec m.\vec n = - \left| {\vec m} \right|\left| {\vec n} \right|$