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
$$\cos \theta ={\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}$$, where $\overrightarrow{a}$ and $\overrightarrow{b}$ 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,
 $\overrightarrow{OP}=\hat{i}+\hat{j}+\hat{k}$
The modulus of the vector r is given by,
 $\Rightarrow \overrightarrow{r}=a\hat{i}+b\hat{j}+c\hat{k}$ is $\left|\overrightarrow{r}\right|=\sqrt{a^2+b^2+c^2}$
The modulus of $\left|\overrightarrow{OP}\right|$ is given by,

 $$\begin{gathered} \Rightarrow \left|\overrightarrow{OP}\right|=\sqrt{1^2+1^2+1^2} \\ \Rightarrow \left|\overrightarrow{OP}\right|=\sqrt{3} \end{gathered}$$
Vector OQ is given by,
 $$\overrightarrow{OQ}=\hat{i}+\hat{j}$$
The modulus of vector OQ is given by,
 $$\begin{gathered} \Rightarrow \left|\overrightarrow{OQ}\right|=\sqrt{1^2+1^2} \\ \Rightarrow \left|\overrightarrow{OQ}\right|=\sqrt{2} \end{gathered}$$
The angle between $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ is given by,
$$\Rightarrow \cos \theta ={\displaystyle \frac{\overrightarrow{OP}.\overrightarrow{OQ}}{\left|\overrightarrow{OP}\right|\left|\overrightarrow{OQ}\right|}}\cdots \left(1\right)$$
Substitute the value of $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ in equation (1)
The dot product of same unit vector is $\hat{i}.\hat{i}=1$ and that of different unit vector is $\hat{i}.\hat{j}=0$
 $$\begin{gathered} \Rightarrow \cos \theta ={\displaystyle \frac{1+1+0}{\left(\sqrt{3}\right)\left(\sqrt{2}\right)}} \\ \Rightarrow \cos \theta ={\displaystyle \frac{2}{\left(\sqrt{3}\right)\left(\sqrt{2}\right)}} \\ \Rightarrow \cos \theta ={\displaystyle \frac{\sqrt{2}}{\sqrt{3}}} \\ \theta ={\cos }^{-1}\left(0.8165\right) \\ \Rightarrow \theta ={35.26}^o \end{gathered}$$
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
 $\left|\overrightarrow{AB}\right|=\overrightarrow{B}-\overrightarrow{A}$
The modulus of the vector $\overrightarrow{r}=a\hat{i}+b\hat{j}+c\hat{k}$ is calculated by taking the square root of the sum of the square of the coefficients and is given by the formula
 $\left|\overrightarrow{r}\right|=\sqrt{a^2+b^2+c^2}$
The modulus of the vectors tells about the magnitude of the vector.
The angle between the two vectors $\overrightarrow{m}$ and $\overrightarrow{n}$ is given by the formula
$$\cos \theta ={\displaystyle \frac{\overrightarrow{m}.\overrightarrow{n}}{\left|\overrightarrow{m}\right|\left|\overrightarrow{n}\right|}}$$
If the angle between the vectors is ${\displaystyle \frac{\pi }{2}}$ , then $\overrightarrow{m}.\overrightarrow{n}=0$
If the angle between the vector is $0$ , then $\overrightarrow{m}.\overrightarrow{n}=\left|\overrightarrow{m}\right|\left|\overrightarrow{n}\right|$ , and
If the angle between the vector is $\pi$ , then $\overrightarrow{m}.\overrightarrow{n}=-\left|\overrightarrow{m}\right|\left|\overrightarrow{n}\right|$