If $\theta $ is the angle between two vector $\widehat i - 2\widehat j + 3\widehat k$ and $3\widehat i - 2\widehat j + \widehat k$, what is the value of $\theta $.

Hint: To solve an angle between the two vectors, the angle between the vectors should be calculated using the trigonometric formula, and the rearrangement of the point value such point is described. Thus, the angle cosine between two vectors can be determined by dividing the dot product of their magnitude and the by-product of the vector angle.

Formula used: If angle between the given vector,
Then,
$\overrightarrow a .\overrightarrow {b\,} = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta $
The above equation is rearranging to solve for an angle $\theta $,
$\cos \theta = \dfrac{{\vec a \bullet \vec b}}{{\left| {\vec a} \right|\left| {\vec b} \right|}}$
Where,
$\theta $ is the angle between $\overrightarrow a $ and $\overrightarrow b $

Complete step-by-step answer:
Let us assume
$
  \overrightarrow a = \widehat i - 2\widehat j + 3\widehat k \\
  \overrightarrow b = 3\widehat i - 2\widehat j + \widehat k \\
 $
Now, the dot product of the two vector is given by,
$\Rightarrow$$\vec a \bullet \vec b = \left( {\hat i - 2\hat j + 3\hat k} \right) \bullet \left( {3\hat i - 2\hat j + \hat k} \right)$
Multiply with respect to $\widehat i$ , $\widehat j$ and $\widehat k$ ,
$\Rightarrow$$\vec a \bullet \vec b = 3 + 4 + 3$
Simplifying the a above equation,
$\Rightarrow$$\vec a \bullet \vec b = 10$ …………….$(1)$
Now we find magnitudes of $\overrightarrow a $and$\overrightarrow b $
Solving the Magnitude of $\overrightarrow a $
We get,
$\Rightarrow$$\overrightarrow a = \sqrt {{1^2} + ( - {2^2}) + {3^2}} $
On simplifying the above equation,
$\Rightarrow$$\left| {\overrightarrow a } \right| = \sqrt {1 + 4 + 9} $
To add the given value of the equation,
$\Rightarrow$$\left| {\overrightarrow a } \right| = \sqrt {14} $…………….$(2)$
Solving Magnitude of$\overrightarrow b $
We get,
$\Rightarrow$$\overrightarrow b = \sqrt {{3^2} + ( - {2^2}) + {1^2}} $
On simplifying the b above equation,
$\Rightarrow$\[\left| {\overrightarrow b } \right| = \sqrt {9 + 4 + 1} \]
To add the given value of the equation,
$\Rightarrow$$\left| {\overrightarrow b } \right| = \sqrt {14} $ ……………… $(3)$
Now,
We know that, the angle of the vector,
$\Rightarrow$$\vec a \bullet \vec b = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta $
To substitute a respective value of the above equation,
We get,
$\Rightarrow$$10 = \sqrt {14} \times \sqrt {14} \cos \theta $
On simplifying,
$\Rightarrow$$10 = 14 \times \cos \theta $
Rearrange the given equation,
$\Rightarrow$$\cos \theta = \dfrac{{10}}{{14}}$
Here,
$\Rightarrow$$\theta = {\cos ^{ - 1}}\left( {\dfrac{5}{7}} \right)$

Hence the answer of this angle between $\overrightarrow a $ and $\overrightarrow b $ is ${\cos ^{ - 1}}\left( {\dfrac{5}{7}} \right)$

Note: In this problem with the use of dot products to be solved in the vector form. A vector's magnitude is its length. The addition of two vectors is achieved by sequentially laying the vector to tail to create a triangle. The scalar product grows as one vector follows the trajectory of the other one. Thus, the scalar product is limit at angle zero.