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Find the angle between the vectors $\hat{i}-2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+\hat{k}$.

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Last updated date: 13th Jun 2024
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Answer
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Hint: The given problem statement states to find the angle between the vectors with the help of the scalar product. You can assume each vector a variable that means one vector will be “a” and another one will be “b”. So, let’s look at the approach of the problem statement.

Complete Complete Step by Step Solution:
The given problem statement is to find the angle between the vectors $\hat{i}-2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+\hat{k}$.
Now, the first thing is to assume each of the vectors, that means, we will let $\hat{i}-2\hat{j}+3\hat{k}$is $\vec{a}$and $3\hat{i}-2\hat{j}+\hat{k}$ is $\vec{b}$.
So, if we write the vectors in a different manner, that means,
$\Rightarrow \vec{a}=1\hat{i}-2\hat{j}+3\hat{k}$
$\Rightarrow \vec{b}=3\hat{i}-2\hat{j}+1\hat{k}$
Now, we will use the scalar product formula, that means, we get,
$\Rightarrow \vec{a}.\vec{b}=|\vec{a}||\vec{b}|\cos \theta $ where ,$\theta $is the angle between $\vec{a}$ and $\vec{b}$.
Now, we will find$\vec{a}.\vec{b}$, that means, we get,
$\Rightarrow \vec{a}.\vec{b}=(1\hat{i}-2\hat{j}+3\hat{k}).(3\hat{i}-2\hat{j}+1\hat{k})$
Now, when we solve this above equation, we get,
$\Rightarrow \vec{a}.\vec{b}=[(1\hat{i}.3\hat{i})+(-2\hat{j}.-2\hat{j})+(3\hat{k}.1\hat{k})]$
We have kept$\hat{i}.\hat{i}$, $\hat{j}.\hat{j}$ and $\hat{k}.\hat{k}$ because all these have the values 1 instead of $\hat{i}.\hat{j}$, $\hat{j}.\hat{k}$ and $\hat{k}.\hat{i}$because all these have the values as 0.
$\Rightarrow \vec{a}.\vec{b}=[(1.3)+(-2.-2)+(3.1)]$
$\Rightarrow \vec{a}.\vec{b}=3+4+3$
Now, when we solve, we get,
$\Rightarrow \vec{a}.\vec{b}=10$
Now, we will find magnitude of $\vec{a}$, that means, we get,
$\Rightarrow \vec{a}=1\hat{i}-2\hat{j}+3\hat{k}$
$\Rightarrow \vec{a}=\sqrt{{{(1)}^{2}}+{{(-2)}^{2}}+{{(3)}^{2}}}$
Now, when we solve, we get,
$\Rightarrow \vec{a}=\sqrt{1+4+9}$
$\Rightarrow \vec{a}=\sqrt{14}$
Similarly, we will find magnitude of $\vec{b}$, that means, we get,
$\Rightarrow \vec{b}=3\hat{i}-2\hat{j}+1\hat{k}$
$\Rightarrow \vec{b}=\sqrt{{{(3)}^{2}}+{{(-2)}^{2}}+{{(1)}^{2}}}$
Now, when we solve, we get,
$\Rightarrow \vec{b}=\sqrt{9+4+1}$
$\Rightarrow \vec{b}=\sqrt{14}$
Now, we will put the respective values in the formula $\vec{a}.\vec{b}=|\vec{a}||\vec{b}|\cos \theta $, we get,
$\Rightarrow 10=\sqrt{14}\sqrt{14}\cos \theta $
Now, we will rearrange the equation, we get,
$\Rightarrow \dfrac{10}{\sqrt{14}\sqrt{14}}=\cos \theta $
As, we now $\sqrt{y}\sqrt{y}=y$, similarly we will apply in the equation and then we will convert it to lowest terms, we get,
$\Rightarrow \dfrac{10}{14}=\cos \theta $
$\Rightarrow \dfrac{5}{7}=\cos \theta $
Now when we rearrange we will get the value of $\theta $, we get,
$\Rightarrow {{\cos }^{-1}}(\dfrac{5}{7})=\theta $
After rearranging the equation, we get,
$\Rightarrow \theta ={{\cos }^{-1}}(\dfrac{5}{7})$

Therefore, the value of $\theta $or the angle between two vectors is ${{\cos }^{-1}}(\dfrac{5}{7})$.

Note:
In the above problem statement, we have used the fine concept of the scalar product. The scalar product is also known as the dot product. The formula used in the scalar product is $\vec{a}.\vec{b}=|\vec{a}||\vec{b}|\cos \theta $. You need to note that the scalar product is commutative as well as distributive.