Answer
64.8k+ views
Hint: In this question, we can use the dot product to find the angle between the vectors. In dot products, we can calculate the magnitude of each vector separately. After that we can find the dot product of the vectors. We can find the angle by using the formula\[\overline A \bullet \overline B = \left| {\overline A } \right|\left| {\overline B } \right|\cos \theta \]and substituting the all calculated values in it.
Complete step by step solution:
First, we can let
\[\overline A = \widehat i + 2\widehat j + 2\widehat k\]
And $\overline B = \widehat i$
We know that the dot product of two vectors is given as-
\[\overline A \bullet \overline B = \left| {\overline A } \right|\left| {\overline B } \right|\cos \theta \] (i)
Where \[\left| {\overline A } \right| = \] magnitude of \[\overline A \] , \[\left| {\overline B } \right| = \] magnitude of \[\overline B \]and \[\theta = \] angle between the two vectors.
Now, let us find the magnitudes of the two vectors separately.
So, the magnitude of \[\left| {\overline A } \right|\]is
\[
\left| {\overline A } \right| = \sqrt {{1^2} + {2^2} + {2^2}} \\
\Rightarrow \left| {\overline A } \right| = \sqrt {1 + 4 + 4} \\
\Rightarrow \left| {\overline A } \right| = \sqrt 9 \\
\Rightarrow \left| {\overline A } \right| = 3 \\
\]
And the magnitude of \[\left| {\overline B } \right|\]is
\[
\left| {\overline B } \right| = \sqrt {{1^2}} \\
\Rightarrow \left| {\overline B } \right| = \sqrt 1 \\
\Rightarrow \left| {\overline B } \right| = 1 \\
\]
Now, let us find the dot product of both vectors, which will be-
$
\overline A \bullet \overline B = \left( {\widehat i + 2\widehat j + 2\widehat k} \right) \bullet \left( {\widehat i} \right) \\
\Rightarrow \overline A \bullet \overline B = 1 \\
$
Now, the angle \[\theta \]will be written as the following equation by using equation (i), we get-
$\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overline A \bullet \overline B }}{{\left| {\overline A } \right|\left| {\overline B } \right|}}} \right)$
Substituting all the values, we get-
$
\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overline A \bullet \overline B }}{{\left| {\overline A } \right|\left| {\overline B } \right|}}} \right) \\
\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{1}{{\left( 1 \right)\left( 3 \right)}}} \right) \\
\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right) \\
\Rightarrow \theta = {\cos ^{ - 1}}\left( {\cos {{70.53}^ \circ }} \right) \\
\Rightarrow \theta = {70.53^ \circ } \\
$
Hence, The angle between the vectors is ${70.53^ \circ }$.
Note: We have to remember the rules of dot product. In dot product, we know that $\widehat i \bullet \widehat i = 1,\widehat j \bullet \widehat j = 1,\widehat k \bullet \widehat k = 1$ . The other products will be zero in dot product. We have to remember that while finding the magnitude all the coefficients will be squared.
We have to remember that we can find the angle between the two vectors by using dot product. In a vector product, we have unit vectors also in the equation, which gives a difficult calculation. Sometimes, we don’t have the value of the unit vector. But in dot products, we only need to find the magnitudes of the vectors separately and the dot product of the vectors.
Complete step by step solution:
First, we can let
\[\overline A = \widehat i + 2\widehat j + 2\widehat k\]
And $\overline B = \widehat i$
We know that the dot product of two vectors is given as-
\[\overline A \bullet \overline B = \left| {\overline A } \right|\left| {\overline B } \right|\cos \theta \] (i)
Where \[\left| {\overline A } \right| = \] magnitude of \[\overline A \] , \[\left| {\overline B } \right| = \] magnitude of \[\overline B \]and \[\theta = \] angle between the two vectors.
Now, let us find the magnitudes of the two vectors separately.
So, the magnitude of \[\left| {\overline A } \right|\]is
\[
\left| {\overline A } \right| = \sqrt {{1^2} + {2^2} + {2^2}} \\
\Rightarrow \left| {\overline A } \right| = \sqrt {1 + 4 + 4} \\
\Rightarrow \left| {\overline A } \right| = \sqrt 9 \\
\Rightarrow \left| {\overline A } \right| = 3 \\
\]
And the magnitude of \[\left| {\overline B } \right|\]is
\[
\left| {\overline B } \right| = \sqrt {{1^2}} \\
\Rightarrow \left| {\overline B } \right| = \sqrt 1 \\
\Rightarrow \left| {\overline B } \right| = 1 \\
\]
Now, let us find the dot product of both vectors, which will be-
$
\overline A \bullet \overline B = \left( {\widehat i + 2\widehat j + 2\widehat k} \right) \bullet \left( {\widehat i} \right) \\
\Rightarrow \overline A \bullet \overline B = 1 \\
$
Now, the angle \[\theta \]will be written as the following equation by using equation (i), we get-
$\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overline A \bullet \overline B }}{{\left| {\overline A } \right|\left| {\overline B } \right|}}} \right)$
Substituting all the values, we get-
$
\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overline A \bullet \overline B }}{{\left| {\overline A } \right|\left| {\overline B } \right|}}} \right) \\
\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{1}{{\left( 1 \right)\left( 3 \right)}}} \right) \\
\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right) \\
\Rightarrow \theta = {\cos ^{ - 1}}\left( {\cos {{70.53}^ \circ }} \right) \\
\Rightarrow \theta = {70.53^ \circ } \\
$
Hence, The angle between the vectors is ${70.53^ \circ }$.
Note: We have to remember the rules of dot product. In dot product, we know that $\widehat i \bullet \widehat i = 1,\widehat j \bullet \widehat j = 1,\widehat k \bullet \widehat k = 1$ . The other products will be zero in dot product. We have to remember that while finding the magnitude all the coefficients will be squared.
We have to remember that we can find the angle between the two vectors by using dot product. In a vector product, we have unit vectors also in the equation, which gives a difficult calculation. Sometimes, we don’t have the value of the unit vector. But in dot products, we only need to find the magnitudes of the vectors separately and the dot product of the vectors.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)