Answer
351.8k+ views
Hint: In order to solve this problem, we need to find the understanding of the term of direction cosine. Direction cosines of the vector are the cosines of the angles between the vector and the three coordinate axes.
The direction cosines are given by l, m, n. The angles made by vectors with x, y and z axes are $ \alpha ,\beta ,\gamma $ respectively. Therefore, the direction cosines of the vector with the x-axis, y-axis and z-axis are given by $ l=\cos \alpha $ , $ m=\cos \beta $ , $ n=\cos \gamma $ .
Complete step-by-step answer:
We are asked to find the direction cosine of the z-axis.
Let's understand what direction cosines are.
Direction cosines of the vector are the cosines of the angles between the vector and the three coordinate axes.
The direction cosines are given by l, m, n.
The angles made by vectors with x, y and z axes are $ \alpha ,\beta ,\gamma $ respectively.
Therefore, the direction cosine of the vector with the x-axis is given by $ l=\cos \alpha $ .
The direction cosine of the vector with y-axis is given by $ m=\cos \beta $ .
The direction cosine of the vector with z-axis is given by $ n=\cos \gamma $ .
We can show the axis as follows,
All the coordinate axes are perpendicular to each other.
We need to find the direction cosine of the z-axis.
As the vector is along the z-axis the angle made by x-axis is $ {{90}^{\circ }} $ .
The angle made by the z-axis is hence zero.
The angle made by the y-axis with the vector is $ {{90}^{\circ }} $ .
Hence, the angles are $ \alpha ={{90}^{\circ }},\beta ={{90}^{\circ }},\gamma ={{0}^{\circ }} $ .
Taking the cosine of all angles we get,
$ l=\cos 90,m=\cos 90,n=\cos 0 $ .
Finding the values, we get,
$ l=0,m=0,n=1 $
Therefore, the direction cosines are $ \left( l,m,n \right)=\left( 0,0,1 \right) $
Note: We need to understand that all the angles are calculated from the positive axis of all the three coordinates. These direction cosines can find the relationship of any vector with all the coordinate axes.We can similarly find the direction cosine of the x-axis. By symmetry, we can see that the direction cosine of the x-axis is $ \left( l,m,n \right)=\left( 1,0,0 \right) $ and that of the y-axis is $ \left( l,m,n \right)=\left( 0,1,0 \right) $ .
The direction cosines are given by l, m, n. The angles made by vectors with x, y and z axes are $ \alpha ,\beta ,\gamma $ respectively. Therefore, the direction cosines of the vector with the x-axis, y-axis and z-axis are given by $ l=\cos \alpha $ , $ m=\cos \beta $ , $ n=\cos \gamma $ .
Complete step-by-step answer:
We are asked to find the direction cosine of the z-axis.
Let's understand what direction cosines are.
Direction cosines of the vector are the cosines of the angles between the vector and the three coordinate axes.
The direction cosines are given by l, m, n.
The angles made by vectors with x, y and z axes are $ \alpha ,\beta ,\gamma $ respectively.
Therefore, the direction cosine of the vector with the x-axis is given by $ l=\cos \alpha $ .
The direction cosine of the vector with y-axis is given by $ m=\cos \beta $ .
The direction cosine of the vector with z-axis is given by $ n=\cos \gamma $ .
We can show the axis as follows,
![seo images](https://www.vedantu.com/question-sets/ceae94e3-84fb-4b49-9c05-262508954f8d5478551099652275870.png)
All the coordinate axes are perpendicular to each other.
We need to find the direction cosine of the z-axis.
As the vector is along the z-axis the angle made by x-axis is $ {{90}^{\circ }} $ .
The angle made by the z-axis is hence zero.
The angle made by the y-axis with the vector is $ {{90}^{\circ }} $ .
Hence, the angles are $ \alpha ={{90}^{\circ }},\beta ={{90}^{\circ }},\gamma ={{0}^{\circ }} $ .
Taking the cosine of all angles we get,
$ l=\cos 90,m=\cos 90,n=\cos 0 $ .
Finding the values, we get,
$ l=0,m=0,n=1 $
Therefore, the direction cosines are $ \left( l,m,n \right)=\left( 0,0,1 \right) $
Note: We need to understand that all the angles are calculated from the positive axis of all the three coordinates. These direction cosines can find the relationship of any vector with all the coordinate axes.We can similarly find the direction cosine of the x-axis. By symmetry, we can see that the direction cosine of the x-axis is $ \left( l,m,n \right)=\left( 1,0,0 \right) $ and that of the y-axis is $ \left( l,m,n \right)=\left( 0,1,0 \right) $ .
Recently Updated Pages
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)