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We study the different phenomena happening around us by using a 3D coordinate system. It has become crucial to study the behaviour of line, area, and volume elements in the 3D Cartesian coordinate system.
The direction ratio tells us the orientation of a line segment with respect to the coordinate axes. We can draw a unique straight line passing through any two points. This line makes a different angle with respect to each coordinate axis. These angles of inclination help us further in our analysis.
How Do We Represent A Line Segment?
Consider the two points E(d1,e1,f1) and F(d2,e2,f2). Let O(0,0,0) be the origin. The position vector of point L will be:
OE=d1i+e1j+f1k
Position vector of point M will be:
OF=d2i+e2j+f2k
The vector equation of line segment EF is given as:
EF=OF–OE
∴ EF=(d2i+e2j+f2k)-(d1i+e1j+f1k)
=(d2–d1)i+(e2–e1)j+(f2–f1)k ………(1)
Direction Cosines Of Line EF
The line segment EF is represented by equation (1). Does this equation tell us anything about the orientation of the line segment with respect to the coordinate axes? The answer is ‘No’. We need to find the angle of inclination of the line with respect to each axis.
As per the diagram below, line makes an angle , , with the x-axis, y-axis, and z-axis respectively.
Let the length of the line segment EF be P. Then,
x=Pcos()
y=Pcos()
z=Pcos()
If we assume l=cos(), m=cos(), n=cos(), then,
x=Pl
y=Pm
z=Pn
Now, we know that,
P2=x2+y2+z2
∴ P2=(Pcos())2+(Pcos())2+(Pcos())2=P2(cos2()+cos2()+cos2())
∴ cos2()+cos2()+cos2()=1
∴ l2+m2+n2=1 ……………..(2)
Here, l=cos(), m=cos(), n=cos() are called direction cosines.
If we want to find the direction cosines of a given vector, we can find it by dividing each coordinate of the vector by the magnitude of the vector.
Direction Cosine Tells The Direction
If a vector K=Kxi+Kyj+Kzk is given, then the angles , , are given as:
=(K.i)/|K|=(KX)/(Kx2+Ky2+Kz2)
=(K.j)/|K|=(KY)/(Kx2+Ky2+Kz2)
=(K.k)/|K|=(KZ)/(Kx2+Ky2+Kz2)
Direction cosine is the cosine of the angle between lines and the coordinate axis.
Example 1: Find the direction cosines of vector K=3i+j–k.
Solution: Given vector is K=3i+j–k.
|K|=32+12+(-1)2=9+1+1=11
Thus, the direction cosines of the given vector are:
cos()=3/11
cos()=1/11
cos()=-1/11
Direction Ratios
Direction ratios of a given vector gives the information about x-, y-, z-components of a given vector.
Direction ratios are helpful in finding the relation between two different lines. We can also find the angle between two vectors if we know the direction ratios.
If the direction ratios of line 1 are (d1,e1,f1) and of line 2 are (d2,e2,f2), then the angle between them can be calculated with the help of the formula given below:
cos()=(d1d2+e1e2+f1f2)/(d12+e12+f12)*(d22+e22+f22)
By knowing cos(), we can calculate .
Direction Ratios Of A Line Passing Through Two Points
If a straight line passes through two points (d1,e1,f1) and (d2,e2,f2), then the direction ratios of such line are (d2–d1, e2–e1, f2–f1).
Direction cosines of this line will be:
cos()=(d2–d1)/(d2–d1)2+(e2–e1)2+(f2–f1)2
cos()=(e2–e1)/(d2–d1)2+(e2–e1)2+(f2–f1)2
cos()=(f2–f1)/(d2–d1)2+(e2–e1)2+(f2–f1)2
Example 2: Find the direction ratios of vector K=i+4j–k. Also, find the angle between vector A=3i+4j-2k and B=i-3j+2k.
Solution:
Direction ratios of K=i+4j–k are (1, 4, -1)
Let the angle between A=3i+4j-2k and B=i-3j+2k be .
Then cos() is given as:
cos()=A.B/|A||B|
∴ cos()=(3i+4j-2k).(i-3j+2k)/(32+42+(-2)2)*(12+(-3)2+(2)2)
∴ cos()=(3-12-4)/(9+16+4)(1+9+4)
∴ cos()=(-13)/2914=-0.645
∴ =130.16°
The relation between direction ratios and direction cosines is:
l/a=m/b=n/c ………(3)
Conclusion
We studied the direction ratios and direction cosines of a line in a 3D Cartesian coordinate system. Direction cosines are the angles made by the line with respect to different axes. Direction ratios are the components of a given vector line.
