HOW TO FIND THE DIRECTION COSINES OF A VECTOR WITH GIVEN RATIOS

Question 1 :

Find the direction cosines of a vector whose direction ratios are

(i) 1 , 2 , 3 (ii) 3 , - 1 , 3 (iii) 0 , 0 , 7

Solution :

(i)  x  =  1, y  =  2  and z  =  3

|r vector|  =  r  =  √(x² + y² + z²)   =  √(1² + 2² + 3²) 

  =  √(1+4+9) 

=  √14

Hence direction cosines are ( 1/√14, 2/√14, 3/√14)

(ii)  x  =  3, y  =  -1  and z  =  3

|r vector|  =  r  =  √(x² + y² + z²)   =  √(3² + (-1)² + 3²) 

  =  √(9+1+9) 

=  √19

Hence direction cosines are ( 3/√19, -1/√19, 3/√19)

(iii)  x  =  0, y  =  0  and z  =  7

|r vector|  =  r  =  √(x² + y² + z²)   =  √(0² + 0² + 7²) 

  =  √49

=  7

Hence direction cosines are ( 0, 0, 1)

Question 2 :

Find the direction cosines and direction ratios of the following vectors.

(i)  3i vector - 4j vector + 8k vector

Solution :

x  =  3, y  =  -4 and z  =  8

|r vector|  =  r  =  √(x² + y² + z²)   =  √(3² + (-4)² + 8²) 

  =  √(9+16+64) 

=  √89

Direction cosines :

(x/r, y/r, z/r)

x/r  =  3/√89

y/r  =  -4/√89

z/r  =  8/√89

Hence direction cosines are ( 3/√89, -4/√89, 8/√89)

Direction ratios :

Direction ratios are (3, -4, 8).

(ii)  3i vector + j vector + k vector

Solution :

x  =  3, y  =  1 and z  =  1

|r vector|  =  r  =  √(x² + y² + z²)   =  √3² + 1² + 1²) 

  =  √(9+1+1) 

=  √11

Direction cosines :

(x/r, y/r, z/r)

x/r  =  3/√11

y/r  =  1/√11

z/r  =  1/√11

Hence direction cosines are ( 3/√11, 1/√11, 1/√11)

Direction ratios :

Direction ratios are (3, 1, 1).

(iii)  j vector

Solution :

x  =  0,  y  =  1 and z  =  0

|r vector|  =  r  =  √(x² + y² + z²)   =  √0² + 1² + 0²) 

  =  √1 

=  1

Direction cosines :

(x/r, y/r, z/r)

x/r  =  0/1  =  0

y/r  =  1/1  =  1

z/r  =  0/1  =  0

Hence direction cosines are ( 0, 1, 0)

Direction ratios :

Direction ratios are (0, 1, 0).

(iv)  5i vector - 3j vector - 48k vector

Solution :

x  =  5,  y  =  -3 and z  =  -48

|r vector|  =  r  =  √(x² + y² + z²)   =  √5² + (-3)² + (-48)² 

  =  √(25 + 9 + 2304) 

=   √2338

Direction cosines :

(x/r, y/r, z/r)

x/r  =  5/√2338

y/r  =  -3/√2338

z/r  =  -48/√2338

Hence direction cosines are

(5/√2338, -3/√2338, -48/√2338)

Direction ratios :

Direction ratios are (5, -3, -48).

(v)  3i vector + 4j vector - 3k vector

Solution :

x  =  3,  y  =  4 and z  =  -3

|r vector|  =  r  =  √(x² + y² + z²)   =  √3² + 4² + (-3)² 

  =  √(9 + 16 + 9) 

=   √34

Direction cosines :

(x/r, y/r, z/r)

x/r  =  3/√34

y/r  =  4/√34

z/r  =  -3/√34

Hence direction cosines are

(3/√34, 4/√34, -3/√34)

Direction ratios :

Direction ratios are (3, 4, -3).

(vi)  i vector - k vector

Solution :

x  =  1,  y  =  0 and z  =  -1

|r vector|  =  r  =  √(x² + y² + z²)   =  √1² + 0² + (-1)² 

  =  √(1 + 0 + 1) 

=   √2

Direction cosines :

(x/r, y/r, z/r)

x/r  =  1/√2

y/r  =  0/√2  =  0

z/r  =  -1/√2

Hence direction cosines are

(1/√2, 0, -1/√2)

Direction ratios :

Direction ratios are (1, 0, -1).

Question 3 :

Find the acute angle which the line joining the points (1 -3, 2) and (3, -5, 1) makes with the coordinate axis.

Solution :

Let 

b - a = (3 - 1)i vector + (-5 + 3)j vector + (1 - 2)k vector

= 2i vector -2j vector - 1k vector

|b - a vector| = r = √(x² + y² + z²) 

r = √(2² + (-2)² + (-1)²) 

= √(4 + 4 + 1) 

= √9

= 3

Direction cosines are

= 2/3, -2/3 and -1/3

Direction angles are,

cos^-1(2/3)

cos^-1(-2/3)

cos^-1(-1/3)

48.2°, 48.2° and 70.5°

Question 4 :

Find the angles which the vector u = 3i - 6j + 2k makes with the coordinate axes.

Solution :

u = √(x² + y² + z²) 

Here x = 3, y = -6 and z = 2

u = √(3² + (-6)² + 2²) 

= √(9 + 36 + 4)

= √49

= 7

• Direction cosines are, 3/7, -6/7 and 2/7
• Direction angles are

cos^-1(3/7), cos^-1(-6/7) and cos^-1(2/7)

Question 5 :

What are direction cosines of the line equally inclined in the axes ?

Solution :

If the line makes acute angles α, β, Γ with the axes, we have α = β = Γ

cos α = l, cos β = m and cos Γ = n

r = √(x² + y² + z²) 

r = √(1² + 1² + 1²)

r = √3

Direction cosines are, 1/√3, 1/√3, 1/√3

Question 6 :

The direction ratios of the line 3x + 1 = 6y − 2 = 1 – z are:

(a) 3, 6, 1 (b) 3, 6, –1

(c) 2, 1, 6 (d) 2, 1, –6

Solution :

3x + 1 = 6y − 2 = 1 – z

[x + (1/3)] / (1/3) = [(y - 1/3)] / (-1/6) = (z - 1) / -1

Direction ratios are 1/3, -1/6 and -1

2, 1, –6

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