Let P be a point in the space with coordinates (x, y, z) and of distance r from the origin.
Let R, S and T be the foots of the perpendiculars drawn from P to the x, y and z axes respectively. Then
∠PRO = 90°
∠PSO = 90°
∠PTO = 90°
OR = x, OS = y, OT = z and OP = r.
OR = x, OS = y, OT = z and OP = r.
(It may be difficult to visualize that ∠PRO = ∠PSO = ∠PTO = 90° in the figure; as they are foot of the perpendiculars to the axes from P; in a three dimensional model we can easily visualize the fact.)
Let α, β, γ be the angles made by the vector OP vector with the positive x, y and z axes respectively. That is,
∠PRO = α, ∠POS = β, ∠POT = γ
In triangle OPR, ∠PRO = 90°, ∠POR = α, OR = x, and OP = r. Hence cos α = OR/OP = x/r |
In a similar way we can find that cos β = y/r and cos γ = z/r .
Direction Angles :
Here the angles α, β, γ are called direction angles of the vector OP vector = r vector.
Direction Cosines :
Cos α, cos β, cos γ are called direction cosines of the vector OP = xi vector + yj vector + zk vector.
Thus (x/r, y/r, z/r), where r = √(x^{2} + y^{2} + z^{2}) are the direction cosines of the vector r vector = xi vector + yj vector + zk vector.
Question 1 :
Verify whether the following ratios are direction cosines of some vector or not.
(i) 1/5 , 3/5 , 4/5
Solution :
In order to verify the given ratios are direction cosines, it has to satisfy the condition given below.
The sum of the squares of the direction cosines of r vector is 1.
That is ,
x^{2} + y^{2} + z^{2 } = 1
r vector = xi vector + yj vector + zk vector
r vector = (1/5)i vector + (3/5)j vector + (4/5)k vector
x^{2} + y^{2} + z^{2 } = (1/5)^{2} + (3/5)^{2} + (4/5)^{2}
= (1/25) + (9/25) + (16/25)
= (1 + 9 + 16)/25
= 26/25
x^{2} + y^{2} + z^{2 } ≠ 1
Hence the direction ratios are not direction cosines of some vector.
(ii) 1/√2, 1/2 , 1/2
Solution :
r vector = (1/√2)i vector + (1/2)j vector + (1/2)k vector
x^{2} + y^{2} + z^{2 } = (1/√2)^{2} + (1/2)^{2} + (1/2)^{2}
= (1/2) + (1/4) + (1/4)
= (2 + 1 + 1)/4
= 1
Hence the direction ratios are the direction cosines of some vector.
(iii) 4/3, 0, 3/4
Solution :
r vector = (4/3)i vector + (0)j vector + (3/4)k vector
x^{2} + y^{2} + z^{2 } = (4/3)^{2} + (0)^{2} + (3/4)^{2}
= (16/9) + 0 + (9/16)
= (256 + 0 + 81)/144
x^{2} + y^{2} + z^{2 } ≠ 1
Hence the direction ratios are not direction cosines of some vector.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
May 22, 24 06:32 AM
May 17, 24 08:12 AM
May 14, 24 08:53 AM