Test: Resolution of Vectors (May 18) - JEE MCQ

Let=  ∴ A_x = 1, A_y = 1, A_z = 1
and
cos α, cos β andcos y are the direction cosines of .

Let OP and OQ represent two vectorsmaking an angle (α + β). Using the parallelogram method of vector addition,
Resultant vector, 
SN is normal to OP and PM is normal to OS.
From the geometry of the figure,
OS² = ON² + SN² = (OP + PN)² + SN² = (A + Bcos(α + β))² + (Bsin(α + β))²

R² = A² + B² + 2ABcos(α + β)

In ΔOSN, SN = OSsinα = Rsinα and in ΔPSN,SN = PSsin(α + β) = Bsin(α + β)
Rsinα = Bsin(α + β) or R/sin(α + β)= B/sinα
Similarly,
PM = Asinα = Bsinβ
A/sinβ = B/sinα
Combining (i) and (ii), we get
R/sin(α + β) = A/sinβ = B/sinα
From eqn, (iii), Rsinβ = Asin(α + β)

Unit vector is vector with magnitude unity but having specific direction.

Value of unit vector is given by:

Where,

 = unit vector

A = vector a

∣A∣ = magnitude of vector a

A vector, its magnitude and the angle between two vectors do not depend on the choice of the orientation of the coordinate axes. So angle between are independent of the orientation of the coordinate axes. But the quantity A_x + B_y depends upon the magnitude of the components along x and y axes, so it will change with change in coordinate axes.

Given , (say) components of along the direction of

Squaring both sides we get 

cos² α + cos² β + cos² Y = 1
(1 - sin² α) + (1 - sin² β) + (1 - sin² Y) = 1
or sin² α + sin² β + sin² Y = 3 - 1 = 2