Find modulus,Direction cosines,Unit vectors of (1)4a-2b+c,(2)a-7b+5c,(3)4a-3b+c.
a+2b_c
To find the modulus, direction cosines, and unit vectors of the given vectors, we will use certain formulas and definitions. Let's start by finding these quantities for each given vector:
1) Vector: 4a - 2b + c
(i) Modulus of the vector:
The modulus (or magnitude) of a vector can be found using the formula:
|V| = √(V₁² + V₂² + V₃² + ... + Vn²)
For this vector, the modulus can be calculated as follows:
|4a - 2b + c| = √((4a)² + (-2b)² + c²) = √(16a² + 4b² + c²)
(ii) Direction cosines of the vector:
The direction cosines of a vector are the cosines of the angles that the vector makes with the coordinate axes. The direction cosines are given by:
cos α = V₁ / |V|
cos β = V₂ / |V|
cos γ = V₃ / |V|
For this vector, the direction cosines are:
cos α = 4a / √(16a² + 4b² + c²)
cos β = -2b / √(16a² + 4b² + c²)
cos γ = c / √(16a² + 4b² + c²)
(iii) Unit vector in the direction of the vector:
The unit vector in the direction of a given vector V can be found by dividing the vector by its modulus:
U = V / |V|
For this vector, the unit vector is:
U = (4a - 2b + c) / √(16a² + 4b² + c²)
2) Vector: a - 7b + 5c
(i) Modulus of the vector:
|a - 7b + 5c| = √(a² + (-7b)² + 5c²)
(ii) Direction cosines of the vector:
cos α = a / √(a² + (-7b)² + 5c²)
cos β = -7b / √(a² + (-7b)² + 5c²)
cos γ = 5c / √(a² + (-7b)² + 5c²)
(iii) Unit vector in the direction of the vector:
U = (a - 7b + 5c) / √(a² + (-7b)² + 5c²)
3) Vector: 4a - 3b + c
(i) Modulus of the vector:
|4a - 3b + c| = √((4a)² + (-3b)² + c²)
(ii) Direction cosines of the vector:
cos α = 4a / √((4a)² + (-3b)² + c²)
cos β = -3b / √((4a)² + (-3b)² + c²)
cos γ = c / √((4a)² + (-3b)² + c²)
(iii) Unit vector in the direction of the vector:
U = (4a - 3b + c) / √((4a)² + (-3b)² + c²)
By using these formulas and definitions, you can easily calculate the modulus, direction cosines, and unit vectors of any given vector.