Find the magnitude and direction of vectors represented by the following pair of components
(a) `\ vec {Ax}` = - 2.3 cm, `\ vec {Ay}` = + 4.1 cm (b) `\ vec {Ax}` = +3.9 cm, `\ vec {Ay}` = -1.8 cm 
(Ans. (a) A= 4.7 and θ = 119.3°, (b) A = 4.3 and Î¸ = 335.2°)

Given:

(a) `\ vec {Ax}` = - 2.3 cm, `\ vec {Ay}` = + 4.1 cm 
(b) `\ vec {Ax}` = +3.9 cm, `\ vec {Ay}` = -1.8 cm  

To Find:

(a) 
Magnitude of vector `\ vec {A}` = A = ?
Direction (angle Î¸) = ?

(b) 
Magnitude of vector `\ vec {A}` = A = ?
Direction (angle Î¸) = ?

Solution:

(a) `\ vec {Ax}` = - 2.3 cm, `\ vec {Ay}` = + 4.1 cm 


The formula to find the magnitude of vector A from its x and y component is 

A  = `sqrt {Ax^2 + Ay^2}`

by putting values

A = `sqrt {(-2.3 cm)^2 + (4.1 cm)^2}`

A = `sqrt {5.29 cm² + 116.81 cm²}`

A = `sqrt {22.1 cm²}`

A = 4.7 cm ------------Ans. 1

Now for direction, we have
Ax = -ve and Ay = +ve So, `\ vec {A}` lies  in 2nd quadrant, Thus the formula for  Î¸ will be

θ = 180° - tan⁻¹(` \frac {Ay}{Ax}`)

θ = 180° - tan⁻¹(`\ frac {4.1 cm}{2.3 cm}`)

θ = 180° - tan⁻¹(-1.783)

θ = 180° - 60.7° 

θ = 119.3° -------------Ans. 2



(b) `\ vec {Ax}` = +3.9 cm, `\ vec {Ay}` = -1.8 cm


The formula to find the magnitude of vector A from its x and y component is 

A  = `sqrt {Ax^2 + Ay^2}`

by putting values

A = `sqrt {(+3.9 cm)^2 + (-1.8 cm)^2}`

A = `sqrt {15.21 cm² + 3.24 cm²}`

A = `sqrt {18.45 cm²}`

A = 4.3 cm ------------Ans. 1


Now for direction, we have
Ax = +ve and Ay = -ve So, `\ vec {A}` lies  in 4nd quadrant, Thus the formula for  Î¸ will be

θ = 360° - tan⁻¹(` \frac {Ay}{Ax}`)

θ = 360° - tan⁻¹(`\ frac {1.8 cm}{3.9 cm}`)

θ = 360° - tan⁻¹(0.462)

θ = 360° - 24.8° 

θ = 335.2° -------------Ans. 2

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