Question

Vector b with magnitude 5 units is added to Vector a =
5i+ 5√3j. If b can assume any direction then
maximum and minimum inclination of a + b from
from x-axis is​

Answer 1

Answer:

The maximum and the minimum magnitude of the resultant two vectors are 17 and 7 units respectively. Then the magnitude of the resultant vector when they act perpendicular to each other is:

Answer 2

Given :  Vector b with magnitude 5 units is added to Vector a = 5i+ 5√3j.  b can assume any direction

To Find :  maximum and minimum inclination of a + b from from x-axis

Solution:

Vector b with magnitude 5 units

Hence vector b =   5Cosα i  +  5Sinαj

Vector a = 5i + 5√3j

a + b  =  5i + 5√3j + 5Cosα i  +  5Sinαj

=> a  + b = (5  + 5Cosα)i+  (5√3 + 5Sinα)j

inclination = tan⁻¹(  (5√3 + 5Sinα) /  (5  + 5Cosα))

5√3 + 5Sinα  is always + ve

5  + 5Cosα can be minimum 0    other wise  + Ve

Hence inclination will be in 1st Quadrant or max at  +ve y axis

Hence maximum inclination =  90°  when cosα = - 1

then a + b  =  5√3j   and    b = -5i

minimum inclination  when sinα  is minimum  = - 1

Hence tan⁻¹ ( (5√3  - 5) / 5 )   = tan⁻¹ (√3  - 1)  = 36.2°  

when b =  - 5j    and  a + b  =  5i + 5(√3 - 1) j

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