How do I find the magnitude of the vector v=21(cos18°i+sin18°j)??

2 Answers
Feb 22, 2018

Given, vec v =21(cos 18 i + sin 18 j)

Angle between 21 cos 18 i and 21 sin 18 j is 90

So, |vec v| = sqrt((21 cos 18)^2 + (21 sin 18)^2) = sqrt(21^2(cos ^2 18 + sin ^2 18))=sqrt(21^2×1)=21

Feb 22, 2018

|| bbv ||=21

Explanation:

If bbv =v_1 bbi +v_2 bbj , then the magnitude of bbv is || bbv ||=sqrt(v_1^2+v_2^2)

bbv =21cos18 bbi +21sin18 bbj .

So

|| bbv ||=sqrt((21cos18)^2+(21sin18)^2)=sqrt(441cos^2 18+441sin^2 18)=sqrt(441(cos^2 18+sin^18))=sqrt441=21