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

Question

1116 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-22T15:21:12

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#

Answer 2 · 2018-02-22T15:21:19

#|| bbv ||=21#

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#