If v v v and u u u are two vectors and r r r is their resultant and r r r is perpendicular to v v v and 1 / 2 1/2 1/2 of u u u then what is angle between v v v and u u u
Solution:
The beginning of the vector r ⃗ \vec{r} r - the beginning of the vector v ⃗ \vec{v} v and the end vector r ⃗ \vec{r} r - end of the vector u ⃗ \vec{u} u
u ⃗ + v ⃗ = r ⃗ \vec {u} + \vec {v} = \vec {r} u + v = r ∣ r ⃗ ∣ = 1 2 ∣ u ⃗ ∣ | \vec {r} | = \frac {1}{2} | \vec {u} | ∣ r ∣ = 2 1 ∣ u ∣
We have a right triangle ABC, we can simply find the sine of the angle alpha ( α − \alpha - α −
the angle between the vectors u ⃗ \vec{u} u and v ⃗ \vec{v} v )
sin α = A C A B = ∣ r ⃗ ∣ ∣ u ⃗ ∣ = 1 2 ∣ u ⃗ ∣ ∣ u ⃗ ∣ = 1 2 \sin \alpha = \frac {A C}{A B} = \frac {| \vec {r} |}{| \vec {u} |} = \frac {\frac {1}{2} | \vec {u} |}{| \vec {u} |} = \frac {1}{2} sin α = A B A C = ∣ u ∣ ∣ r ∣ = ∣ u ∣ 2 1 ∣ u ∣ = 2 1 α = 3 0 0 ( π 6 ) \alpha = 3 0 ^ {0} \left(\frac {\pi}{6}\right) α = 3 0 0 ( 6 π )
Answer: angle between u ⃗ \vec{u} u and v ⃗ \vec{v} v is α = 30 ∘ ( π 6 ) \alpha = 30{}^{\circ}\left(\frac{\pi}{6}\right) α = 30 ∘ ( 6 π )