Consider the two vectors A with arrow = 3 î − ĵ and B with arrow = − î − 4 ĵ.
(a) Calculate A with arrow + B with arrow
î +
ĵ
(b) Calculate A with arrow − B with arrow
î +
ĵ
(c) Calculate
A with arrow + B with arrow
(d) Calculate
A with arrow − B with arrow
(e) Calculate the directions of A with arrow + B with arrow and A with arrow − B with arrow.
A with arrow + B with arrow
° (counterclockwise from the +x axis)
A with arrow − B with arrow
° (counterclockwise from the +x axis)
a)
A ⃗ + B ⃗ = ( 3 ı ^ − ȷ ^ ) + ( − ı ^ − 4 ȷ ^ ) = 2 ı ^ − 5 ȷ ^ \vec A+\vec B=(3 î − ĵ )+(− î − 4 ĵ)=2 î − 5ĵ A + B = ( 3 ı ^ − ȷ ^ ) + ( − ı ^ − 4 ȷ ^ ) = 2 ı ^ − 5 ȷ ^
b)
A ⃗ − B ⃗ = ( 3 ı ^ − ȷ ^ ) − ( − ı ^ − 4 ȷ ^ ) = 4 ı ^ + 3 ȷ ^ \vec A-\vec B=(3 î − ĵ )-(− î − 4 ĵ)=4 î +3ĵ A − B = ( 3 ı ^ − ȷ ^ ) − ( − ı ^ − 4 ȷ ^ ) = 4 ı ^ + 3 ȷ ^
c)
∣ A ⃗ + B ⃗ ∣ = 2 2 + ( − 5 ) 2 = 29 ≈ 5.4 |\vec A+\vec B|=\sqrt{2^2+(-5)^2}=\sqrt{29}\approx 5.4 ∣ A + B ∣ = 2 2 + ( − 5 ) 2 = 29 ≈ 5.4
d)
∣ A ⃗ − B ⃗ ∣ = 4 2 + ( 3 ) 2 = 5 |\vec A-\vec B|=\sqrt{4^2+(3)^2}=5 ∣ A − B ∣ = 4 2 + ( 3 ) 2 = 5
e)
θ + = 360 ° − arctan 5 2 = 291.8 ° \theta_+=360\degree-\arctan\frac{5}{2}=291.8\degree θ + = 360° − arctan 2 5 = 291.8°
θ − = arctan 3 4 = 36.9 ° \theta_-=\arctan\frac{3}{4}=36.9\degree θ − = arctan 4 3 = 36.9°