We are given two vectors in polar form:
Vector A : magnitude ∣ A ⃗ ∣ = 5 |\vec A| = 5 ∣ A ∣ = 5 at angle 30 ∘ 30^\circ 3 0 ∘
Vector B : magnitude ∣ B ⃗ ∣ = 8 |\vec B| = 8 ∣ B ∣ = 8 at angle 120 ∘ 120^\circ 12 0 ∘
We want their resultant vector :
R ⃗ = A ⃗ + B ⃗ . \vec R = \vec A + \vec B. R = A + B .
1. Convert each vector to components
For a vector with magnitude r r r and angle θ \theta θ (in degrees, measured from +x axis):
x = r cos θ , y = r sin θ . x = r\cos\theta, \quad y = r\sin\theta. x = r cos θ , y = r sin θ .
Vector A
A x = 5 cos 30 ∘ , A y = 5 sin 30 ∘ . A_x = 5 \cos 30^\circ, \quad A_y = 5 \sin 30^\circ. A x = 5 cos 3 0 ∘ , A y = 5 sin 3 0 ∘ .
Using exact values:
cos 30 ∘ = 3 / 2 ≈ 0.866 \cos 30^\circ = \sqrt{3}/2 \approx 0.866 cos 3 0 ∘ = 3 /2 ≈ 0.866
sin 30 ∘ = 1 / 2 = 0.5 \sin 30^\circ = 1/2 = 0.5 sin 3 0 ∘ = 1/2 = 0.5
So:
A x = 5 ⋅ 0.866 ≈ 4.33 , A y = 5 ⋅ 0.5 = 2.5. A_x = 5 \cdot 0.866 \approx 4.33, \quad A_y = 5 \cdot 0.5 = 2.5. A x = 5 ⋅ 0.866 ≈ 4.33 , A y = 5 ⋅ 0.5 = 2.5.
Vector B
B x = 8 cos 120 ∘ , B y = 8 sin 120 ∘ . B_x = 8 \cos 120^\circ, \quad B_y = 8 \sin 120^\circ. B x = 8 cos 12 0 ∘ , B y = 8 sin 12 0 ∘ .
Using:
cos 120 ∘ = − 1 / 2 = − 0.5 \cos 120^\circ = -1/2 = -0.5 cos 12 0 ∘ = − 1/2 = − 0.5
sin 120 ∘ = 3 / 2 ≈ 0.866 \sin 120^\circ = \sqrt{3}/2 \approx 0.866 sin 12 0 ∘ = 3 /2 ≈ 0.866
So:
B x = 8 ⋅ ( − 0.5 ) = − 4 , B y = 8 ⋅ 0.866 ≈ 6.93. B_x = 8 \cdot (-0.5) = -4, \quad B_y = 8 \cdot 0.866 \approx 6.93. B x = 8 ⋅ ( − 0.5 ) = − 4 , B y = 8 ⋅ 0.866 ≈ 6.93.
2. Add the components
The resultant components:
R x = A x + B x , R y = A y + B y . R_x = A_x + B_x, \quad R_y = A_y + B_y. R x = A x + B x , R y = A y + B y .
Plugging in values:
R x = 4.33 + ( − 4 ) ≈ 0.33 , R y = 2.5 + 6.93 ≈ 9.43. R_x = 4.33 + (-4) \approx 0.33,\\
R_y = 2.5 + 6.93 \approx 9.43. R x = 4.33 + ( − 4 ) ≈ 0.33 , R y = 2.5 + 6.93 ≈ 9.43.
So the resultant vector in component form is approximately:
R ⃗ ≈ ( 0.33 , 9.43 ) . \vec R \approx (0.33,\; 9.43). R ≈ ( 0.33 , 9.43 ) .
3. Magnitude and direction of the resultant
Magnitude:
∣ R ⃗ ∣ = R x 2 + R y 2 ≈ 0.33 2 + 9.43 2 ≈ 0.11 + 88.94 ≈ 89.05 ≈ 9.43. |\vec R| = \sqrt{R_x^2 + R_y^2}
\approx \sqrt{0.33^2 + 9.43^2}
\approx \sqrt{0.11 + 88.94}
\approx \sqrt{89.05} \approx 9.43. ∣ R ∣ = R x 2 + R y 2 ≈ 0.3 3 2 + 9.4 3 2 ≈ 0.11 + 88.94 ≈ 89.05 ≈ 9.43.
Direction (angle from +x axis):
θ R = tan − 1 ( R y R x ) ≈ tan − 1 ( 9.43 0.33 ) ≈ 88 ∘ . \theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right)
\approx \tan^{-1}\left(\frac{9.43}{0.33}\right) \approx 88^\circ. θ R = tan − 1 ( R x R y ) ≈ tan − 1 ( 0.33 9.43 ) ≈ 8 8 ∘ .
So the resultant is almost vertical, with a small positive x-component:
∣ R ⃗ ∣ ≈ 9.4 units at 88 ∘ . \boxed{|\vec R| \approx 9.4 \text{ units at } 88^\circ.} ∣ R ∣ ≈ 9.4 units at 8 8 ∘ .
4. What the visualization shows
The interactive canvas illustrates:
Vector A (cyan): drawn from the origin at angle θ A \theta_A θ A with length proportional to its magnitude.
Vector B (pink): also from the origin at angle θ B \theta_B θ B .
Resultant vector R (yellow): the sum A ⃗ + B ⃗ \vec A + \vec B A + B , drawn from the origin to the tip of the parallelogram/triangle formed by A and B.
A faint dynamic grid to provide coordinate context.
You can adjust:
The magnitude and angle of each vector and see how:
Their components change.
The resultant direction and length updates in real time.
Mathematically, the visualization implements:
A x = ∣ A ∣ cos ( θ A ) , A y = ∣ A ∣ sin ( θ A ) , B x = ∣ B ∣ cos ( θ B ) , B y = ∣ B ∣ sin ( θ B ) , R x = A x + B x , R y = A y + B y . \begin{aligned}
A_x &= |A|\cos(\theta_A), & A_y &= |A|\sin(\theta_A),\\
B_x &= |B|\cos(\theta_B), & B_y &= |B|\sin(\theta_B),\\
R_x &= A_x + B_x, & R_y &= A_y + B_y.
\end{aligned} A x B x R x = ∣ A ∣ cos ( θ A ) , = ∣ B ∣ cos ( θ B ) , = A x + B x , A y B y R y = ∣ A ∣ sin ( θ A ) , = ∣ B ∣ sin ( θ B ) , = A y + B y .
Then it draws the three vectors on a Cartesian plane (transformed to canvas coordinates) with a dark-slate background and neon colors, and slight animation to keep the scene visually alive.