Question

Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length first then the degrees.

Answer 1

Answer:

Magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

Explanation:

Magnitude of first vector = $|A| = 13\ \text{units}$

Angle = $\theta_1=27^{\circ}$

Magnitude of second vector = $|B| = 11\ \text{units}$

Angle = $\theta_2=45^{\circ}$

x component of first vector

$A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}$

y component of first vector

$A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}$

x component of second vector

$B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}$

y component of first vector

$B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}$

Adding the magnitudes

$C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}$

$C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}$

Magnitude of the sum of the vectors would be

$|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}$

The direction would be

$\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}$

The magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$