Question

find the direction of the vector sum \\(\vec{a} + \vec{b}\\).
vector a is 63.5 m long in a \\(90^{\circ}\\) direction.
vector b is 101 m long in a \\(57.0^{\circ}\\) direction.

Explanation:

Step1: Resolve Vector A into components

Vector A has magnitude \( A = 63.5 \, \text{m} \) and direction \( \theta_A = 90^\circ \).
The x - component of \( \vec{A} \), \( A_x = A\cos\theta_A = 63.5\cos(90^\circ)=0 \)
The y - component of \( \vec{A} \), \( A_y = A\sin\theta_A = 63.5\sin(90^\circ)=63.5 \, \text{m} \)

Step2: Resolve Vector B into components

Vector B has magnitude \( B = 101 \, \text{m} \) and direction \( \theta_B = 57.0^\circ \)
The x - component of \( \vec{B} \), \( B_x = B\cos\theta_B = 101\cos(57.0^\circ)\approx101\times0.5446\approx55.0 \, \text{m} \)
The y - component of \( \vec{B} \), \( B_y = B\sin\theta_B = 101\sin(57.0^\circ)\approx101\times0.8387\approx84.7 \, \text{m} \)

Step3: Find the components of the resultant vector \( \vec{R}=\vec{A}+\vec{B} \)

The x - component of the resultant vector, \( R_x = A_x + B_x=0 + 55.0 = 55.0 \, \text{m} \)
The y - component of the resultant vector, \( R_y = A_y + B_y=63.5+84.7 = 148.2 \, \text{m} \)

Step4: Find the direction of the resultant vector

The direction \( \theta \) of the resultant vector is given by \( \tan\theta=\frac{R_y}{R_x} \)
\( \tan\theta=\frac{148.2}{55.0}\approx2.6945 \)
\( \theta=\arctan(2.6945)\approx69.7^\circ \)

Answer:

The direction of the vector sum \( \vec{A}+\vec{B} \) is approximately \( \boldsymbol{69.7^\circ} \) (measured from the positive x - axis).