Question

find the magnitude of the vector sum \\(vec{a} + vec{b}\\).
vector a is 1.23 m long in a 42.9° direction.
vector b is 5.58 m long in a 32.5° direction.

Explanation:

Step1: Resolve Vector A into components

The x - component of vector \(\vec{A}\) is given by \(A_x = A\cos\theta_A\), where \(A = 1.23\space m\) and \(\theta_A=42.9^{\circ}\).
\(A_x=1.23\cos(42.9^{\circ})\approx1.23\times0.734 = 0.903\space m\)
The y - component of vector \(\vec{A}\) is given by \(A_y = A\sin\theta_A\)
\(A_y = 1.23\sin(42.9^{\circ})\approx1.23\times0.679=0.835\space m\)

Step2: Resolve Vector B into components

The x - component of vector \(\vec{B}\) is given by \(B_x = B\cos\theta_B\), where \(B = 5.58\space m\) and \(\theta_B = 32.5^{\circ}\)
\(B_x=5.58\cos(32.5^{\circ})\approx5.58\times0.843 = 4.704\space m\)
The y - component of vector \(\vec{B}\) is given by \(B_y=B\sin\theta_B\)
\(B_y = 5.58\sin(32.5^{\circ})\approx5.58\times0.537 = 2.996\space m\)

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

The x - component of the resultant vector \(R_x=A_x + B_x\)
\(R_x=0.903 + 4.704=5.607\space m\)
The y - component of the resultant vector \(R_y=A_y + B_y\)
\(R_y=0.835+2.996 = 3.831\space m\)

Step4: Calculate the magnitude of the resultant vector \(\vec{R}\)

The magnitude of a vector with components \(R_x\) and \(R_y\) is given by \(R=\sqrt{R_x^{2}+R_y^{2}}\)
\(R=\sqrt{(5.607)^{2}+(3.831)^{2}}=\sqrt{31.44 + 14.68}=\sqrt{46.12}\approx6.79\space m\)

Answer:

The magnitude of the vector sum \(\vec{A}+\vec{B}\) is approximately \(\boldsymbol{6.79\space m}\)