Chapter 1, Problem 1.1P

If A represents a vector two units in length directed due west, B represents a vector three units in length directed due north, and A + B = C — D and 2B - A = C + D, find the magnitudes and directions of C and D. Take north as the positive y direction.

To determine

The magnitude and direction of vectors C and D.

Answer to Problem 1.1P

The magnitude and direction of vectors

  $\begin{array}{l}\overrightarrow{C}=4.5{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_{y}C=4.5\text{North}\\ \overrightarrow{D}=2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x+1.5{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_{y}D=2.5\text{East}-\text{North East}\end{array}$

Explanation of Solution

Given Information:

Vector A is 2 units directed west, and B is 3 units directed north. Also,

  $\begin{array}{l}\text{<mover accent="true"><mo>A</mo><mo>→</mo></mover>}+\text{<mover accent="true"><mo>B</mo><mo>→</mo></mover>}=\text{<mover accent="true"><mo>C</mo><mo>→</mo></mover>}-\text{<mover accent="true"><mo>D</mo><mo>→</mo></mover>}\\ 2\text{<mover accent="true"><mo>B</mo><mo>→</mo></mover>}-\text{<mover accent="true"><mo>A</mo><mo>→</mo></mover>}=\text{<mover accent="true"><mo>C</mo><mo>→</mo></mover>}+\text{<mover accent="true"><mo>D</mo><mo>→</mo></mover>}\end{array}$

Calculation:

From the given information

  $\begin{array}{l}\text{<mover accent="true"><mo>A</mo><mo>→</mo></mover>}=-2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x\\ \text{<mover accent="true"><mo>B</mo><mo>→</mo></mover>}=3{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\\ \text{<mover accent="true"><mo>C</mo><mo>→</mo></mover>}-\text{<mover accent="true"><mo>D</mo><mo>→</mo></mover>}=\left(-2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x\right)+\left(3{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\right)\\ =-2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x+3{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\left(1\right)\\ \text{<mover accent="true"><mo>C</mo><mo>→</mo></mover>}+\text{<mover accent="true"><mo>D</mo><mo>→</mo></mover>}=2\left(3{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\right)-\left(-2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x\right)\\ =2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x+6{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\left(2\right)\end{array}$

Adding equation (1) and (2)

  $\begin{array}{l}2\text{<mover accent="true"><mo>C</mo><mo>→</mo></mover>}=\left(-2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x+2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x\right)+\left(3{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y+6{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\right)\\ =9{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\\ \Rightarrow \text{<mover accent="true"><mo>C</mo><mo>→</mo></mover>}=4.5{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\end{array}$

Subtracting equation (1) from equation (2)

  $\begin{array}{c}2\text{<mover accent="true"><mo>D</mo><mo>→</mo></mover>}=\left(2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x+2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x\right)+\left(6{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y-3{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\right)\\ =4{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x+3{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\\ \text{<mover accent="true"><mo>D</mo><mo>→</mo></mover>}=2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x+1.5{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_y\end{array}$

So, the magnitude of vector C is 4.5 and direction is towards North.

The magnitude of vector D is $\begin{array}{l}D=\sqrt{2^2+{1.5}^2}\\ =2.5\end{array}$

Vector D is having a magnitude 2.5 and is directed towards East North-East.

Conclusion:

The magnitude and direction of vectors are

  $\begin{array}{l}\overrightarrow{C}=4.5{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_{y}C=4.5\text{North}\\ \overrightarrow{D}=2{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_x+1.5{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_{y}D=2.5\text{East}-\text{North East}\end{array}$