Question

hw 2 begin date: 8/29/2025 12:01:00 am due date: 9/5/2025 11:59:00 pm end date: 12/5/2025 4:59:00 pm
problem 6: (9% of assignment value)
vector a has a given magnitude of |a| = 2.5 and points due east. vector b, of unknown magnitude, points directly north. the sum a + b makes an angle of θ = 42 degrees north of east. refer to the figure.
part (a)
enter an expression for the magnitude of vector b in terms of the magnitude of a, |a|, and the angle θ, using any required trigonometric functions.
part (b)
what is the value of the magnitude of the difference of vectors a and b, |a - b|?
part (c)
what is the value of the magnitude of the sum of vectors a and b, |a + b|?

Explanation:

Step1: Analyze vector components

Since vectors $\vec{A}$ and $\vec{B}$ are perpendicular (east - north), we can use trigonometry for $\vec{A}+\vec{B}$. In a right - triangle formed by $\vec{A}$, $\vec{B}$ and $\vec{A}+\vec{B}$, $\tan\theta=\frac{|\vec{B}|}{|\vec{A}|}$.

Step2: Solve for $|\vec{B}|$

Rearranging the equation $\tan\theta=\frac{|\vec{B}|}{|\vec{A}|}$ gives $|\vec{B}| = |\vec{A}|\tan\theta$.

Step3: For part (b)

The magnitude of $\vec{A}-\vec{B}$: Using the Pythagorean theorem, since $\vec{A}$ and $\vec{B}$ are perpendicular, $|\vec{A}-\vec{B}|=\sqrt{|\vec{A}|^{2}+|\vec{B}|^{2}}$. Substitute $|\vec{B}| = |\vec{A}|\tan\theta$ into it. Given $|\vec{A}| = 2.5$ and $\theta = 42^{\circ}$, $|\vec{B}|=2.5\tan42^{\circ}\approx2.5\times0.9004 = 2.251$. Then $|\vec{A}-\vec{B}|=\sqrt{2.5^{2}+2.251^{2}}=\sqrt{6.25 + 5.067}=\sqrt{11.317}\approx3.36$.

Step4: For part (c)

The magnitude of $\vec{A}+\vec{B}$: Using the Pythagorean theorem, $|\vec{A}+\vec{B}|=\sqrt{|\vec{A}|^{2}+|\vec{B}|^{2}}$. Since $|\vec{B}| = |\vec{A}|\tan\theta$, $|\vec{A}+\vec{B}|=\sqrt{|\vec{A}|^{2}+|\vec{A}|^{2}\tan^{2}\theta}=|\vec{A}|\sqrt{1 + \tan^{2}\theta}$. Substitute $|\vec{A}| = 2.5$ and $\theta = 42^{\circ}$, $|\vec{A}+\vec{B}|=2.5\sqrt{1+\tan^{2}42^{\circ}}=2.5\sqrt{1 + 0.8107}=2.5\sqrt{1.8107}\approx2.5\times1.3456\approx3.36$.

Answer:

Part (a): $|\vec{B}| = |\vec{A}|\tan\theta$
Part (b): $\approx3.36$
Part (c): $\approx3.36$