ORCID iD icon

Back to the class
Quiz 14
1. Find $\vec{x}-3\vec{y}$ where $\vec{x}=\lt 2,1 \gt$ and $\vec{y}=\lt 1, -1 \gt$.
Solution: First notice that $$3\vec{y}=-3\lt 1,-1\gt = \lt -3, 3 \gt.$$ Now we see that $$\vec{x}-3\vec{y}=\lt 2,1\gt + \lt -3,3 \gt = \lt 2+(-3), 1+3 \gt = \lt -1, 4 \gt. $$ 2. Find the magnitude and direction of the resultant of $\vec{A}$ and $\vec{B}$ where $\vec{A}$ has magnitude $53$ and $\theta=101^{\circ}$ and $\vec{B}$ has magnitude $47$ and $\theta=76^{\circ}$.
Solution: We draw the picture for $\vec{A}$:

and we draw the picture for $\vec{B}$:

From this we see that $\vec{A}$ has $y$-component $$y=30\sin(21^{\circ}) = 10.75$$ and $x$-component $$x=30\cos(21^{\circ})=28.01.$$ Also $\vec{B}$ has $y$-component $$y=47\sin(76^{\circ})=45.6,$$ and has $x$-component $$x=47\cos(76^{\circ})=11.37.$$ Thus we may write $\vec{A}=\lt 28.01, 10.75 \gt$ and $\vec{B}=\lt 11.37, 45.6\gt$. So we see that their resultant is $$\vec{A}+\vec{B}=\lt 28.01, 10.75 \gt + \lt 11.37, 45.6 \gt = \lt 39.38, 56.35 \gt.$$ The magnitude of this resultant it $$\mathrm{magnitude}(\vec{A}+\vec{B})=\sqrt{39.38^2 + 56.35^2}=68.7466$$ and the angle is $$\theta = \tan^{-1} \left( \dfrac{56.35}{39.38} \right)=55.05^{\circ}.$$

3. Add the vectors in the diagram. Find the magnitude and direction of the resultant.

Solution: First draw the two relevant triangles: the left one is

and draw the right one:

The first triangle has $x$-component $$x=30\cos(21^{\circ})=28.01,$$ and $y$-component $$y=30\sin(21^{\circ})=10.75.$$ The second triangle has $x$-component $$x=19\cos(12^{\circ})=18.58$$ and $y$-component $$y=19\sin(12^{\circ})=3.95.$$ Therefore the resultant of the two vectors is $$\lt 28.01, 10.75 \gt + \lt 18.58, 3.95 \gt = \lt 46.59, 14.7 \gt.$$ The magnitude of this vector is $$\mathrm{magnitude}(\lt 46.59, 14.7 \gt) = \sqrt{46.59^2 + 14.7^2}=48.854$$ and its direction is $$\theta = \tan^{-1} \left( \dfrac{14.7}{46.59} \right) = 17.51^{\circ}.$$

4. Two forces of $87 N$ and $62 N$ act on right angles. Find the resultant, its magnitude, and angle it makes with the smaller force.
Solution: First draw this:

Clearly the horizontal vector is $\lt 64 ,0 \gt$ and the vertical vector is $\lt 0, 85 \gt$. Therefore the resultant is $$\lt 64, 0 \gt + \lt 0, 85 \gt = \lt 64, 85 \gt.$$ The magnitude of this vector is $$\mathrm{magnitude}(\lt 64, 85 \gt) = \sqrt{64^2 + 85^2} = 106.4$$ and its direction is $$\theta = \tan^{-1} \left( \dfrac{85}{64} \right) = 53.02^{\circ}.$$

5. Airplace traveling at $500 \dfrac{\mathrm{km}}{\mathrm{hr}}$ bearing $32^{\circ}$ south of east. Wind velocity is $45 \dfrac{\mathrm{km}}{\mathrm{hr}}$ $20^{\circ}$ north of west. Find the resultant velocity of the plane with respect to the ground. What is the actual direction relative to due east?
Solution: First draw the diagram:

The wind vector (the left one) has $x$-component $$x=45\cos(20^{\circ})=42.29$$ and $y$-component $$y=45\sin(20^{\circ})=15.39.$$ The airplane vector (the right one) has $x$-component $$x=500\cos(32^{\circ})=424$$ and $y$-component $$y=500\sin(32^{\circ})=265.$$ Therefore their resultant is $$\lt 42.29, 15.39 \gt + \lt 424, 265 \gt = \lt 466.29, 280.39 \gt.$$ The resultant velocity of the plane with respect to the ground is the $x$-component of the resultant, i.e. $466.29 \dfrac{\mathrm{km}}{\mathrm{hr}}$ and the angle relative to due east is given by $$\theta = \tan^{-1} \left( \dfrac{280.39}{466.29} \right) = 31.02^{\circ}.$$