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1. Find $\vec{x}-3\vec{y}$ where $\vec{x}=\lt 2,1 \gt$ and $\vec{y}=\lt 1, -1 \gt$.

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.

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.

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?

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 $$\theta = \tan^{-1} \left( \dfrac{280.39}{466.29} \right) = 31.02^{\circ}.$$