Exercises — Vector addition — triangle law, parallelogram law

WHAT tool? Same direction means $\theta =0^{\circ }$. WHY $\theta =0$? The angle between them is zero — they are parallel arrows pointing one way. $R=\sqrt{A^2+B^2+2AB\cos 0^{\circ }}=\sqrt{A^2+B^2+2AB}=\sqrt{(A+B{)}^2}=A+B.$ $R=6+9=\boxed{15\text{m/s}}.$ Sanity: parallel vectors just add like ordinary numbers — this is the maximum possible resultant.

WHAT tool? Opposite means $\theta =180^{\circ }$, and $\cos 180^{\circ }=-1$. $R=\sqrt{A^2+B^2-2AB}=\sqrt{(A-B{)}^2}=|A-B|.$ $R=|6-9|=\boxed{3\text{m/s}}.$ Direction: the resultant points along the larger vector — here the $9\text{m/s}$ one. WHY? After cancelling $6$ of the opposing $9$, only $3$ of the bigger vector survives, pointing its way. This is the minimum possible resultant.

WHAT tool? $\theta =90^{\circ }$, and $\cos 90^{\circ }=0$ kills the cross term. $R=\sqrt{A^2+B^2+0}=\sqrt{25+144}=\sqrt{169}=\boxed{13\text{N}}.$ WHY it collapses to Pythagoras: the two forces are perpendicular, so they form the two legs of a right triangle and $R$ is the hypotenuse.

Magnitude. $\cos 60^{\circ }=\frac{1}{2}$: $R=\sqrt{4^2+3^2+2(4)(3)(0.5)}=\sqrt{16+9+12}=\sqrt{37}\approx \boxed{6.08}.$ Direction. $\sin 60^{\circ }=0.8660$: $\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\frac{3(0.8660)}{4+3(0.5)}=\frac{2.598}{5.5}=\mathrm{0.4724.}$ $\alpha =\arctan (0.4724)\approx \boxed{25.3^{\circ }}.$ WHY arctan? $\tan \alpha$ tells us the ratio (opposite over adjacent) of the right triangle formed by dropping a vertical from the tip of $\vec{R}$; $\arctan$ asks "which angle has that tangent?" and hands back $\alpha$.

Magnitude. $\cos 120^{\circ }=-0.5$ (the cross term now subtracts): $R=\sqrt{49+25+2(7)(5)(-0.5)}=\sqrt{49+25-35}=\sqrt{39}\approx \boxed{6.24}.$ Direction. $\sin 120^{\circ }=0.8660$: $\tan \alpha =\frac{5(0.8660)}{7+5(-0.5)}=\frac{4.330}{4.5}=0.9623\Rightarrow \alpha =\arctan (0.9623)\approx \boxed{43.9^{\circ }}.$ WHY the denominator shrank: $B\cos \theta =-2.5$ points backwards along $\vec{A}$, so the horizontal reach $R_x=A+B\cos \theta$ is only $4.5$, not $7+5$. The resultant tilts more steeply → larger $\alpha$.

$R=\sqrt{8^2+8^2+0}=\sqrt{128}=8\sqrt{2}\approx \boxed{11.31}.$ $\tan \alpha =\frac{8\sin 90^{\circ }}{8+8\cos 90^{\circ }}=\frac{8}{8}=1\Rightarrow \alpha =\boxed{45^{\circ }}.$ WHY $45^{\circ }$? Equal vectors → the resultant bisects the angle between them, and half of $90^{\circ }$ is $45^{\circ }$. A symmetry gift.

WHAT we do: treat the magnitude formula as an equation for the unknown $\theta$. $R^2=A^2+B^2+2AB\cos \theta \Rightarrow 144=36+64+96\cos \theta .$ $144=100+96\cos \theta \Rightarrow \cos \theta =\frac{44}{96}=\mathrm{0.4583.}$ $\theta =\arccos (0.4583)\approx \boxed{62.7^{\circ }}.$ WHY $\arccos$? We know $\cos \theta$ and want the angle back — $\arccos$ is the "undo" button for cosine. Since $R=12$ is between the minimum $|8-6|=2$ and the maximum $8+6=14$, a valid angle must exist. ✔

WHAT "perpendicular to $\vec{A}$" means: the resultant has no component along $\vec{A}$, i.e. $R_x=0$. $R_x=A+B\cos \theta =0\Rightarrow \cos \theta =-\frac{A}{B}=-\frac{10}{6}=-\mathrm{1.667.}$ But $\cos \theta$ can never be less than $-1$! So no such angle exists — it is impossible for a $6$-unit vector to swing a $10$-unit vector all the way to a right angle. $\boxed{\text{Impossible}}$. WHY impossible geometrically: to cancel all $10$ units of horizontal reach, $\vec{B}$ would need a backward component of $10$, but its whole length is only $6$. Look at the picture below — the tip of $\vec{B}$ can never slide far enough left.

Set up along $\vec{B}$ this time. Put $\vec{B}$ on the x-axis; then $\vec{A}$ makes angle $\theta$ with it. The component of the resultant along $\vec{B}$ must vanish: $B+A\cos \theta =0\Rightarrow \cos \theta =-\frac{B}{A}.$ For $A=5,B=4$: $\cos \theta =-0.8\Rightarrow \theta =\arccos (-0.8)\approx \boxed{143.1^{\circ }}.$ Why it is possible here: we need a backward reach of $B=4$, and $\vec{A}$ has length $5\ge 4$, so it can supply it. (Contrast L3.2, where the roles made it impossible.)

WHY components, not the pair formula? The two-vector formula only handles two arrows at once. With three, the clean route is resolving each into x and y and summing separately. $R_x=5+4\cos 90^{\circ }+3\cos 180^{\circ }=5+0-3=2\text{N}.$ $R_y=5\sin 0^{\circ }+4\sin 90^{\circ }+3\sin 180^{\circ }=0+4+0=4\text{N}.$ $R=\sqrt{R_x^2+R_y^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\approx \boxed{4.47\text{N}}.$ $\tan \alpha =\frac{R_y}{R_x}=\frac{4}{2}=2\Rightarrow \alpha =\arctan (2)\approx \boxed{63.4^{\circ }}\text{ above }+x.$ Sign check: $R_x>0,R_y>0$ → resultant sits in the first quadrant, so $\alpha$ measured anticlockwise from $+x$ is correct as is.

This is vector addition in disguise: true velocity $=$ boat velocity $+$ river velocity, and they are at $90^{\circ }$. $R=\sqrt{4^2+3^2}=\sqrt{25}=\boxed{5\text{km/h}}.$ $\tan \alpha =\frac{3}{4}=0.75\Rightarrow \alpha =\arctan (0.75)\approx \boxed{36.9^{\circ }}\text{ downstream of straight-across}.$ WHY $90^{\circ }$: the boat aims perfectly across while the current pushes sideways along the bank — the two velocities are perpendicular, so it is the $3$–$4$–$5$ right triangle again.

Key idea from Subtraction of vectors and the difference vector: subtracting $\vec{B}$ means adding $-\vec{B}$, which points $180^{\circ }$ opposite. So the effective angle inside the magnitude formula becomes $180^{\circ }-60^{\circ }=120^{\circ }$, giving $\cos 120^{\circ }=-0.5$: $|\vec{A}-\vec{B}|=\sqrt{25+25+2(25)(-0.5)}=\sqrt{50-25}=\sqrt{25}=\boxed{5}.$ Beautiful check: with two equal sides of length $5$ and a $60^{\circ }$ gap, the three points form an equilateral triangle, so the third side (the difference) is also $5$. ✔

Set $R=A$ and $B=A$ in the magnitude formula: $A^2=A^2+A^2+2A^2\cos \theta \Rightarrow 0=A^2+2A^2\cos \theta .$ $\Rightarrow \cos \theta =-\frac{1}{2}\Rightarrow \theta =\boxed{120^{\circ }}.$ WHY it makes sense: three equal-length arrows at $120^{\circ }$ apart form a closed loop (like the Mercedes star). Two of them added tip-to-tail must therefore equal the reverse of the third — same length $A$. ✔

Step 1 — resolve the known resultant into components. $R_x=R\cos 30^{\circ }=10(0.8660)=8.660,R_y=R\sin 30^{\circ }=10(0.5)=5.$ Step 2 — peel off $\vec{A}$ to expose $\vec{B}$'s components ($\vec{B}=\vec{R}-\vec{A}$): $B_x=R_x-A=8.660-8=0.660,B_y=R_y-0=5.$ Step 3 — recombine into magnitude and angle. $B=\sqrt{0.660^2+5^2}=\sqrt{0.4356+25}=\sqrt{25.44}\approx \boxed{5.04}.$ $\theta =\arctan \frac{B_y}{B_x}=\arctan \frac{5}{0.660}=\arctan (7.576)\approx \boxed{82.5^{\circ }}\text{ from }\vec{A}.$ Sign check: $B_x>0$ and $B_y>0$ → $\vec{B}$ lives in quadrant I, so $\theta$ is a plain first-quadrant angle just under $90^{\circ }$. Look at the figure — $\vec{B}$ is a short arrow pointing almost straight up.

(a) Maximum at $\theta =0^{\circ }$: $R_{\max }=9+5=\boxed{14}$. Minimum at $\theta =180^{\circ }$: $R_{\min }=|9-5|=\boxed{4}$. Every achievable resultant lies in $[4,14]$. (b) Solve for the angle: $8^2=9^2+5^2+2(9)(5)\cos \theta \Rightarrow 64=106+90\cos \theta .$ $\cos \theta =\frac{64-106}{90}=\frac{-42}{90}=-0.4667\Rightarrow \theta =\arccos (-0.4667)\approx \boxed{117.8^{\circ }}.$ Check: $8$ is inside $[4,14]$, so a valid $\theta$ exists — and it is obtuse, consistent with the negative cosine. ✔