Exercises — Direction cosines and direction ratios

WHAT to check first: direction cosines must be a unit vector, i.e. their squares add to $1$. $6^2+(-2{)}^2+3^2=36+4+9=49\ne 1.$ So $(6,-2,3)$ are direction ratios, not direction cosines.

WHY divide by the length: the direction ratios point the right way but are too long. Shrinking the arrow to length $1$ keeps the direction and fixes the length. The length is $\sqrt{6^2+(-2{)}^2+3^2}=\sqrt{49}=7.$ Direction cosines (taking the $+$ sign): $\left(\frac{6}{7},\frac{-2}{7},\frac{3}{7}\right).$ Check: $\frac{36}{49}+\frac{4}{49}+\frac{9}{49}=\frac{49}{49}=1$ ✓

WHAT the DCs are: they are literally the cosines of these angles. $l=\cos 90^{\circ }=0,m=\cos 60^{\circ }=\frac{1}{2},n=\cos 30^{\circ }=\frac{\sqrt{3}}{2}.$ $l=0$ means the line is perpendicular to the $x$-axis (leans zero toward the front wall). Verify: $0^2+{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=0+\frac{1}{4}+\frac{3}{4}=1.$ ✓

Step 1 — DRs from two points. $\overrightarrow{AB}$ points along the line, so its components are direction ratios: $(5-2,7-3,11-(-1))=(3,4,12).$ Step 2 — length. $\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13.$ Step 3 — normalise. Divide each ratio by $13$: $\left(\frac{3}{13},\frac{4}{13},\frac{12}{13}\right).$ Check: $\frac{9+16+144}{169}=\frac{169}{169}=1$ ✓ (See Distance between points in 3D — that  ​ is the distance $AB$.)

WHY use the DR-angle formula: the two triples are unit-vector components once normalised, and the dot product of two unit vectors is the cosine of the angle. The DR version bakes the normalising into one fraction. Numerator (dot product): $2\cdot 4+2\cdot 1+1\cdot 8=8+2+8=18.$ Magnitudes: $\sqrt{2^2+2^2+1^2}=\sqrt{9}=3,\sqrt{4^2+1^2+8^2}=\sqrt{81}=9.$ Cosine: $\cos \theta =\frac{18}{3\cdot 9}=\frac{18}{27}=\frac{2}{3}.$ $\theta ={\cos }^{-1}\frac{2}{3}\approx 48.19^{\circ }.$

WHAT "equal angles" means: $l=m=n$. Call the common value $t$. WHY the identity solves it: it is one equation linking all three: $t^2+t^2+t^2=1\Rightarrow 3t^2=1\Rightarrow t=\pm \frac{1}{\sqrt{3}}.$ The angle: $\alpha ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)\approx 54.74^{\circ }$ (or the opposite direction ${\cos }^{-1}\left(-\frac{1}{\sqrt{3}}\right)\approx 125.26^{\circ }$). Picture: this is the diagonal of a cube from a corner — it leans the same amount toward each wall (see figure below).

WHY the identity, not something else: we know two of three components of a unit vector; the length-1 constraint pins the third (up to sign). $n^2=1-l^2-m^2=1-\frac{1}{4}-\frac{1}{4}=\frac{1}{2}.$ $n=\pm \frac{1}{\sqrt{2}}.$ Both signs are real lines — the two opposite directions along the same line: $\gamma ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ }\text{or}\gamma =135^{\circ }.$

WHICH condition: perpendicular $\iff$ dot product of DRs $=0$ (no need to normalise — a zero dot product stays zero after scaling). $\lambda \cdot 1+(-3)\cdot 2+2\cdot \lambda =0$ $\lambda -6+2\lambda =0\Rightarrow 3\lambda =6\Rightarrow \lambda =2.$

Step 1 — DRs of each line (components of $\overrightarrow{PQ}$ and $\overrightarrow{RS}$): $\overrightarrow{PQ}=(3-1,4-(-1),-2-2)=(2,5,-4),$ $\overrightarrow{RS}=(3-0,5-3,6-2)=(3,2,4).$ Step 2 — perpendicularity test (dot product $=0$): $2\cdot 3+5\cdot 2+(-4)\cdot 4=6+10-16=0.$ Since the dot product is $0$, the lines are perpendicular. ∎

Step 1 — DRs of the two sides meeting at $A$. The interior angle at $A$ sits between $\overrightarrow{AB}$ and $\overrightarrow{AC}$, so use those directions (both leaving $A$). $\overrightarrow{AB}=(-1-3,1-5,2-(-4))=(-4,-4,6),$ $\overrightarrow{AC}=(-5-3,-5-5,-2-(-4))=(-8,-10,2).$ Step 2 — DCs of $\overrightarrow{AB}$. Length $\sqrt{(-4{)}^2+(-4{)}^2+6^2}=\sqrt{16+16+36}=\sqrt{68}=2\sqrt{17}$. $\left(\frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\right)=\left(\frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}\right).$ Step 3 — angle at $A$. $\overrightarrow{AB}\cdot \overrightarrow{AC}=(-4)(-8)+(-4)(-10)+(6)(2)=32+40+12=84.$ $|\overrightarrow{AC}|=\sqrt{64+100+4}=\sqrt{168}=2\sqrt{42}.$ $\cos A=\frac{84}{(2\sqrt{17})(2\sqrt{42})}=\frac{84}{4\sqrt{714}}=\frac{21}{\sqrt{714}}\approx \mathrm{0.786.}$ So $\angle A\approx {\cos }^{-1}(0.786)\approx 38.2^{\circ }$.

Set-up: $l=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$, $m=\cos 60^{\circ }=\frac{1}{2}$. Use length-1 constraint for $n$: $n^2=1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\Rightarrow n=\pm \frac{1}{2}.$ Acute angle with $z$: take $n=+\frac{1}{2}$, so $\gamma ={\cos }^{-1}\frac{1}{2}=60^{\circ }$. One set of DRs: any triple proportional to $\left(\frac{1}{\sqrt{2}},\frac{1}{2},\frac{1}{2}\right)$. Multiply by $2$: $\left(\sqrt{2},1,1\right)$. Check DCs: $\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1$ ✓

Translate words to symbols. "Equally inclined to $x$ and $y$" $\Rightarrow l=m$. "Angle $60^{\circ }$ with $z$" $\Rightarrow n=\cos 60^{\circ }=\frac{1}{2}$. Apply the identity to find $l$: $l^2+l^2+{\left(\frac{1}{2}\right)}^2=1\Rightarrow 2l^2=\frac{3}{4}\Rightarrow l^2=\frac{3}{8}\Rightarrow l=\pm \sqrt{\frac{3}{8}}=\pm \frac{\sqrt{6}}{4}.$ Direction cosines: $\left(\pm \frac{\sqrt{6}}{4},\pm \frac{\sqrt{6}}{4},\frac{1}{2}\right).$ Check: $\frac{6}{16}+\frac{6}{16}+\frac{1}{4}=\frac{12}{16}+\frac{4}{16}=\frac{16}{16}=1$ ✓

WHY this works: ${\sin }^2\theta =1-{\cos }^2\theta$ turns every sine into a known cosine, and we already know the cosines sum-of-squares to $1$. ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =(1-{\cos }^2\alpha )+(1-{\cos }^2\beta )+(1-{\cos }^2\gamma ).$ $=3-({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma )=3-1=2.■$

Every problem here is the same picture — a length-1 arrow in a room. If someone hands you how far it leans (cosines), the leftover lean is forced by "the arrow is length 1" ($l^2+m^2+n^2=1$). If someone hands you two points, subtract to get the arrow, then shrink it to length 1. If someone hands you two arrows, dot them and divide by lengths to get the angle. That's the entire chapter.