Exercises — Relation between direction cosines - l² + m² + n² = 1

What we test: does $l^2+m^2+n^2=1$?

• (a) $\frac{1}{9}+\frac{4}{9}+\frac{4}{9}=\frac{9}{9}=1$ ✓ — valid.
• (b) $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\ne 1$ ✗ — invalid.
• (c) $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$ ✓ — valid.

Answer: (a) and (c) only.

Use the identity to solve for the missing cosine: $m^2=1-{\left(\frac{6}{7}\right)}^2-{\left(\frac{2}{7}\right)}^2=1-\frac{36}{49}-\frac{4}{49}=\frac{49-40}{49}=\frac{9}{49}.$ So $m=\pm \frac{3}{7}$. Since we're told $m>0$, $m=\frac{3}{7}$.

Step 1 — direction ratios = $B-A$ (the arrow from $A$ to $B$): $(4-2,2-(-1),5-3)=(2,3,2).$ Step 2 — magnitude: $\sqrt{2^2+3^2+2^2}=\sqrt{4+9+4}=\sqrt{17}.$ Step 3 — divide to normalise: $l=\frac{2}{\sqrt{17}},m=\frac{3}{\sqrt{17}},n=\frac{2}{\sqrt{17}}.$ Check: $\frac{4+9+4}{17}=\frac{17}{17}=1$ ✓

$l=\cos 60^{\circ }=\frac{1}{2}$, $n=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$. $m^2=1-\frac{1}{4}-\frac{1}{2}=\frac{1}{4}\Rightarrow m=\pm \frac{1}{2}.$ So $\beta =60^{\circ }$ or $120^{\circ }$ (both directions of the line are allowed).

Magnitude $=\sqrt{1+4+4}=\sqrt{9}=3$. $l=\frac{1}{3},m=-\frac{2}{3},n=\frac{2}{3}.$ Check: $\frac{1+4+4}{9}=1$ ✓ (Sign of $m$ is kept — normalising never changes signs.)

Equal angles $\Rightarrow l=m=n$. Substitute into the identity: $3l^2=1\Rightarrow l=\pm \frac{1}{\sqrt{3}}.$ The common angle $\alpha$ satisfies $\cos \alpha =\frac{1}{\sqrt{3}}$, so $\alpha =\arccos \frac{1}{\sqrt{3}}\approx 54.74^{\circ }$. The $\pm$ gives the two opposite orientations (one at $\approx 54.74^{\circ }$, the mirror at $\approx 125.26^{\circ }$).

Test the identity numerically: ${\cos }^{2}50^{\circ }+{\cos }^{2}60^{\circ }+{\cos }^{2}70^{\circ }\approx 0.4132+0.2500+0.1170=\mathrm{0.7802.}$ This is $\ne 1$, so the triple fails the constraint — no such line exists. The relation is a filter on possible angle-triples.

Start from the identity ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$. Replace each cosine-squared using ${\cos }^2\theta =1-{\sin }^2\theta$: $(1-{\sin }^2\alpha )+(1-{\sin }^2\beta )+(1-{\sin }^2\gamma )=1.$ $3-({\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma )=1\Rightarrow {\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2.■$

The angle between two lines uses the dot product of the two unit direction vectors: $\cos \theta =l_1{l}_2+m_1{m}_2+n_1{n}_2.$ $\cos \theta =\frac{1}{3}\cdot \frac{2}{3}+\frac{2}{3}\cdot \left(-\frac{1}{3}\right)+\frac{2}{3}\cdot \frac{2}{3}=\frac{2}{9}-\frac{2}{9}+\frac{4}{9}=\frac{4}{9}.$ So $\theta =\arccos \frac{4}{9}\approx 63.6^{\circ }$. Since $\cos \theta \ne 0$, they are not perpendicular.

In the $xy$-plane the line is perpendicular to the $z$-axis, so $n=\cos 90^{\circ }=0$. $l=\cos 30^{\circ }=\frac{\sqrt{3}}{2}$. $m^2=1-\frac{3}{4}-0=\frac{1}{4}\Rightarrow m=\pm \frac{1}{2}.$ Answer: $\left(\frac{\sqrt{3}}{2},\frac{1}{2},0\right)$ or $\left(\frac{\sqrt{3}}{2},-\frac{1}{2},0\right)$ — the two lines symmetric about the $x$-axis.

A line perpendicular to both the $x$-axis and the $y$-axis must point along the $z$-axis. Formally, its ratios are the cross-product $(1,0,0)\times (0,1,0)=(0,0,1)$. Magnitude $=1$, so direction cosines are $(0,0,\pm 1)$. Check: $0+0+1=1$ ✓

$l=\cos 90^{\circ }=0$, and $m=n=\cos \beta$. Substitute: $0+{\cos }^2\beta +{\cos }^2\beta =1\Rightarrow 2{\cos }^2\beta =1\Rightarrow \cos \beta =\pm \frac{1}{\sqrt{2}}.$ So $\beta =45^{\circ }$ or $135^{\circ }$. The line lies in the $yz$-plane, bisecting the $y$- and $z$-axes.

Direction cosines: $l=\cos 60^{\circ }=\frac{1}{2}$, $m=\cos 60^{\circ }=\frac{1}{2}$. $n^2=1-\frac{1}{4}-\frac{1}{4}=\frac{1}{2}\Rightarrow n=\pm \frac{1}{\sqrt{2}}.$ Positive $z$ ⟹ $n=+\frac{1}{\sqrt{2}}$. Then $\vec{r}=10(l,m,n)$ because a vector = (its length) × (its unit direction): $\vec{r}=\left(10\cdot \frac{1}{2},10\cdot \frac{1}{2},10\cdot \frac{1}{\sqrt{2}}\right)=\left(5,5,\frac{10}{\sqrt{2}}\right)=(5,5,5\sqrt{2}).$ Check length: $\sqrt{25+25+50}=\sqrt{100}=10$ ✓

Equal magnitudes $\Rightarrow |l|=|m|=|n|$, so $l^2=m^2=n^2$. The identity gives $3l^2=1$, hence $|l|=\frac{1}{\sqrt{3}}$ and ${\cos }^{-1}\frac{1}{\sqrt{3}}\approx 54.74^{\circ }$ (or its supplement $125.26^{\circ }$ when the cosine is negative). Each of $l,m,n$ independently takes sign $+$ or $-$: that's $2\times 2\times 2=8$ sign patterns (orientations). But a line and its reverse (all signs flipped) are the same line. Pairing each pattern with its negation halves the count: $8/2=4$ distinct lines — the four body-diagonals of a cube. $■$

• L1: valid ⟺ $l^2+m^2+n^2=1$; missing cosine from $\pm \sqrt{1-{\text{(others)}}^2}$.
• L2: ratios → cosines by dividing by magnitude; signs preserved.
• L3: equal-angle line has $\cos =\frac{1}{\sqrt{3}}$ ($\approx 54.74^{\circ }$); $\sum {\sin }^2=2$.
• L4: angle between lines $=\arccos (l_1{l}_2+m_1{m}_2+n_1{n}_2)$; keep $\pm$.
• L5: vector $=$ length $\times$ unit direction; a line $=$ two opposite orientations.