Exercises — Angle between two lines

WHAT we do: look only at the denominators, ignore the point. WHY: in symmetric form the numbers under the fractions are exactly the components of the direction arrow; the numerators only tell us a point the line passes through, which does not affect orientation. $\vec{b}=(2,-3,5)$ (The point $(4,-1,0)$ is irrelevant to any angle question.)

Test: perpendicular $⟺$ dot product $=0$. (a) $1\cdot 0+0\cdot 1+0\cdot 0=0$ → perpendicular ✅ (b) $1\cdot 1+1\cdot 1+0=2\ne 0$ → not perpendicular (in fact identical directions, angle $0^{\circ }$). Look at figure below: (a) is the x-axis vs the y-axis — a clean right angle.

Numerator (dot product): $3(1)+(-1)(2)+2(4)=3-2+8=9$. Magnitudes: $|\overrightarrow{b_1}|=\sqrt{9+1+4}=\sqrt{14}$, $|\overrightarrow{b_2}|=\sqrt{1+4+16}=\sqrt{21}$. Combine: $\cos \theta =\frac{|9|}{\sqrt{14}\sqrt{21}}=\frac{9}{\sqrt{294}}=\frac{9}{7\sqrt{6}}\approx 0.5249$ $\theta ={\cos }^{-1}(0.5249)\approx 58.3^{\circ }$

Read off: $\overrightarrow{b_1}=(2,2,-1)$, $\overrightarrow{b_2}=(4,1,8)$. Dot: $2(4)+2(1)+(-1)(8)=8+2-8=2$. Magnitudes: $|\overrightarrow{b_1}|=\sqrt{4+4+1}=3$, $|\overrightarrow{b_2}|=\sqrt{16+1+64}=\sqrt{81}=9$. $\cos \theta =\frac{|2|}{3\cdot 9}=\frac{2}{27}\approx 0.0741\Rightarrow \theta \approx 85.8^{\circ }$

Direction cosines are already unit arrows (their length is 1), so the denominator equals 1 — we just take the dot product (see Direction cosines and direction ratios). $\cos \theta =\left|\frac{1}{\sqrt{2}}\cdot 0+\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}+0\cdot \frac{1}{\sqrt{2}}\right|=\left|\frac{1}{2}\right|=\frac{1}{2}$ $\theta ={\cos }^{-1}\frac{1}{2}=60^{\circ }$

Perpendicular ⇒ dot product $=0$ (with nonzero lengths, only the numerator must vanish; $\cos 90^{\circ }=0$). $2k+k(-2)+3(4)=0\Rightarrow 2k-2k+12=0\Rightarrow 12=0.$ This is impossible — the $k$-terms cancel, leaving $12=0$. Conclusion: there is no value of $k$ that makes them perpendicular. The $x$- and $y$-contributions always cancel, so the dot product is stuck at $12$ regardless of $k$. Recognising "no solution" is the whole point of this level.

Cross product $\overrightarrow{b_1}\times \overrightarrow{b_2}$: for two identical vectors this is the zero vector $(0,0,0)$, because there's no plane spanned. $\sin \theta =\frac{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}=\frac{0}{3}=0\Rightarrow \theta =0^{\circ }.$ Why $\sin$ is the sharper tool here: near $\theta =0^{\circ }$, $\cos \theta$ is flat (its graph is nearly horizontal), so a tiny rounding error in the dot product barely moves the angle — small errors hide. But $\sin \theta$ changes steeply near $0^{\circ }$, so it magnifies and exposes any deviation. Use $\sin$ (cross product) to detect near-parallel; use $\cos$ (dot) to detect near-perpendicular.

$\overrightarrow{b_1}=(2,-4,6)$, $\overrightarrow{b_2}=(-1,2,-3)$. Check proportionality: $\frac{2}{-1}=\frac{-4}{2}=\frac{6}{-3}=-2$. All equal ⇒ $\overrightarrow{b_1}=-2\overrightarrow{b_2}$: same direction line (just scaled and reversed). Angle: parallel lines make angle $\boxed{0^{\circ }}$. Confirm with the formula: $\cos \theta =\frac{|2(-1)+(-4)(2)+6(-3)|}{\sqrt{4+16+36}\sqrt{1+4+9}}=\frac{|-2-8-18|}{\sqrt{56}\sqrt{14}}=\frac{28}{\sqrt{784}}=\frac{28}{28}=1.$ The modulus rescued us: even though $\overrightarrow{b_1}=-2\overrightarrow{b_2}$ points "backwards," $|\cos \theta |=1$ still reports $0^{\circ }$ (same line), not $180^{\circ }$.

Cross product (see Cross product of vectors): $\overrightarrow{b_1}\times \overrightarrow{b_2}=\left(\begin{array}{c}2\cdot 1-2\cdot 2\\ 2\cdot 2-1\cdot 1\\ 1\cdot 2-2\cdot 2\end{array}\right)=\left(\begin{array}{c}-2\\ 3\\ -2\end{array}\right).$ $|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{4+9+4}=\sqrt{17}.$ $\sin \theta =\frac{\sqrt{17}}{3\cdot 3}=\frac{\sqrt{17}}{9}\approx 0.4581\Rightarrow \theta \approx 27.3^{\circ }.$ Cross-check with cosine: $\cos \theta =\frac{|8|}{9}=\frac{8}{9}$. Test the identity ${\sin }^2+{\cos }^2=1$: ${\left(\frac{\sqrt{17}}{9}\right)}^2+{\left(\frac{8}{9}\right)}^2=\frac{17+64}{81}=\frac{81}{81}=1.✅$ Both routes give $\theta \approx 27.3^{\circ }$ — the two tools agree, as they must.

WHY the cross product: $\overrightarrow{b_1}\times \overrightarrow{b_2}$ is by construction perpendicular to both inputs — exactly the arrow we need. $\vec{d}=(1,0,1)\times (0,1,-1)=\left(\begin{array}{c}0(-1)-1(1)\\ 1(0)-1(-1)\\ 1(1)-0(0)\end{array}\right)=(-1,1,1).$ Angle with $(1,1,1)$: $\cos \theta =\frac{|-1+1+1|}{\sqrt{1+1+1}\sqrt{1+1+1}}=\frac{1}{3}.$ $\theta ={\cos }^{-1}\frac{1}{3}\approx 70.5^{\circ }.$

Direction vectors: diagonal $\overrightarrow{b_1}=(1,1,1)$, edge $\overrightarrow{b_2}=(1,0,0)$. Dot: $1(1)+1(0)+1(0)=1$. Magnitudes: $\sqrt{3}$ and $1$. $\cos \theta =\frac{|1|}{\sqrt{3}\cdot 1}=\frac{1}{\sqrt{3}}\approx 0.5774\Rightarrow \theta \approx 54.7^{\circ }.$ This is the famous "cube-diagonal angle." See the figure: the red diagonal leans $\approx 54.7^{\circ }$ off each edge.

Set up $\cos 60^{\circ }=\frac{1}{2}$: $\frac{|1(1)+1(-1)+p(0)|}{\sqrt{1+1+p^2}\sqrt{1+1+0}}=\frac{|0|}{\sqrt{2+p^2}\cdot \sqrt{2}}=0.$ The numerator is always $0$ (the first two components cancel and $p$ multiplies zero). So $\cos \theta =0$ for every $p$, meaning the angle is always $90^{\circ }$, never $60^{\circ }$. Conclusion: no value of $p$ gives $60^{\circ }$. The two lines are perpendicular for all $p$ — a structural fact the algebra reveals instantly.

$\vec{u},\vec{v}$: $\cos \theta =\frac{|1|}{1\cdot \sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \theta =45^{\circ }.$ $\vec{u},\vec{w}$: $\cos \theta =\frac{|1|}{1\cdot \sqrt{3}}=\frac{1}{\sqrt{3}}\Rightarrow \theta \approx 54.7^{\circ }.$ $\vec{v},\vec{w}$: $\cos \theta =\frac{|1+1+0|}{\sqrt{2}\cdot \sqrt{3}}=\frac{2}{\sqrt{6}}\approx 0.8165\Rightarrow \theta \approx 35.3^{\circ }.$ Most spread apart: the pair $(\vec{u},\vec{w})$ at $\approx 54.7^{\circ }$ — the widest of the three.