Exercises — Angle between line and plane

WHAT: plug into the master formula. $\vec{b}\cdot \vec{n}=(1)(0)+(0)(0)+(0)(1)=0$. WHY it matters: a dot product of $0$ means $\sin \theta =0$, so $\theta =0^{\circ }$. The line runs flat along the plane (it is parallel to it), like a stick lying on the ground. Its length: $|\vec{b}|=1$, $|\vec{n}|=1$, but the numerator already killed it. $\sin \theta =\frac{|0|}{1\cdot 1}=0\Rightarrow \boxed{\theta =0^{\circ }}.$

Both arrows point straight up — they are parallel. When $\vec{b}\parallel \vec{n}$ the line is perpendicular to the plane (the stick is the flagpole). $\vec{b}\cdot \vec{n}=0+0+10=10$, $|\vec{b}|=5$, $|\vec{n}|=2$. $\sin \theta =\frac{|10|}{5\cdot 2}=\frac{10}{10}=1\Rightarrow \boxed{\theta =90^{\circ }}.$ WHAT IT LOOKS LIKE: the stick stands bolt upright out of the ground — its shadow is a single point, so the angle to the ground is the full right angle.

The plane $z=0$ is written $0x+0y+1z+0=0$, so its normal is $\vec{n}=(0,0,1)$ (see Equation of a plane). Here $\vec{b}=(1,1,0)$. $\vec{b}\cdot \vec{n}=0+0+0=0\Rightarrow \sin \theta =0\Rightarrow \boxed{\theta =0^{\circ }}$. The line has no vertical part, so it lies flat on the floor.

$\vec{b}=(2,-1,2)$, $\vec{n}=(3,4,12)$. $\vec{b}\cdot \vec{n}=(2)(3)+(-1)(4)+(2)(12)=6-4+24=26$. $|\vec{b}|=\sqrt{4+1+4}=3$, $|\vec{n}|=\sqrt{9+16+144}=\sqrt{169}=13$. $\sin \theta =\frac{|26|}{3\cdot 13}=\frac{26}{39}=\frac{2}{3}\Rightarrow \boxed{\theta ={\sin }^{-1}\frac{2}{3}\approx 41.8^{\circ }}.$

Reading the line's denominators gives $\vec{b}=(2,3,6)$ (see Direction ratios and direction cosines); reading the plane's coefficients gives $\vec{n}=(2,1,-2)$. $\vec{b}\cdot \vec{n}=(2)(2)+(3)(1)+(6)(-2)=4+3-12=-5$. WHY the bars: a line direction is unsigned, so a negative dot product is fine — take $|-5|=5$ to keep $\sin \theta \ge 0$. $|\vec{b}|=\sqrt{4+9+36}=7$, $|\vec{n}|=\sqrt{4+1+4}=3$. $\sin \theta =\frac{5}{21}\Rightarrow \boxed{\theta ={\sin }^{-1}\frac{5}{21}\approx 13.8^{\circ }}.$

$\vec{n}=(1,1,1)$. Notice $\vec{b}=\vec{n}$ — same direction, so line ⟂ plane. $\vec{b}\cdot \vec{n}=1+1+1=3$, $|\vec{b}|=|\vec{n}|=\sqrt{3}$. $\sin \theta =\frac{3}{\sqrt{3}\cdot \sqrt{3}}=\frac{3}{3}=1\Rightarrow \boxed{\theta =90^{\circ }}.$

WHAT: parallel to plane means the line lies flat along it, i.e. $\vec{b}\perp \vec{n}$, i.e. $\vec{b}\cdot \vec{n}=0$. $\vec{n}=(2,-3,1)$. $\vec{b}\cdot \vec{n}=(2)(2)+(k)(-3)+(3)(1)=4-3k+3=7-3k.$ Set to zero: $7-3k=0\Rightarrow \boxed{k=\frac{7}{3}}$. Check the logic: with $k=7/3$ the numerator of $\sin \theta$ vanishes, so $\theta =0^{\circ }$ — parallel. (This is the same $\vec{b}\cdot \vec{n}=0$ test behind Coplanarity of lines.)

$\vec{n}=(1,1,1)$, $\sin 30^{\circ }=\frac{1}{2}$. $\vec{b}\cdot \vec{n}=1+1+c=2+c$, $|\vec{b}|=\sqrt{2+c^2}$, $|\vec{n}|=\sqrt{3}$. $\frac{1}{2}=\frac{|2+c|}{\sqrt{2+c^2}\sqrt{3}}.$ Square both sides (safe since everything is $\ge 0$): $\frac{1}{4}=\frac{(2+c{)}^2}{3(2+c^2)}\Rightarrow 3(2+c^2)=4(2+c{)}^2.$ $6+3c^2=4(4+4c+c^2)=16+16c+4c^2$. $0=c^2+16c+10$. Solve: $c=\frac{-16\pm \sqrt{256-40}}{2}=-8\pm \sqrt{54}=-8\pm 3\sqrt{6}$. Since $c>0$: $\boxed{c=-8+3\sqrt{6}\approx -0.652}$ — but that is negative, so no positive $c$ works. WHY: $\sqrt{6}\approx 2.449$, $3\sqrt{6}\approx 7.35<8$, so both roots are negative. The problem's demand ($c>0$ and $30^{\circ }$) is impossible; the only real solutions are $c\approx -0.652$ and $c\approx -15.35$. This teaches you to check feasibility, not just push algebra.

Two conditions are needed (parallel is not enough — the line must also touch):

1. Parallel test: $\vec{b}\cdot \vec{n}=(2)(1)+(3)(-2)+(4)(1)=2-6+4=0$. ✓ (θ = 0°, line parallel to plane).
2. Point test: does the point $(1,2,3)$ satisfy the plane? $1-2(2)+3=1-4+3=0$. ✓ Both hold, so $\boxed{\text{the line lies entirely in the plane}}$.

Step 1 (WHAT): the line ⟂ first plane, so its direction is that plane's normal: $\vec{b}=(3,-1,2)$. Step 2: second plane has normal $\vec{n}=(1,1,1)$. Step 3 (angle): $\vec{b}\cdot \vec{n}=3-1+2=4$, $|\vec{b}|=\sqrt{9+1+4}=\sqrt{14}$, $|\vec{n}|=\sqrt{3}$. $\sin \theta =\frac{|4|}{\sqrt{14}\sqrt{3}}=\frac{4}{\sqrt{42}}\approx 0.6172\Rightarrow \boxed{\theta \approx 38.1^{\circ }}.$ The starting point $(1,0,-1)$ is a distractor — the angle depends only on directions.

Step 1 — line direction: $\vec{b}=B-A=(4-2,4-1,7-3)=(2,3,4)$. Step 2 — plane normal: $P,Q,R$ all have $z=0$, so the plane is the floor $z=0$, normal $\vec{n}=(0,0,1)$. Step 3: $\vec{b}\cdot \vec{n}=0+0+4=4$, $|\vec{b}|=\sqrt{4+9+16}=\sqrt{29}$, $|\vec{n}|=1$. $\sin \theta =\frac{4}{\sqrt{29}}\approx 0.7428\Rightarrow \boxed{\theta \approx 47.98^{\circ }}.$

$\vec{b}=(1,2,2)$, $|\vec{b}|=\sqrt{1+4+4}=3$. Plane 1: ${\vec{n}}_1=(2,2,1)$, $\vec{b}\cdot {\vec{n}}_1=2+4+2=8$, $|{\vec{n}}_1|=3$. $\sin {\theta }_1=\frac{8}{9}\approx 0.889$. Plane 2: ${\vec{n}}_2=(1,2,2)=\vec{b}$! So the line ⟂ plane 2, $\sin {\theta }_2=1$, ${\theta }_2=90^{\circ }$. $\sin {\theta }_2=1>\sin {\theta }_1=\frac{8}{9}\Rightarrow \boxed{{\theta }_2>{\theta }_1}.$ Since $\sin$ is increasing on $[0^{\circ },90^{\circ }]$, comparing sines directly compares the angles.

Two constraints on $\vec{b}$:

1. Line in plane $2x+y-z=3$ ⇒ $\vec{b}\perp (2,1,-1)$: $2p+q-r=0$.
2. Perpendicular to $(1,-1,1)$: $p-q+r=0$. Add the two equations: $3p=0\Rightarrow p=0$. Then from (2): $-q+r=0\Rightarrow r=q$. Pick $q=1$: $\vec{b}=(0,1,1)$. Check (1): $0+1-1=0$ ✓. Angle with plane $x-y+2z=0$: $\vec{n}=(1,-1,2)$. $\vec{b}\cdot \vec{n}=0-1+2=1$, $|\vec{b}|=\sqrt{2}$, $|\vec{n}|=\sqrt{6}$. $\sin \theta =\frac{|1|}{\sqrt{2}\sqrt{6}}=\frac{1}{\sqrt{12}}=\frac{1}{2\sqrt{3}}\approx 0.2887\Rightarrow \boxed{\theta \approx 16.78^{\circ }}.$

Direction cosines are just the components of a unit direction arrow, so $|\vec{b}|=1$ and $|\vec{n}|=1$ already (see Direction ratios and direction cosines). The master formula $\sin \theta =\frac{|\vec{b}\cdot \vec{n}|}{|\vec{b}||\vec{n}|}$ has denominator $1\cdot 1=1$, and $\vec{b}\cdot \vec{n}=\alpha l+\beta m+\gamma n$. $\therefore \boxed{\sin \theta =|\alpha l+\beta m+\gamma n|.}$ Numeric sanity check: take $\vec{b}=(2,3,6)/7$ and $\vec{n}=(2,1,-2)/3$ from L2.2. Then $\alpha l+\beta m+\gamma n=\frac{4+3-12}{21}=\frac{-5}{21}$, and $|-5/21|=5/21$ — matching L2.2 exactly. ✓

$\vec{n}=(2,-1,2)$, $\vec{b}\cdot \vec{n}=(1)(2)+(-2)(-1)+(2)(2)=2+2+4=8$. $|\vec{b}|=\sqrt{1+4+4}=3$, $|\vec{n}|=\sqrt{4+1+4}=3$. Angle from normal uses $\cos$ (angle directly between two arrows): $\cos \gamma =\frac{|8|}{3\cdot 3}=\frac{8}{9}\Rightarrow \gamma ={\cos }^{-1}\frac{8}{9}\approx 27.27^{\circ }.$ Angle from surface uses $\sin$: $\sin \theta =\frac{8}{9}\Rightarrow \theta ={\sin }^{-1}\frac{8}{9}\approx 62.73^{\circ }.$ Verify: $\theta +\gamma \approx 62.73^{\circ }+27.27^{\circ }=90^{\circ }$ ✓ — the complementary relationship, proven live.