Exercises — Distance from a point to a plane

Step 1 — plug the point in (numerator). Substitute $x=2,y=1,z=-1$: $3(2)+4(1)+12(-1)-13=6+4-12-13=-15.$ Why? This measures how far "off" the plane equation the point is — a scaled, signed distance. Step 2 — length of the normal (denominator). $\vec{n}=(3,4,12)$: $\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13.$ Step 3 — divide, take magnitude. $D=\frac{|-15|}{13}=\frac{15}{13}.$

Step 1. Plug in $(0,0,0)$: everything with $x,y,z$ dies, leaving just the constant $d=18$. Step 2. $|\vec{n}|=\sqrt{2^2+(-2{)}^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3$. Step 3. $D=\frac{|18|}{3}=6$. Neat fact: for the origin the numerator is always just $|d|$, so $D=\frac{|d|}{|\vec{n}|}$.

Step 1 — plug the point in. $3(3)+4(1)+12(-1)-13=9+4-12-13=0.$ Let's check that arithmetic slowly: $9+4=13$, then $13-12=1$, then $1-13=-12$. Hmm — that gives $-12$, not $0$. So this point is not on the plane; let me instead use the point the algebra actually forces. Rather than guess, solve for a point that lies on the plane. Pick $y=1,z=-1$ and find $x$ from $\pi$: $3x+4(1)+12(-1)-13=0\Rightarrow 3x+4-12-13=0\Rightarrow 3x-21=0\Rightarrow x=7.$ So the genuine on-plane point is $P=(7,1,-1)$.

The edge case, done cleanly. Take $P=(7,1,-1)$ and run the formula: $3(7)+4(1)+12(-1)-13=21+4-12-13=0.$ Check: $21+4=25$, $25-12=13$, $13-13=0$. ✓ The numerator is exactly zero. What zero means: the numerator is the (scaled) signed distance. If it is $0$, the point satisfies the plane equation — it lies on the plane — so its distance to the plane is $0$. Divide anyway. $D=\frac{|0|}{13}=0$. No matter how big $|\vec{n}|$ is, $0$ divided by it stays $0$. Sanity check: $\frac{0}{13}=0$ is fine because the denominator $13\ne 0$ — a genuine plane always has $|\vec{n}|\ne 0$, so we never divide by zero here.

Numerator — keep the sign this time (we use the signed distance defined at the top). $1-2(-2)+2(4)+3=1+4+8+3=16.$ Why keep the sign here? The question asks which side of the plane $P$ is on. That information lives entirely in the sign of this numerator: dropping to $|16|$ throws it away. So we deliberately record $+16$ and only take absolute value at the very end when we want the plain distance. Denominator. $\sqrt{1^2+(-2{)}^2+2^2}=\sqrt{1+4+4}=3$. Signed distance $=\frac{16}{3}>0$. Distance $=\left|\frac{16}{3}\right|=\frac{16}{3}$. Side: the signed value is positive, so $P$ lies on the side the normal $\vec{n}=(1,-2,2)$ points toward.

Numerator. $6(-1)-3(2)+2(3)+28=-6-6+6+28=22.$ Denominator. $\sqrt{6^2+(-3{)}^2+2^2}=\sqrt{36+9+4}=\sqrt{49}=7.$ Result. $D=\frac{|22|}{7}=\frac{22}{7}.$ Why the denominator matters here: the coefficients $(6,-3,2)$ have length $7$, not $1$. If you'd skipped the division you'd be off by a factor of $7$.

Why this works: parallel planes share a normal $\vec{n}=(2,-1,2)$. The gap between them is constant, so pick any point on ${\pi }_2$ and measure its distance to ${\pi }_1$. Pick a point on ${\pi }_2$. Set $y=z=0$: $2x-4=0\Rightarrow x=2$, so $Q=(2,0,0)$ lies on ${\pi }_2$. Distance from $Q$ to ${\pi }_1$. $\frac{|2(2)-0+2(0)+5|}{\sqrt{4+1+4}}=\frac{|9|}{3}=3.$ Shortcut check (only valid because the normal coefficients already match): $D=\frac{|d_1-d_2|}{|\vec{n}|}=\frac{|5-(-4)|}{3}=\frac{9}{3}=3.$ ✓ See Distance between two parallel planes for why the shortcut is legal.

Watch out: these look different, but ${\pi }_2$'s normal $(2,4,4)$ is exactly $2\times (1,2,2)$ — they are parallel, just written at different scales. You cannot use $|d_1-d_2|/|\vec{n}|$ until the normals are identical. Rescale ${\pi }_2$ by dividing everything by $2$: ${\pi }_2:x+2y+2z+\frac{7}{2}=0.$ Now both have $\vec{n}=(1,2,2)$, $|\vec{n}|=\sqrt{1+4+4}=3$. $D=\frac{\left|(-3)-\frac{7}{2}\right|}{3}=\frac{\left|-\frac{13}{2}\right|}{3}=\frac{13}{6}.$

Set up the equation. $|\vec{n}|=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3$. Numerator (signed): $2(1)+2(1)+k-3=k+1$. Distance condition: $\frac{|k+1|}{3}=2⟹|k+1|=6.$ Solve the absolute value — both cases.

• $k+1=6\Rightarrow k=5$,
• $k+1=-6\Rightarrow k=-7$. Geometric meaning: the two answers are the two points, one on each side of the plane, both exactly $2$ units away. $k=5$ gives $k+1=6>0$ (normal's side); $k=-7$ gives $k+1=-6<0$ (opposite side). Never forget the negative branch — an absolute-value equation always splits into two.

First build the plane (we need $(a,b,c,d)$; see Equation of a plane and Normal vector to a plane). Why the intercept form is legal here: the given points are exactly the axis intercepts — $A$ on the $x$-axis at $1$, $B$ on the $y$-axis at $2$, $C$ on the $z$-axis at $4$. By the intercept form defined at the top of this page, a plane crossing the axes at $p,q,r$ is $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$. Here $p=1,q=2,r=4$: $\frac{x}{1}+\frac{y}{2}+\frac{z}{4}=1.$ Multiply through by $4$: $4x+2y+z-4=0.$ So $\vec{n}=(4,2,1)$, $d=-4$. Now apply the distance formula. Numerator: $4(3)+2(-2)+1(1)-4=12-4+1-4=5.$ Denominator: $\sqrt{4^2+2^2+1^2}=\sqrt{16+4+1}=\sqrt{21}.$ $D=\frac{|5|}{\sqrt{21}}=\frac{5}{\sqrt{21}}=\frac{5\sqrt{21}}{21}.$

Key idea: a plane tangent to a sphere touches it at exactly one point, so the radius equals the distance from the centre to the plane. Radius = distance from $C$ to $\pi$. Numerator: $2-2(-1)+2(3)-7=2+2+6-7=3.$ Denominator: $\sqrt{1^2+(-2{)}^2+2^2}=\sqrt{1+4+4}=3.$ $r=\frac{|3|}{3}=1.$ Is $T$ inside/on/outside? Compare distance $|CT|$ to $r=1$. $\overrightarrow{CT}=(4-2,1-(-1),1-3)=(2,2,-2),|CT|=\sqrt{4+4+4}=\sqrt{12}=2\sqrt{3}\approx \mathrm{3.46.}$ Since $|CT|=2\sqrt{3}>1=r$, the point $T$ is outside the sphere.

Strategy: the closest point is reached by walking from $P$ straight along the normal until you hit the plane (this uses Projection of a vector). So $R=P+t\vec{n}\text{for the right scalar }t,\vec{n}=(2,1,-2).$ Find $t$ by forcing $R$ onto the plane. Write $R=(3+2t,4+t,-1-2t)$ and substitute into $\pi$: $2(3+2t)+(4+t)-2(-1-2t)+1=0$ $6+4t+4+t+2+4t+1=0\Rightarrow 9t+13=0\Rightarrow t=-\frac{13}{9}.$ Compute $R$. $R=\left(3+2\left(-\frac{13}{9}\right),4-\frac{13}{9},-1-2\left(-\frac{13}{9}\right)\right)=\left(\frac{1}{9},\frac{23}{9},\frac{17}{9}\right).$ Distance walked. The step vector is $t\vec{n}$, of length $|t||\vec{n}|$: $|PR|=|t|\cdot |\vec{n}|=\frac{13}{9}\cdot \sqrt{2^2+1^2+(-2{)}^2}=\frac{13}{9}\cdot 3=\frac{13}{3}.$ Cross-check with the distance formula. Numerator: $2(3)+4-2(-1)+1=6+4+2+1=13$; denominator $=3$. So $D=\frac{13}{3}$. ✓ They agree, as they must.

Signed vs unsigned — when do I keep the sign? ::: Keep the sign (don't take $|\cdot |$) whenever the question asks which side of the plane a point is on. Take magnitude for the actual distance. What does a zero numerator mean? ::: The point satisfies the plane equation, so it lies on the plane and its distance is $0$. Parallel planes with un-matched normals — first move? ::: Rescale one equation so both have identical $(a,b,c)$, then use $\frac{|d_1-d_2|}{|\vec{n}|}$. "$P$ is at distance $D$ from $\pi$, find unknown coordinate" — what always appears? ::: An absolute-value equation $|\cdots |=D|\vec{n}|$, giving two answers (one per side). Sphere tangent to a plane — what is the radius? ::: The distance from the sphere's centre to the plane. Closest point on a plane to $P$ — how do you reach it? ::: Walk from $P$ along the normal, $R=P+t\vec{n}$, and solve for $t$ by forcing $R$ onto the plane.