Point/Line/Plane (3D) — Distances & Intersections

Vectors, Geometry

Let's Do Math!

Intro: We parse vector forms and show the algebra step by step: substitute, simplify, solve for the parameter if needed, and normalize by the plane’s normal length.

Worked example

$P=(2,−1,4)$; plane $2x − 2y + 1z − 11 = 0$.

Identify plane coefficients: $$A=2,\;B=-2,\;C=1,\;D=-11.$$

Compute numerator $|Ax_0+By_0+Cz_0+D|$: $$|2\cdot2+(-2)\cdot(-1)+1\cdot4-11|=|4+2+4-11|=|{-1}|=1.$$

Compute denominator $\sqrt{A^2+B^2+C^2}$: $$\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3.$$

Distance formula: $$d=\dfrac{1}{3}.$$

Conclusion: $$\boxed{d=\tfrac{1}{3}}.$$

Note: The sign inside the absolute value gives the **signed** distance if divided by $\|\mathbf{n}\|$, but the distance itself is nonnegative.

Line: $\mathbf{r}(t)=\langle 1,2,0\rangle + t\,\langle 2,−1,3\rangle$. Plane: $x+2y−z=7$.

Extract data: line point $\mathbf{r}_0=(1,2,0)$, direction $\mathbf{d}=(2,-1,3)$; plane normal $\mathbf{n}=(1,2,-1)$ and $D=-7$ so plane is $\mathbf{n}\cdot\mathbf{r}=7$.

Parametric coordinates from the line: $$x=1+2t,\quad y=2- t,\quad z=0+3t=3t.$$

Substitute into plane equation $x+2y-z=7$: $$(1+2t)+2(2-t)-3t=7.$$

Simplify: $$1+2t+4-2t-3t=7\;\Rightarrow\;5-3t=7.$$

Solve for $t$: $$-3t=2\;\Rightarrow\; t=-\tfrac{2}{3}.$$

Find the intersection point by plugging $t$ back into the line:

$$x=1+2\Big(-\tfrac{2}{3}\Big)=1-\tfrac{4}{3}=-\tfrac{1}{3},$$

$$y=2-\Big(-\tfrac{2}{3}\Big)=2+\tfrac{2}{3}=\tfrac{8}{3},$$

$$z=3\Big(-\tfrac{2}{3}\Big)=-2.$$

Therefore the intersection point is $$\boxed{\left(-\tfrac{1}{3},\;\tfrac{8}{3},\;-2\right)}.$$

Quick check in the plane: $$x+2y-z=-\tfrac{1}{3}+2\cdot\tfrac{8}{3}-(-2)= -\tfrac{1}{3}+\tfrac{16}{3}+2=\tfrac{15}{3}+2=5+2=7\;\checkmark$$

Line $\mathbf{r}(t)=\langle 0,0,1\rangle+t\,\langle 1,2,−1\rangle$, plane $x+2y−z=5$ and variant $x+2y−z=0$.

Line direction $\mathbf{d}=(1,2,-1)$; plane normal $\mathbf{n}=(1,2,-1)$. Since $\mathbf{d}\cdot\mathbf{n}=1+4+1=6\ne0$, the line is **not** parallel and will intersect.

Intersect with $x+2y−z=5$: substitute $x=t$, $y=2t$, $z=1- t$:

$$t+2(2t)-(1-t)=5\;\Rightarrow\;t+4t-1+t=5\;\Rightarrow\;6t=6\Rightarrow t=1,$$

Intersection point: $(1,2,0)$.

Parallel scenario criterion: if instead $\mathbf{d}\cdot\mathbf{n}=0$ then the line is parallel to the plane. In that case:

• If also $\mathbf{n}\cdot\mathbf{r}_0\ne 7$ (or the plane’s right-hand side), **no intersection** (distinct parallel).

• If $\mathbf{n}\cdot\mathbf{r}_0$ **equals** the plane’s RHS, the **entire line lies in the plane** (infinitely many intersection points).

FAQs

How to detect if a line is parallel to a plane?

Let n be the plane’s normal and d the line direction. If $n·d = 0$, the line is parallel to the plane. Then check whether the line’s base point satisfies the plane: if yes → infinitely many points (line lies in plane); if no → no intersection.

Signed vs. unsigned distance for point→plane?

The signed value is $(Ax0+By0+Cz0+D)/‖n‖$, where $n=(A,B,C)$. The distance reported here is the absolute value of that $(always ≥ 0)$.

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