Porisma

Vectors: planes, normals and distances

← Back to learning units · Complete chapter reading and print view · Open reference notes separately

By the end, you can

  • Describe planes using normals or two directions
  • Find point-to-plane distance
  • Relate angles and intersections to vector products

about 40 minutes for the lesson · practise at your own pace.

The central idea

A plane can be written n⋅r=c\mathbf n\cdot\mathbf r=c, where its nonzero normal vector is perpendicular to every direction in the plane. Equivalently, two independent in-plane directions give a parametric equation. The distance from P is ∣n⋅p−c∣/∣n∣|\mathbf n\cdot\mathbf p-c|/|\mathbf n|.

Derive a parametric plane

Choose a fixed point A with position a and two nonparallel nonzero directions b,c parallel to the plane. A displacement in the plane is AR→=λb+μc\overrightarrow{AR}=\lambda\mathbf b+\mu\mathbf c. Triangle addition gives OR→−OA→=AR→\overrightarrow{OR}-\overrightarrow{OA}=\overrightarrow{AR}, hence

r=a+λb+μc,λ,μ∈R.\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c,\qquad\lambda,\mu\in\mathbb R.

This equation is not unique: another base point or independent pair of directions can describe the same plane. A drawn parallelogram is only a finite patch of an infinite plane and does not bound the real parameters. Parallel directions would not span a plane.

Coordinate planes and the source room analogy

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Two schematic views share x and y axes along the floor and z vertically. The floor lies in the xy-plane, left wall in xz, and right wall in yz. The second panel labels the same surfaces as floor, left wall and right wall. Finite patches illustrate infinite coordinate planes; decorative furniture is omitted.
Two schematic views share x and y axes along the floor and z vertically. The floor lies in the xy-plane, left wall in xz, and right wall in yz. The second panel labels the same surfaces as floor, left wall and right wall. Finite patches illustrate infinite coordinate planes; decorative furniture is omitted.
Two schematic views share x and y axes along the floor and z vertically. The floor lies in the xy-plane, left wall in xz, and right wall in yz. The second panel labels the same surfaces as floor, left wall and right wall. Finite patches illustrate infinite coordinate planes; decorative furniture is omitted.
Open original diagram (new tab)

Two directions and one normal

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Left: plane patch contains A and R; position arrows a from O to A and r from O to R, in-plane directions b and c, and displacement AR=lambda b+mu c. Right: a nonzero normal n is perpendicular to in-plane AB, giving (b−a) dot n=0 and r dot n=a dot n.
Left: plane patch contains A and R; position arrows a from O to A and r from O to R, in-plane directions b and c, and displacement AR=lambda b+mu c. Right: a nonzero normal n is perpendicular to in-plane AB, giving (b−a) dot n=0 and r dot n=a dot n.
Left: plane patch contains A and R; position arrows a from O to A and r from O to R, in-plane directions b and c, and displacement AR=lambda b+mu c. Right: a nonzero normal n is perpendicular to in-plane AB, giving (b−a) dot n=0 and r dot n=a dot n.
Open original diagram (new tab)

Method map: One plane, three equivalent forms

Convert the same plane between three forms. Compare the methods, then try a worked case.

  1. Parametric → scalar product

    Check that the spanning directions are independent. Their cross product must be nonzero.

    r=a+su+tv\mathbf r=\mathbf a+s\mathbf u+t\mathbf v

    Find the normalPerpendicular to both directions

    n=u×v≠0\mathbf n=\mathbf u\times\mathbf v\ne\mathbf0. If the cross product is zero, these directions do not span a plane.

    Fix its positionKeep the matching constant

    d=a⋅nd=\mathbf a\cdot\mathbf n, so r⋅n=d\mathbf r\cdot\mathbf n=d. Scaling n\mathbf n requires scaling dd by the same nonzero factor.

    The parameters s and t are independent real numbers. A different point or independent spanning pair can describe the same plane.

  2. Scalar product ↔ Cartesian

    Expand a dot product, or read the Cartesian coefficients as a normal.

    r⋅(nx,ny,nz)=d\mathbf r\cdot(n_x,n_y,n_z)=d

    ExpandCartesian equation

    nxx+nyy+nzz=dn_xx+n_yy+n_zz=d. The normal’s components are the coefficients of x, y and z.

    RecogniseScalar-product equation

    With r=(x,y,z)\mathbf r=(x,y,z) and n=(nx,ny,nz)\mathbf n=(n_x,n_y,n_z), collect the terms as r⋅n=d\mathbf r\cdot\mathbf n=d.

    If all three coefficients are zero, the equation describes all space when d=0, or no points when d≠0; it is not a plane.

  3. Cartesian → parametric

    Solve for a coordinate with a nonzero coefficient and let the other two be free parameters.

    nxx+nyy+nzz=d,n≠0n_xx+n_yy+n_zz=d,\quad\mathbf n\ne\mathbf0

    Choose free coordinatesAvoid division by zero

    If the coefficient of z is zero, solve for x or y instead. A nonzero normal guarantees at least one valid choice.

    Verify the whole familySame plane, new parameters

    Check that the base point satisfies the equation and the two independent directions are perpendicular to the normal. Substitute arbitrary parameters to verify the full family.

    Do not expect to recover the original base point and directions. Equivalent parametrizations need not look identical.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Parametric → scalar product

Start with r=(1,2,3)+s(1,2,1)+t(1,3,4)\mathbf r=(1,2,3)+s(1,2,1)+t(1,3,4). Which vector is perpendicular to both directions? Predict a scalar-product equation before opening the working.

Show reasoning
  1. 1. Find a nonzero normal

    Cross the two independent spanning directions. Reversing the order reverses the normal’s sign, which is equally valid.

    n=(1,2,1)×(1,3,4)=(5,−3,1)\mathbf n=(1,2,1)\times(1,3,4)=(5,-3,1)
  2. 2. Use a point on the plane

    The normal alone does not fix the plane’s position. Use the base point a=(1,2,3)\mathbf a=(1,2,3) to find the constant.

    d=a⋅n=5−6+3=2d=\mathbf a\cdot\mathbf n=5-6+3=2
  3. 3. Write scalar-product form

    Every displacement within the plane is perpendicular to the normal.

    r⋅(5,−3,1)=2\mathbf r\cdot(5,-3,1)=2
Continue with the lesson question

Scalar product → Cartesian

The same plane has r⋅(5,−3,1)=2\mathbf r\cdot(5,-3,1)=2. What equation do the coordinates (x,y,z)(x,y,z) satisfy?

Show reasoning
  1. 1. Expand the dot product

    Write r=(x,y,z)\mathbf r=(x,y,z). The coefficients are the components of the normal, not a point on the plane.

    5x−3y+z=25x-3y+z=2
  2. 2. Check the original point

    Substitute the original base point; keep the constant consistent if the normal is rescaled.

    5(1)−3(2)+3=25(1)-3(2)+3=2
Continue with the lesson question

Cartesian → parametric

Return from 5x−3y+z=25x-3y+z=2 to a parametric plane. Will it need exactly the same point and directions as the starting equation?

Show reasoning
  1. 1. Choose two free coordinates

    The coefficient of z is nonzero. Set x=s and y=t, then solve for z. For another plane, solve for a coordinate with a nonzero coefficient.

    x=s,y=t,z=2−5s+3tx=s,\quad y=t,\quad z=2-5s+3t
  2. 2. Collect the two directions

    The point and directions differ from the original representation, but describe the same plane. Both directions are perpendicular to the original normal.

    r=(0,0,2)+s(1,0,−5)+t(0,1,3)\mathbf r=(0,0,2)+s(1,0,-5)+t(0,1,3)
  3. 3. Verify the complete family

    Substitute arbitrary s,t, not just one point. The free parameters cancel.

    5s−3t+(2−5s+3t)=25s-3t+(2-5s+3t)=2
Continue with the lesson question

What if the directions are dependent?

Illustration: r=(1,2,3)+s(1,2,1)+t(2,4,2)\mathbf r=(1,2,3)+s(1,2,1)+t(2,4,2). Does two-parameter notation guarantee a plane?

Show reasoning
  1. 1. Check independence first

    The second direction is twice the first, so their cross product is zero. This family traces a line, not a plane; do not use the zero vector as a plane normal.

    r=(1,2,3)+(s+2t)(1,2,1)\mathbf r=(1,2,3)+(s+2t)(1,2,1)
Source conversion flow: parametric equation leads to scalar-product form using the cross product of independent directions and d=a dot n. Scalar-product and Cartesian equations convert in both directions by expanding or recognizing the dot product.
Open original diagram (new tab)

Method map: Classify a line and a plane in two checks

First test the direction. Only a zero direction test needs a point test. The line direction and plane normal must both be nonzero.

  1. Test the line direction

    Does the line direction have a nonzero dot product with the plane normal?

    r=a+λm\mathbf r=\mathbf a+\lambda\mathbf m

    r⋅n=d\mathbf r\cdot\mathbf n=d

    Yes · nonzeroExactly one intersection

    When m⋅n≠0\mathbf m\cdot\mathbf n\ne0, solve

    λ=d−a⋅nm⋅n.\lambda=\frac{d-\mathbf a\cdot\mathbf n}{\mathbf m\cdot\mathbf n}.

    Substitute this parameter into the line to find the intersection point.

    No · zeroContinue to the point test

    When m⋅n=0\mathbf m\cdot\mathbf n=0, the line direction is parallel to the plane. This alone does not distinguish containment from a disjoint line.

    Never divide by a zero direction test.

  2. Only if zero: test a line point

    Use this check only when the direction test is zero. Does the line point lie in the plane?

    m⋅n=0\mathbf m\cdot\mathbf n=0

    a⋅n=d\mathbf a\cdot\mathbf n=d

    Yes · equalThe whole line lies in the plane

    a⋅n=d\mathbf a\cdot\mathbf n=d: every parameter gives a plane point, so there are infinitely many common points.

    No · unequalParallel and disjoint

    a⋅n≠d\mathbf a\cdot\mathbf n\ne d: substitution is inconsistent, so there are no common points.

    A point on the plane proves containment only after the direction test has returned zero.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Start with this line

Illustration: predict the number of common points.

x+2y−z=3x+2y-z=3 r=λ(2,−1,1)\mathbf r=\lambda(2,-1,1)

If the line crosses, must the intersection parameter be positive?

Show reasoning
  1. 1. Test the direction

    Read a nonzero normal from the plane’s coefficients. The nonzero dot product guarantees exactly one crossing.

    n=(1,2,−1)m⋅n=2−2−1=−1\begin{gathered}\mathbf n=(1,2,-1)\\\mathbf m\cdot\mathbf n=2-2-1\\=-1\end{gathered}
  2. 2. Find the parameter

    Substitute the line into the plane. A line uses every real parameter; a negative value is valid.

    2λ−2λ−λ=3−λ=3λ=−3\begin{gathered}2\lambda-2\lambda-\lambda=3\\-\lambda=3\\\lambda=-3\end{gathered}
  3. 3. Check the intersection

    Use the parameter in the line, then verify the point in the plane.

    P=(−6,3,−3)−6+6+3=3\begin{gathered}P=(-6,3,-3)\\-6+6+3=3\end{gathered}

Change the direction

Keep the plane and origin. Change only the direction.

x+2y−z=3x+2y-z=3 r=λ(2,−1,0)\mathbf r=\lambda(2,-1,0)

Does a zero dot product prove containment?

Show reasoning
  1. 1. The direction is now parallel

    Read the same normal from the plane. The zero dot product requires a point test. Do not divide by it.

    n=(1,2,−1)m⋅n=2−2=0\begin{gathered}\mathbf n=(1,2,-1)\\\mathbf m\cdot\mathbf n=2-2=0\end{gathered}
  2. 2. Test the base point

    The origin is not in this plane. Since the direction is parallel to the plane, no point of the line can enter it.

    a⋅n=0≠3\mathbf a\cdot\mathbf n=0\ne3
  3. 3. Check the whole line

    Substitution gives a contradiction for every parameter: the line and plane are disjoint.

    2λ−2λ=0≠32\lambda-2\lambda=0\ne3

Move the base point

Keep the plane and direction. Change only the base point.

x+2y−z=3x+2y-z=3 r=(3,0,0)+λ(2,−1,0)\begin{aligned}\mathbf r&=(3,0,0)\\&+\lambda(2,-1,0)\end{aligned}

How many common points now?

Show reasoning
  1. 1. Reuse the direction test

    The plane and direction are unchanged, so the dot product is still zero.

    n=(1,2,−1)m⋅n=0\begin{gathered}\mathbf n=(1,2,-1)\\\mathbf m\cdot\mathbf n=0\end{gathered}
  2. 2. Test the new base point

    This point is in the plane. Together with the zero direction test, that proves containment.

    a⋅n=3\mathbf a\cdot\mathbf n=3
  3. 3. Verify every parameter

    The parameters cancel and the plane equation is always satisfied. There are infinitely many common points.

    (3+2λ)−2λ=3\begin{gathered}(3+2\lambda)-2\lambda\\=3\end{gathered}

Compare the three geometric cases

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Three source schematic roles, also repeated in the printed summary table: a line crosses the plane at P; a parallel line through A is disjoint from the plane; a line through A lies entirely in the plane. Normal n and direction m are shown. Only the first case has m dot n nonzero.
Three source schematic roles, also repeated in the printed summary table: a line crosses the plane at P; a parallel line through A is disjoint from the plane; a line through A lies entirely in the plane. Normal n and direction m are shown. Only the first case has m dot n nonzero.
Three source schematic roles, also repeated in the printed summary table: a line crosses the plane at P; a parallel line through A is disjoint from the plane; a line through A lies entirely in the plane. Normal n and direction m are shown. Only the first case has m dot n nonzero.
Open original diagram (new tab)

Acute and obtuse angles to the normal

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Two schematic source cases share a plane, its normal n, the line and a dashed in-plane projection direction. Left: m points upward along the line, making an acute theta to n; alpha=90 degrees−theta. Right: m is reversed, making theta obtuse; alpha=theta−90 degrees. The line–plane angle stays the same, giving the absolute dot-product sine formula.
Two schematic source cases share a plane, its normal n, the line and a dashed in-plane projection direction. Left: m points upward along the line, making an acute theta to n; alpha=90 degrees−theta. Right: m is reversed, making theta obtuse; alpha=theta−90 degrees. The line–plane angle stays the same, giving the absolute dot-product sine formula.
Two schematic source cases share a plane, its normal n, the line and a dashed in-plane projection direction. Left: m points upward along the line, making an acute theta to n; alpha=90 degrees−theta. Right: m is reversed, making theta obtuse; alpha=theta−90 degrees. The line–plane angle stays the same, giving the absolute dot-product sine formula.
Open original diagram (new tab)

Construct the perpendicular foot

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Source schematic: P is off the plane; a line r=p+lambda n through P meets it at the perpendicular foot F. Its direction is parallel to normal n. A dashed continuation passes beyond F, and a right-angle marker shows perpendicularity.
Source schematic: P is off the plane; a line r=p+lambda n through P meets it at the perpendicular foot F. Its direction is parallel to normal n. A dashed continuation passes beyond F, and a right-angle marker shows perpendicularity.
Source schematic: P is off the plane; a line r=p+lambda n through P meets it at the perpendicular foot F. Its direction is parallel to normal n. A dashed continuation passes beyond F, and a right-angle marker shows perpendicularity.
Open original diagram (new tab)

Projection length and signed displacement

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Four source roles: a perspective plane with A in it and P above its perpendicular foot; the same projection as a side view with height h; a positive-d plane reached from origin O along n; a negative-d plane reached opposite n. Distance uses an absolute value, whereas signed displacement is d/|n|.
Four source roles: a perspective plane with A in it and P above its perpendicular foot; the same projection as a side view with height h; a positive-d plane reached from origin O along n; a negative-d plane reached opposite n. Distance uses an absolute value, whereas signed displacement is d/|n|.
Four source roles: a perspective plane with A in it and P above its perpendicular foot; the same projection as a side view with height h; a positive-d plane reached from origin O along n; a negative-d plane reached opposite n. Distance uses an absolute value, whereas signed displacement is d/|n|.
Open original diagram (new tab)

Parallel planes and a common intersection line

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Left: distinct parallel plane patches with parallel normals n1,n2. Right: nonparallel plane patches meet on a straight line through P. Their normals, called a,b in the source intersection method, are n1,n2 here; the common-line direction is their cross product. Screen positions are schematic.
Left: distinct parallel plane patches with parallel normals n1,n2. Right: nonparallel plane patches meet on a straight line through P. Their normals, called a,b in the source intersection method, are n1,n2 here; the common-line direction is their cross product. Screen positions are schematic.
Left: distinct parallel plane patches with parallel normals n1,n2. Right: nonparallel plane patches meet on a straight line through P. Their normals, called a,b in the source intersection method, are n1,n2 here; the common-line direction is their cross product. Screen positions are schematic.
Open original diagram (new tab)

Plane angles from acute or obtuse normal angles

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Two cross-sections show the same pair of intersecting planes and their perpendicular normals. With an acute normal angle theta, the plane angle alpha equals theta. Reversing one normal makes theta obtuse and alpha=pi−theta. Supplementary/equal angles give the absolute dot-product cosine formula.
Two cross-sections show the same pair of intersecting planes and their perpendicular normals. With an acute normal angle theta, the plane angle alpha equals theta. Reversing one normal makes theta obtuse and alpha=pi−theta. Supplementary/equal angles give the absolute dot-product cosine formula.
Two cross-sections show the same pair of intersecting planes and their perpendicular normals. With an acute normal angle theta, the plane angle alpha equals theta. Reversing one normal makes theta obtuse and alpha=pi−theta. Supplementary/equal angles give the absolute dot-product cosine formula.
Open original diagram (new tab)

Explore

For the plane x+2y+2z=6x+2y+2z=6, compare lines from the origin with directions (2,−1,0)(2,-1,0) and (1,0,0)(1,0,0). Use dot products to predict whether each crosses the plane before substituting. Move the base point of the first line to (6,0,0)(6,0,0) without changing its direction and predict what changes. Explain how a contradiction, an identity and a single parameter solution represent the three geometric outcomes.

Related interactive example

Explore this related example, then return to the prediction task above.

A point and its perpendicular foot

Switch parts of the scene on one at a time and read how each is found. Drag to turn the picture.

Open the interactive view to explore this geometry in 3D.

Open the full interactive problem →

Worked example

The plane x+2y+2z=9x+2y+2z=9 has normal (1,2,2)(1,2,2) with magnitude 3. The origin is distance ∣0−9∣/3=3|0-9|/3=3 from the plane. Its perpendicular foot is (1,2,2)(1,2,2).

The normal vector is perpendicular to the plane; it is not a direction lying in the plane.

Check your understanding

A point is P=(2,1,3)P=(2,1,3) and a plane is Π:x+2y+2z=1\Pi:x+2y+2z=1. Find the perpendicular foot F, the distance from P to the plane and the reflection Q of P in the plane. Verify the plane, direction and midpoint conditions. Explain why writing the plane as −x−2y−2z=−1-x-2y-2z=-1 must give the same results.

Mastery learning objectives

At the end of this chapter, I should be able to:

Formulate and convert plane equations

Formulate a parametric plane equation using a point and two directions parallel to the plane but not to each other.

Formulate a scalar-product plane equation using a point and a nonzero perpendicular normal.

Convert a parametric plane equation to scalar-product form.

Convert scalar-product and Cartesian plane equations in both directions.

Relate a point to a plane

Determine whether a given point lies on a given plane.

Find the position vector of the perpendicular foot from a point to a plane.

Find the position vector of a point reflected about a plane.

Find the perpendicular or shortest distance from a point to a plane.

Relate a line to a plane

Determine and describe the relationship between a line and a plane in three dimensions.

Find their point of intersection when there is a unique intersection.

Calculate the angle between a line and a plane when they intersect at a point.

Relate two planes

Determine and describe the relationship between two distinct planes in three dimensions.

Find the perpendicular distance between parallel planes.

Determine an equation of the intersection line of two intersecting planes using a graphing calculator or analytically.

Find the angle between two planes using their respective normals, taking the acute angle or its boundary value.

Choose a practice question

The source common-question map retains its printed Lecture / Discussion / Review numbering, including “Hence”, “verify” and “last part”. A dash represents a blank source cell. Use the practice list to open the incorporated questions; this map records the source references rather than certifying completion of any unreviewed pages.

Common question typeLecture examplesDiscussion questionsReview problems (follow up)
Equation of a plane: parametric, scalar product, Cartesian28, 29, 30, 42(ii)1, 3(iii), 8(iv), 9(a) & (b), 10(iv)1(iv), 2(ii), 5(ii) & (iv), 6(a)(i) & 6(b)(ii), 7(iii), 9(iii) & (v), 10(iii) & (v), 11(iii)(hence), 12(i), 13(iii)
Line–plane relationship: one crossing, no intersection, or contained line31, 32, 33, 42(vi)2, 3(i), 5(i), 8(ii) & (iii), 9(c), 10(ii)4(iii), 5(i), 6(b)(i), 7(ii), 8(i) & (ii), 10(ii), 11(i), 12(ii), 13(i), 14(iii)
Angle between a line and a plane3423(ii), 9(i), 11(ii), 14(ii), 15(i)
Whether a point lies on a plane35—1(i), 6(a)(i)(verify)
Perpendicular foot from a point to a plane364(i), 10(ii)1(iii), 2(iv), 3(iii), 5(iii), 6(a)(ii), 9(ii)
Perpendicular distance from a point to a plane374(ii)3(i), 8(iii), 13(iv)
Distance between two parallel planes38—8(iv), 10(iv), 13(iv)(hence)
Intersection line of two nonparallel planes39, 406, 7(ii), 8(i)1(ii), 4(ii), 7(i), 9(iv)
Angle between two planes41, 42(iii)6, 7(i), 10(iv)(hence)4(i)
Vectors (Basic): ratio theorem—4(iii)3(iii)(hence), 6(a)(ii)(hence), 9(ii)(hence), 13(ii)
Vectors (Basic): distance using dot or cross product—3(ii), 7(iii), 10(iii)5(iv)(hence), 11(iii), 15(iii)
Vectors (Basic): triangle or parallelogram area——14(v)
Vectors (Lines): vector, Cartesian and parametric line equations—5(ii), 10(i)2(iii), 10(i), 12(last part), 15(ii)
Vectors (Lines): relationship between two lines——2(i), 4(iv), 14(i)
Prior topics: transformations——8(v)

Source reference correction: the area row prints Review 14(v), but Review 14 has only (i)–(iv). The triangle-area task is in (iv), following the reflected-ray construction. The table retains the printed reference; use that final part of Question 14.

Practice questions

Practice section 1 · Questions 1–6

Build a parametric plane from lines or points · Use a point and a normal to write a plane · +4 more

Practice section 2 · Questions 7–12

Find the acute angle between a line and a plane · Check a point by substituting into a plane · +4 more

Practice section 3 · Questions 13–18

Use two plane normals and their common base point · Measure the acute angle using two plane normals · +4 more

Practice section 4 · Questions 19–24

Drop a normal from a point and reflect through its foot · Reflect a line using its fixed intersection and one reflected point · +4 more

Practice section 5 · Questions 25–30

Project a pyramid shadow under two different lighting models · Build an intersection line and a perpendicular plane · +4 more

Practice section 6 · Questions 31–36

Use normals for parallel planes and lines inside a plane · Recover plane constants from an intersection line · +4 more

Practice section 7 · Questions 37–40

Reflect a light ray using the plane normal · Find the locus of a fixed-ratio point between a line and a plane · +2 more

More practice

Use geometry to choose a direction, normal or projection, then verify the resulting point in every required plane. Distinguish a line–plane angle from an angle between normals.

Ready to try this on your own?

For “Vectors: planes, normals and distances”, try a question without the working and explain your method. This records your confidence, saved in this browser only.

Ready to practise?Chapter practice