Vectors: planes, normals and distances

Move within a plane

Choose a point A with position vector $\mathbf a$ and two nonzero, nonparallel directions $\mathbf b,\mathbf c$ parallel to the plane. For any point R in the plane,

$$\begin{aligned}\overrightarrow{AR}&=\lambda\mathbf b+\mu\mathbf c,\\\overrightarrow{OR}-\overrightarrow{OA}&=\lambda\mathbf b+\mu\mathbf c,\\\mathbf r&=\mathbf a+\lambda\mathbf b+\mu\mathbf c,\quad\lambda,\mu\in\mathbb R.\end{aligned}$$

The drawn parallelogram is a finite representation of an infinite plane; any patch shape can be used. Real parameters are not restricted to its edges. Changing the base point or independent directions gives another equation of the same plane. Parallel directions span only a line. The source QR code links to a parametric-plane activity; this source pointer is retained, without claiming a review of the external activity.

Use a normal instead of two directions

For A and B in a plane with nonzero normal $\mathbf n$, the in-plane displacement is perpendicular to $\mathbf n$:

$$\begin{aligned}\overrightarrow{AB}\cdot\mathbf n&=0,\\(\mathbf b-\mathbf a)\cdot\mathbf n&=0,\\\mathbf b\cdot\mathbf n&=\mathbf a\cdot\mathbf n=d.\end{aligned}$$

The same value is obtained at any point R, giving the scalar-product form

$$\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n.$$

For two independent in-plane directions $\mathbf p,\mathbf q$, a normal is $\mathbf p\times\mathbf q$. If the plane passes through the origin, $\mathbf a\cdot\mathbf n=0$. The converse also holds when $\mathbf n\ne\mathbf0$.

The source lists four ways to determine a unique plane: a nonzero normal and a point; two nonparallel directions and a point; two intersecting, nonparallel lines; or three non-collinear points. A zero normal gives no plane constraint, and three collinear points do not determine a unique plane.

For a line-plane angle, first compare the line direction with the plane normal, then take the complementary acute angle. For two planes, compare their normals.

Convert between plane equations

Write $\mathbf r=(x,y,z)$ and the nonzero normal $\mathbf n=(n_1,n_2,n_3)$. Expanding $\mathbf r\cdot\mathbf n=d$ gives

$$n_1x+n_2y+n_3z=d.$$

These are equivalent forms of the same plane; choose the one convenient for the task. The source summary gives the conversion routes:

• Parametric $\mathbf r=\mathbf a+\lambda\mathbf m_1+\mu\mathbf m_2$: find $\mathbf n=\mathbf m_1\times\mathbf m_2$, then $d=\mathbf a\cdot\mathbf n$.
• Scalar product $\mathbf r\cdot\mathbf n=d$: expand the dot product to obtain the Cartesian equation.
• Cartesian $n_1x+n_2y+n_3z=d$: read off $\mathbf n=(n_1,n_2,n_3)$ to recover the scalar-product form.

The directions must be independent and the normal nonzero. Multiplying both the normal and $d$ by the same nonzero scalar preserves the plane. Scaling only one side generally changes it.

Three forms of each coordinate plane

All parameters below are real.

$xy$

Parametric vector form:

$$\mathbf r=\lambda\mathbf i+\mu\mathbf j$$

Scalar-product form:

$$\mathbf r\cdot\mathbf k=0$$

Cartesian form:

$$z=0$$

$yz$

Parametric vector form:

$$\mathbf r=\lambda\mathbf j+\mu\mathbf k$$

Scalar-product form:

$$\mathbf r\cdot\mathbf i=0$$

Cartesian form:

$$x=0$$

$xz$

Parametric vector form:

$$\mathbf r=\lambda\mathbf i+\mu\mathbf k$$

Scalar-product form:

$$\mathbf r\cdot\mathbf j=0$$

Cartesian form:

$$y=0$$

Classify a line and a plane

Test the line direction against the plane normal first. A nonzero dot product gives one intersection. A zero dot product needs a second test: substitute a point from the line to distinguish containment from a disjoint parallel line.

Let the line be $\mathbf r=\mathbf a+\lambda\mathbf m$ and the plane $\mathbf r\cdot\mathbf n=d$, with $\mathbf m\ne\mathbf0$ and $\mathbf n\ne\mathbf0$.

$\mathbf m\cdot\mathbf n\ne0$

Point test: No additional test needed

Relationship: One crossing

Common points: Exactly one

$\mathbf m\cdot\mathbf n=0$

Point test:

$$\mathbf a\cdot\mathbf n\ne d$$

Relationship: Parallel and disjoint

Common points: None

$\mathbf m\cdot\mathbf n=0$

Point test:

$$\mathbf a\cdot\mathbf n=d$$

Relationship: Line lies in plane

Common points: Infinitely many

Solve the one-crossing case

For the one-crossing case, follow the source’s three steps. First write the intersection position as $\overrightarrow{OP}=\mathbf a+\lambda\mathbf m$. Then substitute it into the plane:

$$\begin{aligned}(\mathbf a+\lambda\mathbf m)\cdot\mathbf n&=d,\\\mathbf a\cdot\mathbf n+\lambda(\mathbf m\cdot\mathbf n)&=d,\\\lambda&=\frac{d-\mathbf a\cdot\mathbf n}{\mathbf m\cdot\mathbf n}.\end{aligned}$$

Finally substitute this parameter back into the line to obtain P. The source summary’s last instruction says “find $\Pi$”; the required result is the intersection point P, not the plane.

Interpret a zero direction test

When $\mathbf m\cdot\mathbf n=0$, do not divide by it. Substitution instead gives either a contradiction (no intersection) or an identity (every line point is in the plane). The source uses “parallel” for both an in-plane direction and a disjoint parallel line; the point test distinguishes those cases. A nonparallel line cannot be disjoint from an infinite plane in three-dimensional Euclidean space.

Derive the angle between a line and a plane

The source defines $\alpha$ using the acute angle between the line and its projection in the plane. For nonzero direction $\mathbf m$ and normal $\mathbf n$, first find the directed-vector angle $\theta\in[0,\pi]$:

$$\cos\theta=\frac{\mathbf m\cdot\mathbf n}{|\mathbf m||\mathbf n|}.$$

The source refers back to its earlier vector-angle section 12.2. It gives two cases, each with a complement method and a trigonometric derivation.

Acute $\theta$. Here $\alpha=\pi/2-\theta$. Using $\cos(\pi/2-\alpha)=\sin\alpha$ gives

$$\sin\alpha=\frac{\mathbf m\cdot\mathbf n}{|\mathbf m||\mathbf n|}.$$

Obtuse $\theta$. Here $\alpha=\theta-\pi/2$. Expand the cosine of a sum:

$$\begin{aligned}\cos(\pi/2+\alpha)&=\cos(\pi/2)\cos\alpha-\sin(\pi/2)\sin\alpha\\&=-\sin\alpha.\end{aligned}$$

Thus $-\sin\alpha=(\mathbf m\cdot\mathbf n)/(|\mathbf m||\mathbf n|)$, and the negative dot product must be negated before taking the nonnegative line–plane angle. Both cases combine as

$$\alpha=\sin^{-1}\!\left(\frac{|\mathbf m\cdot\mathbf n|}{|\mathbf m||\mathbf n|}\right).$$

Reversing either vector does not change the geometric line–plane angle. The source’s word “acute” describes its general examples; the formula also includes the boundary cases $\alpha=0$ for a direction parallel to the plane and $\alpha=\pi/2$ for a perpendicular line. In the latter case the projected direction is zero, so the projection-angle description is understood by this perpendicular convention. Neither zero vector defines the required direction or normal.

Explain the three possible intersection equations

For $\Pi:x+2y+2z=6$, classify the lines $l_1:\mathbf r=(0,0,0)+t(2,-1,0)$, $l_2:\mathbf r=(6,0,0)+t(2,-1,0)$ and $l_3:\mathbf r=(0,0,0)+t(1,0,0)$. Predict the outcome using direction and point tests, then interpret the equation obtained by substitution.

After your attempt: compare the reasoning

With normal $\mathbf n=(1,2,2)$, the first two directions have dot product zero with the normal.

The origin is off the plane, so $l_1$ is parallel and disjoint: substitution gives $0=6$.

The base point $(6,0,0)$ is in the plane, so all of $l_2$ lies in it: substitution gives $6=6$.

For $l_3$, the direction dot product is 1, so there is one crossing. Substitution gives $t=6$, at $(6,0,0)$.

An inconsistent equation, an identity and a unique parameter correspond to different geometric situations.

Compare a line angle with its normal angle

A line has direction $\mathbf m=(1,0,\sqrt3)$ and a plane has normal $\mathbf n=(0,0,1)$. Find their angle. Repeat for line directions $(1,0,0)$ and $(0,0,1)$. Explain why blindly reporting the angle to the normal would exchange the two boundary cases.

After your attempt: compare the reasoning

For $(1,0,\sqrt3)$ the normal angle is $30^\circ$, so the line–plane angle is $60^\circ$.

The dot-product cosine gives the angle to the normal; take its complement for the plane.

For $(1,0,0)$ the normal angle is $90^\circ$ but the line–plane angle is $0^\circ$; for $(0,0,1)$ the normal angle is $0^\circ$ but the line–plane angle is $90^\circ$.

Use $\sin\alpha=|\mathbf m\cdot\mathbf n|/(|\mathbf m||\mathbf n|)$ for the line–plane angle.

A direction test alone says nothing about whether a parallel line is contained in the plane.

Check whether a point lies on the plane

For membership, substitute the point into the plane equation. For a perpendicular foot, move from the point along the normal until you reach the plane. For distance, take the length of that displacement; its parameter alone is not a distance.

Follow the source’s two steps: convert to scalar-product form $\mathbf r\cdot\mathbf n=d$ if needed, then substitute the point’s position vector $\mathbf p$. Equality $\mathbf p\cdot\mathbf n=d$ means the point lies on the plane.

Find the perpendicular foot

For a point P and plane $\Pi:\mathbf r\cdot\mathbf n=d$, the source constructs F in five steps:

1. Form the normal line $l_{PF}:\mathbf r=\mathbf p+\lambda\mathbf n$.
2. Identify F as the intersection of this line and the plane.
3. Write $\overrightarrow{OF}=\mathbf p+\lambda\mathbf n$ for its parameter.
4. Impose $(\mathbf p+\lambda\mathbf n)\cdot\mathbf n=d$ and solve for $\lambda$.
5. Substitute this parameter into the line to find F.

Check the foot and its boundary case

Since $\mathbf n\ne\mathbf0$, the denominator $\mathbf n\cdot\mathbf n$ is positive. Equivalently,

$$\lambda=\frac{d-\mathbf p\cdot\mathbf n}{|\mathbf n|^2},\qquad\mathbf f=\mathbf p+\lambda\mathbf n.$$

The source assumes P is off the plane. The same formula also covers P in the plane, when $\lambda=0$ and F=P.

Find the distance by projection or by its foot

Method 1: projection length. Choose any A in the plane and use the unit normal $\widehat{\mathbf n}=\mathbf n/|\mathbf n|$. The source takes the projection of $\overrightarrow{PA}$:

$$\begin{aligned}h&=\left|\overrightarrow{PA}\cdot\widehat{\mathbf n}\right|\\&=\frac{|(\mathbf a-\mathbf p)\cdot\mathbf n|}{|\mathbf n|}\\&=\frac{|d-\mathbf p\cdot\mathbf n|}{|\mathbf n|}.\end{aligned}$$

Its absolute value is a distance, so reversing PA to AP does not change h. The unit-normal hat and absolute-value bars are essential; native extraction can lose them.

When P is the origin O, the source obtains

$$\begin{aligned}h_O&=\left|\overrightarrow{OA}\cdot\widehat{\mathbf n}\right|\\&=\left|\overrightarrow{OA}\cdot\frac{\mathbf n}{|\mathbf n|}\right|\\&=\frac{|d|}{|\mathbf n|}.\end{aligned}$$

The signed perpendicular displacement from O toward the plane along the chosen normal is $d/|\mathbf n|$. The source separates $d>0$ (the plane is reached along $\mathbf n$) and $d<0$ (it is reached in the opposite direction). If $d=0$, the plane passes through O. Reversing the normal and d preserves the plane and distance but reverses this orientation-dependent signed value.

Method 2: perpendicular-foot length. Find F by the preceding five steps, then calculate $h=|\overrightarrow{PF}|$. This is the source’s alternative to projection.

The source remark recommends finding a convenient plane point by observation: $(0,2,0)$ lies on $\mathbf r\cdot(1,2,3)=4$. Any point used for projection must satisfy the plane equation; the base point of an unrelated line need not do so.

Decide how two planes meet

Compare the normals before solving simultaneous equations. Parallel normals require a constant check to distinguish coincident and disjoint planes. Nonparallel normals give an intersection line: find a common point and a direction perpendicular to both normals.

The source considers two distinct planes. If their nonzero normals are parallel, the planes are parallel; otherwise they intersect in a straight line. In symbols, parallel normals satisfy $\mathbf n_1=k\mathbf n_2$ for a nonzero real k. For nonparallel normals no such k exists.

Distinguish coincident planes

If distinctness is not given, parallel normals can also describe the same plane. For $\mathbf r\cdot\mathbf n_1=d_1$ and $\mathbf r\cdot\mathbf n_2=d_2$, when $\mathbf n_1=k\mathbf n_2$ the planes coincide exactly when $d_1=kd_2$. Check both the normal and constant; a normal comparison alone does not prove disjointness.

The source’s two-plane relationship activity is retained as an external pointer from its QR code; external activity content has not been reviewed here.

Separate parallel planes by projection

Their separation is the perpendicular distance between them. Choose one point in each plane and project the joining displacement onto a unit common normal. Alternatively, first write both equations with the same normal $\mathbf n$ and then use

$$h=\frac{|d_1-d_2|}{|\mathbf n|}.$$

Matching the normals requires rescaling the constants too. Coincident planes have zero separation.

Find the line common to nonparallel planes

For planes with normals $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$, the source gives two methods.

Method 1: simultaneous equations using the calculator. Expand to

$$\begin{aligned}a_1x+a_2y+a_3z&=d_1,\\b_1x+b_2y+b_3z&=d_2.\end{aligned}$$

Use the source calculator application PlySmlt2 to express two variables in terms of the third, then read a parametric equation. Choose a free variable that the equations permit; not every coordinate must be a valid free choice in every problem.

Method 2: a common point and cross product. If P is known to belong to both planes, their common line is perpendicular to both normals and therefore has direction $\mathbf m=\mathbf a\times\mathbf b$:

$$\mathbf r=\mathbf p+\lambda\mathbf m,\quad\lambda\in\mathbb R.$$

The source remark requires a point common to both planes, not merely a point in one. Verify that point in both equations. Nonparallel normals ensure $\mathbf m\ne\mathbf0$; parallel normals would give a zero cross product and cannot supply a line direction. The source’s intersection-line activity is retained from its second QR code, without a claim of external-content review.

Derive the acute angle between two planes

Use their normals, not arbitrary in-plane directions. The source refers to the earlier vector-angle formula in section 12.2:

$$\cos\theta=\frac{\mathbf n_1\cdot\mathbf n_2}{|\mathbf n_1||\mathbf n_2|}.$$

When $\theta$ is acute, the plane angle is $\alpha=\theta$. When $\theta$ is obtuse, the smaller plane angle is $\alpha=\pi-\theta$. The second case has two printed methods: take that supplementary angle directly, or use

$$\begin{aligned}\cos(\pi-\alpha)&=\cos\pi\cos\alpha+\sin\pi\sin\alpha\\&=-\cos\alpha.\end{aligned}$$

Hence the obtuse case gives $\alpha=\cos^{-1}[-(\mathbf n_1\cdot\mathbf n_2)/(|\mathbf n_1||\mathbf n_2|)]$. Both orientations combine as

$$\alpha=\cos^{-1}\!\left(\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}\right).$$

The cross-sectional diagrams use equal and supplementary angles to relate the plane angle to the normal angle. Reversing either normal leaves $\alpha$ unchanged. Although the source says “acute”, the formula also covers perpendicular planes at $\pi/2$ and parallel/coincident planes at 0; both normals must remain nonzero. This is a cosine formula for two planes, whereas a line–plane angle uses the sine formula derived earlier.

Use a common line to meet a third plane

Find the common line of $\Pi_1:x+y=2$ and $\Pi_2:y+z=3$. Verify a point and direction for your answer in both planes. Then decide whether that line meets $\Pi_3:x+z=5$, and find the point if it does.

After your attempt: compare the reasoning

Choose $y=t$. Then $x=2-t$ and $z=3-t$, so the common line is

$\mathbf r=(2,0,3)+t(-1,1,-1).$

The base point satisfies both plane equations. The direction is perpendicular to both normals $(1,1,0)$ and $(0,1,1)$; their cross product is $(1,-1,1)$, the negative of our equally valid direction. Substitution into the third plane gives $5-2t=5$, so $t=0$ and the intersection is $(2,0,3)$. The sequential method first finds a line from two planes, then intersects that line with the third plane.

Review the source mastery objectives twice

The source provides separate blank self-assessment columns headed “At the end of lecture” and “At the end of tutorial”.

Revisit every objective at both checkpoints; a completed lesson is not by itself evidence of mastery.

Reflection of a point is explicitly included in the original objectives: find its perpendicular foot F and use the image position $2\mathbf f-\mathbf a$.

The source describes the plane angle by its normals; choose the smaller angle as derived above.