Vector Lines, Planes, and 3D Geometry

Overview

This branch develops the Vector II layer of the topic. The main shift is from vectors as movements to vectors as a language for describing geometric objects in three dimensions.

The central idea is:

a line or plane is a set of points generated by directions or selected by constraints.

Lines in 3D

A line in 3D is determined by one point and one direction.

If $\mathbf{a}$ is the position vector of a fixed point on the line and $\mathbf{m}$ is a direction vector, then

$$\boxed{\mathbf{r}=\mathbf{a}+\lambda \mathbf{m},\lambda \in \mathbb{R}}$$

Here $\mathbf{r}$ is the position vector of a general point on the line.

Caption: A point on a line is reached by starting from a fixed point and moving along one direction vector.

Line Through Two Points

If a line passes through $A$ and $B$, then a direction vector is

$$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}.$$

Hence a vector equation is

$$\mathbf{r}=\overrightarrow{OA}+\lambda \overrightarrow{AB}.$$

Caption: For a line through $A$ and $B$, first form the displacement $\overrightarrow{AB}$, then use it as the direction vector of the line.

The same line can be written in many ways. You may choose a different fixed point or a non-zero scalar multiple of the direction vector.

Cartesian Form of a Line

From

$$\mathbf{r}=\left(\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right)+\lambda \left(\begin{array}{c}l\\ m\\ n\end{array}\right),$$

we get

$$x=x_0+\lambda l,y=y_0+\lambda m,z=z_0+\lambda n.$$

When $l,m,n$ are non-zero, this can be written as

$$\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}.$$

Use the parametric form when a component of the direction vector is zero.

Relationships Between Lines

For two lines

$${\ell }_1:\mathbf{r}=\mathbf{a}+\lambda \mathbf{m},{\ell }_2:\mathbf{r}=\mathbf{b}+\mu \mathbf{n},$$

start by comparing $\mathbf{m}$ and $\mathbf{n}$.

Caption: Line relationships in 3D are classified by direction first, then by whether the two line equations share a point.

Parallel or Identical

If $\mathbf{m}$ and $\mathbf{n}$ are scalar multiples, the lines have parallel directions.

Equivalently, for non-zero 3D direction vectors, you may test

$$\mathbf{m}\times \mathbf{n}=0.$$

This works because the vector product has magnitude

$$|\mathbf{m}\times \mathbf{n}|=|\mathbf{m}||\mathbf{n}|\sin \theta ,$$

so it is zero when the direction vectors are parallel or anti-parallel.

This test only tells you that the directions are parallel. It does not yet tell you whether the two lines are the same line.

Then test whether a point from one line lies on the other line.

• If yes, the lines are identical.
• If no, the lines are distinct and parallel.

Intersecting or Skew

If $\mathbf{m}$ and $\mathbf{n}$ are not scalar multiples, solve

$$\mathbf{a}+\lambda \mathbf{m}=\mathbf{b}+\mu \mathbf{n}.$$

• If a consistent pair $(\lambda ,\mu )$ exists, the lines intersect.
• If the equations are inconsistent, the lines are skew.

Skew lines are possible only in 3D geometry. They are non-parallel lines that do not meet.

Planes in 3D

A plane can be described in two main ways.

Parametric Form

If a plane passes through a fixed point with position vector $\mathbf{a}$ and has two non-parallel direction vectors ${\mathbf{m}}_1$ and ${\mathbf{m}}_2$, then

$$\boxed{\mathbf{r}=\mathbf{a}+\lambda {\mathbf{m}}_1+\mu {\mathbf{m}}_2,\lambda ,\mu \in \mathbb{R}}$$

The two directions must be non-parallel. Otherwise the expression generates only a line.

Normal Form

If $\mathbf{n}$ is perpendicular to the plane and ${\mathbf{r}}_0$ is the position vector of a fixed point on the plane, then

$$\boxed{(\mathbf{r}-{\mathbf{r}}_0)\cdot \mathbf{n}=0}$$

This says that the vector from the fixed point to any point on the plane lies inside the plane, so it is perpendicular to the normal vector.

Equivalently,

$$\boxed{\mathbf{r}\cdot \mathbf{n}=D}$$

for some constant $D$.

Caption: Parametric form generates points using in-plane directions; normal form tests membership using a perpendicular direction.

Cartesian Plane Equation

If

$$\mathbf{r}=\left(\begin{array}{c}x\\ y\\ z\end{array}\right),\mathbf{n}=\left(\begin{array}{c}a\\ b\\ c\end{array}\right),$$

then

$$\mathbf{r}\cdot \mathbf{n}=D$$

becomes

$$ax+by+cz=D.$$

The vector

$$\left(\begin{array}{c}a\\ b\\ c\end{array}\right)$$

is a normal vector to the plane.

Relationships Involving Planes

Line and Plane

For

$$\ell :\mathbf{r}=\mathbf{a}+\lambda \mathbf{m},\Pi :\mathbf{r}\cdot \mathbf{n}=D,$$

the key test is

$$\mathbf{m}\cdot \mathbf{n}.$$

This test compares the direction of the line with the normal direction of the plane.

• If $\mathbf{m}\cdot \mathbf{n}\ne 0$, then $\mathbf{m}$ has a component in the normal direction. The line is not parallel to the plane, so it must cut the plane at exactly one point.
• If $\mathbf{m}\cdot \mathbf{n}=0$, then $\mathbf{m}$ is perpendicular to $\mathbf{n}$. This means the line direction is parallel to the plane, but it does not yet tell us whether the line is outside the plane or contained in it.

Caption: A line can cross a plane, be parallel to it, or lie entirely on it; the direction test and point test separate the cases.

To distinguish the last two cases, check whether the fixed point on the line lies on the plane:

$$\mathbf{a}\cdot \mathbf{n}=D.$$

The full decision structure is:

• If $\mathbf{m}\cdot \mathbf{n}\ne 0$, the line intersects the plane at one point.
• If $\mathbf{m}\cdot \mathbf{n}=0$ and $\mathbf{a}\cdot \mathbf{n}\ne D$, the line is parallel to the plane but not on it.
• If $\mathbf{m}\cdot \mathbf{n}=0$ and $\mathbf{a}\cdot \mathbf{n}=D$, the line lies entirely on the plane.

This is why the point test matters. The condition $\mathbf{m}\cdot \mathbf{n}=0$ only says that the line direction is parallel to the plane; it does not decide whether the line itself is inside the plane.

If the line intersects the plane, the intersection point is found by substituting

$$\mathbf{r}=\mathbf{a}+\lambda \mathbf{m}$$

into

$$\mathbf{r}\cdot \mathbf{n}=D$$

and solving for $\lambda$.

Two Planes

For

$${\Pi }_1:\mathbf{r}\cdot {\mathbf{n}}_1=D_1,{\Pi }_2:\mathbf{r}\cdot {\mathbf{n}}_2=D_2,$$

compare the normal vectors.

The normal vector controls the orientation of a plane. Therefore, the first question is not whether the two equations look similar, but whether their normal vectors point in the same direction.

• If ${\mathbf{n}}_1$ and ${\mathbf{n}}_2$ are not parallel, the planes are not parallel. In 3D, two non-parallel planes meet in a line.
• If ${\mathbf{n}}_1$ and ${\mathbf{n}}_2$ are parallel, the planes have the same orientation. They may be distinct parallel planes, or they may be the same plane written in different forms.

Caption: Two non-parallel normal vectors give intersecting planes; parallel normal vectors give parallel or identical planes.

To test whether the normal vectors are parallel, check whether one is a scalar multiple of the other:

$${\mathbf{n}}_2=k{\mathbf{n}}_1$$

for some non-zero scalar $k$. Equivalently, in three dimensions,

$${\mathbf{n}}_1\times {\mathbf{n}}_2=0.$$

The vector-product test is useful when the scalar multiple is not immediately obvious.

If the normal vectors are parallel, compare the whole plane equations consistently. For example, if

$${\mathbf{n}}_2=k{\mathbf{n}}_1,$$

then compare whether

$$D_2=k{D}_1.$$

• If $D_2=k{D}_1$, the two equations represent the same plane.
• If $D_2\ne k{D}_1$, the planes are distinct and parallel.

This constant check is essential. It is not enough to say that the normals are parallel, because parallel normals only tell us that the planes have the same orientation.

If the normal vectors are not parallel, the planes intersect in a line. To find the line of intersection, solve the two plane equations simultaneously. There will be one free parameter, because the common set is a line rather than a single point.

A practical workflow is:

1. Compare ${\mathbf{n}}_1$ and ${\mathbf{n}}_2$.
2. If they are parallel, compare the constants using the same scalar multiple.
3. If they are not parallel, solve the two plane equations and express the common points using a parameter.

Example:

$${\Pi }_1:x+2y-z=3,{\Pi }_2:2x+4y-2z=6$$

represent the same plane, because the second equation is exactly twice the first.

But

$${\Pi }_1:x+2y-z=3,{\Pi }_2:2x+4y-2z=7$$

represent distinct parallel planes, because the left side has been multiplied by $2$ but the constant has not.

For non-parallel planes, such as

$${\Pi }_1:x+y+z=4,{\Pi }_2:x-y+z=2,$$

the normals are not scalar multiples. Subtracting the equations gives $2y=2$, so $y=1$. Substituting into ${\Pi }_1$ gives $x+z=3$. Letting $z=\lambda$ gives the line

$$\mathbf{r}=\left(\begin{array}{c}3\\ 1\\ 0\end{array}\right)+\lambda \left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right).$$

So the intersection of two non-parallel planes is not just a visual line; it is the set of all points satisfying both plane equations.

Three Planes

Three-plane problems are simultaneous-equation problems in geometric language. Each plane gives one linear equation in $x,y,z$, and the question is asking for the common solution set of all three equations.

For

$${\Pi }_1:a_{1}x+b_{1}y+c_{1}z=d_1,{\Pi }_2:a_{2}x+b_{2}y+c_{2}z=d_2,{\Pi }_3:a_{3}x+b_{3}y+c_{3}z=d_3,$$

the coefficient vectors

$$\left(\begin{array}{c}{a}_1\\ b_1\\ c_1\end{array}\right),\left(\begin{array}{c}{a}_2\\ b_2\\ c_2\end{array}\right),\left(\begin{array}{c}{a}_3\\ b_3\\ c_3\end{array}\right)$$

are the normal vectors of the three planes. The algebra and geometry should agree:

• A unique solution means the three planes meet at one point.
• One free parameter means the common solution set is a line.
• Two free parameters means the common solution set is a plane, usually because the equations are dependent.
• An inconsistent system means there is no common point.

This is why row reduction or simultaneous solving is not just algebraic manipulation. It is a way of identifying the dimension of the intersection.

The safest workflow is:

1. Write the three plane equations in standard linear form.
2. Solve the simultaneous system systematically.
3. Watch for contradictions, such as $0=5$.
4. Count the number of free parameters if the system is consistent.
5. Translate the algebraic result back into geometry.

The common outcomes are:

Algebraic result	Geometry
unique solution $(x,y,z)$	three planes meet at one point
one parameter remains	three planes share a common line
two parameters remain	the three equations describe the same plane, or reduce to one plane condition
contradiction occurs	no common point

Caption: Solving three plane equations reveals whether the common solution set has zero, one, or two free parameters, or is empty.

Examples of what the algebra is telling you:

• If solving gives $x=1$, $y=2$, $z=3$, the planes meet at the single point $(1,2,3)$.
• If solving gives $x=2-\lambda$, $y=1+\lambda$, $z=\lambda$, the common set is a line.
• If the system reduces to only $x+2y-z=3$, then the common set is a plane.
• If the system produces $0=4$, the three planes have no common point.

Be careful with the phrase “the planes intersect pairwise”. Three planes may intersect two at a time without sharing one common point. The question is usually asking whether there is a point, line, or plane common to all three planes.

For exam work, the conclusion should be stated in both forms:

• algebraically, by giving the point, line equation, plane equation, or contradiction
• geometrically, by saying how the three planes are arranged

For example, if the solution is

$$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}2\\ 1\\ 0\end{array}\right)+\lambda \left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right),\lambda \in \mathbb{R},$$

then the geometric conclusion is that the three planes share a common line, not merely that “there are infinitely many solutions”.

Angles

Angle Between Two Lines

If the lines have direction vectors ${\mathbf{m}}_1$ and ${\mathbf{m}}_2$, then the acute angle $\theta$ between them satisfies

$$\cos \theta =\frac{|{\mathbf{m}}_1\cdot {\mathbf{m}}_2|}{|{\mathbf{m}}_1||{\mathbf{m}}_2|}.$$

Caption: The angle between two lines is computed from their direction vectors; the absolute value gives the acute angle.

The position of the lines does not affect the angle. Only their directions matter. This is why the formula uses direction vectors rather than position vectors of points on the lines.

The absolute value appears because a line has no preferred orientation. Reversing a direction vector changes ${\mathbf{m}}_1\cdot {\mathbf{m}}_2$ to its negative, but it does not change the geometric line. Therefore, H2 questions usually ask for the acute angle between two lines:

$$0^{\circ }\le \theta \le 90^{\circ }.$$

If a question asks for the obtuse angle instead, find the acute angle first and subtract from $180^{\circ }$.

A reliable workflow is:

1. extract one direction vector from each line
2. compute ${\mathbf{m}}_1\cdot {\mathbf{m}}_2$
3. compute $|{\mathbf{m}}_1|$ and $|{\mathbf{m}}_2|$
4. substitute into the cosine formula
5. check whether the question wants the acute or obtuse angle

Common traps:

• Do not use a point vector such as $\mathbf{a}$ as the line direction.
• Do not forget the absolute value if the acute angle is required.
• Do not assume two skew lines cannot have an angle. Their angle is still defined by their direction vectors.

Angle Between Two Planes

The angle between two planes is the angle between their normal vectors:

$$\cos \theta =\frac{|{\mathbf{n}}_1\cdot {\mathbf{n}}_2|}{|{\mathbf{n}}_1||{\mathbf{n}}_2|}.$$

Caption: The angle between two planes is found from their normal vectors; the side view shows why the acute plane angle matches the acute angle between normals.

This formula works because the orientation of a plane is controlled by its normal vector. Instead of trying to measure the angle between two slanted surfaces directly, compare the directions perpendicular to those surfaces.

For a plane written as

$$ax+by+cz=D,$$

a normal vector is

$$\mathbf{n}=\left(\begin{array}{c}a\\ b\\ c\end{array}\right).$$

So the calculation is usually:

1. extract the normal vector from each plane equation
2. compute ${\mathbf{n}}_1\cdot {\mathbf{n}}_2$
3. compute $|{\mathbf{n}}_1|$ and $|{\mathbf{n}}_2|$
4. use the cosine formula
5. take the acute angle unless the question explicitly asks otherwise

The absolute value has the same role as in the angle-between-lines formula. A normal vector can be reversed without changing the plane, so $\mathbf{n}$ and $-\mathbf{n}$ describe the same plane orientation. The acute angle between planes should not depend on which normal direction was chosen.

If the angle between the normal vectors is obtuse, the acute angle between the planes is its supplement:

$${\theta }_{\text{planes}}=180^{\circ }-{\theta }_{\text{normals}}.$$

Using

$$\frac{|{\mathbf{n}}_1\cdot {\mathbf{n}}_2|}{|{\mathbf{n}}_1||{\mathbf{n}}_2|}$$

builds this acute-angle convention directly into the calculation.

Common traps:

• Do not use direction vectors lying in the planes unless they are chosen very carefully.
• Do not confuse the line of intersection with a normal vector.
• Do not forget that parallel planes have angle $0^{\circ }$, because their normal vectors are parallel.
• Do not report an obtuse angle when the question asks for the angle between two planes in the usual acute sense.

Angle Between a Line and a Plane

The angle between a line and a plane is the angle between the line and its projection onto the plane. In the figure, the line meets the plane at $P$, the point $Q$ lies on the line, and $N$ is the projection of $Q$ onto the plane. The angle $\theta$ is the angle between $PQ$ and $PN$.

Caption: The line-plane angle is measured between the line and its projection on the plane; the angle with the normal is complementary.

This definition matters because the plane contains infinitely many lines through $P$. The projection direction gives the smallest angle between the given line and any line lying in the plane.

If the line has direction vector $\mathbf{m}$ and the plane has normal vector $\mathbf{n}$, then the normal vector is perpendicular to the plane. Therefore, the angle between $\mathbf{m}$ and $\mathbf{n}$ is the complement of the line-plane angle:

$$\text{angle between }\mathbf{m}\text{ and }\mathbf{n}=90^{\circ }-\theta .$$

Using the dot product on $\mathbf{m}$ and $\mathbf{n}$ gives

$$\cos (90^{\circ }-\theta )=\frac{|\mathbf{m}\cdot \mathbf{n}|}{|\mathbf{m}||\mathbf{n}|}.$$

Since $\cos (90^{\circ }-\theta )=\sin \theta$, the working formula is

$$\boxed{\sin \theta =\frac{|\mathbf{m}\cdot \mathbf{n}|}{|\mathbf{m}||\mathbf{n}|}}$$

A reliable workflow is:

1. identify the line direction vector $\mathbf{m}$
2. identify a normal vector $\mathbf{n}$ to the plane
3. compute $\mathbf{m}\cdot \mathbf{n}$
4. substitute into the sine formula
5. report the acute line-plane angle $\theta$

For example, if

$$\ell :\mathbf{r}=\mathbf{a}+\lambda \mathbf{m}$$

and

$$\Pi :ax+by+cz=D,$$

then use

$$\mathbf{m}\text{and}\mathbf{n}=\left(\begin{array}{c}a\\ b\\ c\end{array}\right).$$

Special cases are useful checks:

• If $\mathbf{m}\cdot \mathbf{n}=0$, the line direction is parallel to the plane, so the line-plane angle is $0^{\circ }$.
• If $\mathbf{m}$ is parallel to $\mathbf{n}$, the line is perpendicular to the plane, so the line-plane angle is $90^{\circ }$.

Common traps:

• Do not use the cosine formula directly with $\mathbf{m}$ and $\mathbf{n}$ as the final answer unless you want the angle between the line and the normal.
• Do not confuse a line lying in the plane with a line merely parallel to the plane; both have line-plane angle $0^{\circ }$, but their positions are different.
• Do not forget the absolute value when the acute angle is required.

Distances

Distances in vector geometry are perpendicular distances.

The useful first question is always: perpendicular to what?

• For point-to-plane distance, use the plane normal direction.
• For point-to-line distance, use the perpendicular component away from the line direction.

Point to Plane

A point-to-plane distance is the shortest distance from a point to a plane. The shortest path is perpendicular to the plane, so the normal vector is the key object.

Suppose the plane is

$$\mathbf{r}\cdot \mathbf{n}=D$$

and the external point $P$ has position vector $\mathbf{p}$. Let $A$ be any point on the plane, with position vector $\mathbf{a}$. Then

$$\overrightarrow{AP}=\mathbf{p}-\mathbf{a}$$

is a vector from the plane to the point.

Caption: The point-plane distance is the perpendicular component of a point-to-plane vector in the normal direction.

The vector $\overrightarrow{AP}$ is usually not perpendicular to the plane. It can be split into:

• a component parallel to the plane
• a component perpendicular to the plane

Only the perpendicular component gives the distance. In the figure, $H$ is the foot of the perpendicular from $P$ to the plane, so the distance is $PH$.

If $\hat{\mathbf{n}}$ is a unit normal vector, then the perpendicular component of $\overrightarrow{AP}$ in the normal direction is

$$d=|\overrightarrow{AP}\cdot \hat{\mathbf{n}}|.$$

But the plane equation usually gives a normal vector $\mathbf{n}$ that is not necessarily a unit vector. Since

$$\hat{\mathbf{n}}=\frac{\mathbf{n}}{|\mathbf{n}|},$$

we have

$$d=\frac{|\overrightarrow{AP}\cdot \mathbf{n}|}{|\mathbf{n}|}.$$

Now use the fact that $A$ lies on the plane. Therefore

$$\mathbf{a}\cdot \mathbf{n}=D.$$

So

$$(\mathbf{p}-\mathbf{a})\cdot \mathbf{n}=\mathbf{p}\cdot \mathbf{n}-D.$$

Hence the standard formula is

$$\boxed{d=\frac{|\mathbf{p}\cdot \mathbf{n}-D|}{|\mathbf{n}|}}$$

A reliable workflow is:

1. write the plane in the form $\mathbf{r}\cdot \mathbf{n}=D$
2. identify the point position vector $\mathbf{p}$
3. compute $\mathbf{p}\cdot \mathbf{n}-D$
4. take the absolute value
5. divide by $|\mathbf{n}|$

Common traps:

• Do not use the length of $\mathbf{p}-\mathbf{a}$ unless that vector is perpendicular to the plane.
• Do not forget to divide by $|\mathbf{n}|$ if $\mathbf{n}$ is not a unit vector.
• Do not omit the absolute value; distance must be non-negative.
• If the plane is written as $ax+by+cz+d=0$, rewrite it as $ax+by+cz=-d$ or use the equivalent formula carefully.

Point to Line

A point-to-line distance is the shortest distance from a point to a line. The shortest segment must be perpendicular to the line.

For a line

$$\mathbf{r}=\mathbf{a}+\lambda \mathbf{m}$$

and a point $P$ with position vector $\mathbf{p}$, the vector

$$\overrightarrow{AP}=\mathbf{p}-\mathbf{a}$$

joins a known point $A$ on the line to the point $P$.

Caption: The point-line distance is the perpendicular height from the point to the line.

In the figure, $H$ is the foot of the perpendicular from $P$ to the line. The required distance is $PH$, not the length of the slanted vector $\overrightarrow{AP}$.

There are two common ways to compute this distance.

Method 1: subtract the parallel component.

1. Form $\overrightarrow{AP}$, or equivalently $\mathbf{p}-\mathbf{a}$.
2. Project $\overrightarrow{AP}$ onto the line direction $\mathbf{m}$.
3. Subtract that parallel component.
4. Take the magnitude of the remaining perpendicular component.

Method 2: use the vector product area formula.

The magnitude

$$\left|(\mathbf{p}-\mathbf{a})\times \mathbf{m}\right|$$

is the area of the parallelogram formed by $\overrightarrow{AP}$ and $\mathbf{m}$. The same area is also

$$\text{base}\times \text{height}=|\mathbf{m}|d.$$

Therefore

$$\boxed{d=\frac{\left|(\mathbf{p}-\mathbf{a})\times \mathbf{m}\right|}{|\mathbf{m}|}}$$

Equivalently,

$$d=\frac{\left|\overrightarrow{AP}\times \mathbf{m}\right|}{|\mathbf{m}|}.$$

This formula works in 3D because the vector product measures the area generated by two vectors. It is often faster than explicitly finding the foot $H$.

Common traps:

• Do not use $|\mathbf{p}-\mathbf{a}|$ unless $\mathbf{p}-\mathbf{a}$ is already perpendicular to the line.
• Do not forget to divide by $|\mathbf{m}|$; the vector product gives area, not height.
• Any point $A$ on the line may be used, but $\mathbf{m}$ must be a direction vector of the line.
• If the question asks for the foot $H$ as well as the distance, use the projection method rather than only the vector-product formula.

Enrichment: Skew-Line Distance

The current 9758 summary excludes formal shortest-distance calculations between skew lines. If this appears in older notes, treat it as enrichment rather than core revision material.

Conceptually, the shortest distance between skew lines is along a direction perpendicular to both line directions, often found using a vector product.

Core Examples

Example 1: Line Through Two Points

Let $A(1,2,3)$ and $B(4,0,5)$. Then

$$\overrightarrow{AB}=\left(\begin{array}{c}4\\ 0\\ 5\end{array}\right)-\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)=\left(\begin{array}{c}3\\ -2\\ 2\end{array}\right).$$

A vector equation of the line is

$$\mathbf{r}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)+\lambda \left(\begin{array}{c}3\\ -2\\ 2\end{array}\right).$$

Example 2: Plane Through a Point and Two Directions

If a plane passes through

$$\mathbf{a}=\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right)$$

and is parallel to

$${\mathbf{m}}_1=\left(\begin{array}{c}2\\ 1\\ 0\end{array}\right),{\mathbf{m}}_2=\left(\begin{array}{c}0\\ 1\\ 3\end{array}\right),$$

then

$$\mathbf{r}=\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right)+\lambda \left(\begin{array}{c}2\\ 1\\ 0\end{array}\right)+\mu \left(\begin{array}{c}0\\ 1\\ 3\end{array}\right).$$

Example 3: Line-Plane Intersection Test

For

$$\mathbf{r}=\left(\begin{array}{c}1\\ 2\\ 0\end{array}\right)+\lambda \left(\begin{array}{c}1\\ -1\\ 2\end{array}\right)$$

and

$$\mathbf{r}\cdot \left(\begin{array}{c}2\\ 1\\ -1\end{array}\right)=4,$$

we compute

$$\left(\begin{array}{c}1\\ -1\\ 2\end{array}\right)\cdot \left(\begin{array}{c}2\\ 1\\ -1\end{array}\right)=-1.$$

Since this is non-zero, the line intersects the plane at exactly one point.

Common Errors

• Treating the vector equation of a line as unique.
• Using two parallel direction vectors to define a plane.
• Confusing a direction vector in a plane with a normal vector to the plane.
• Saying a line is parallel to a plane from $\mathbf{m}\cdot \mathbf{n}=0$ without checking whether it lies in the plane.
• Forgetting that two non-parallel lines in 3D may be skew.
• Thinking two non-parallel planes meet at one point. They meet in a line.
• Using the angle between a line and a plane normal when the question asks for the angle between the line and the plane.
• Treating distance formulas as arbitrary rather than as perpendicular projections.

Revision Checklist

• Can I explain why one parameter generates a line?
• Can I explain why two non-parallel parameters generate a plane?
• Can I convert between parametric and normal forms of a plane?
• Can I read a normal vector directly from $ax+by+cz=D$?
• Can I classify two lines as parallel, identical, intersecting, or skew?
• Can I classify a line and a plane using both the direction test and the point test?
• Can I classify two planes by comparing their normal vectors?
• Can I choose the correct vectors for line-line, plane-plane, and line-plane angles?
• Can I explain distance formulas using perpendicular components or projections?

Related

• Vectors
• Vector Foundations and Products