Linear Algebra and Vector Geometry 201-SN4-RE • CEGEP, Montreal

Revision sheet: lines and planes in space (201-SN4-RE)

This sheet does not repeat the distance formulas: they are in your notes. It deals with what decides the mark: recognizing in one second whether a vector is a direction vector or a normal, knowing which test separates two relative positions, and writing up an intersection the way a 201-SN4-RE grader expects it.

It is written for CEGEP students in the Science program, and for Social Science students in the mathematics profile who take linear algebra. Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A line is given by a POINT and a DIRECTION, a plane by a POINT and a NORMAL. Almost every mark lost in this chapter comes from using the normal vector as if it were a direction vector, or the other way round.

This chapter is part of 201-SN4-RE, Linear Algebra and Vector Geometry

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Direction vector
Normal vector
Parametric equations of a line
Symmetric equations
Scalar equation of a plane
Skew lines
Line-plane relative position
Point-to-plane distance
Point-to-line distance
Angle between two planes
Orthogonal projection

The essentials

Four ways of writing, and what each one tells you

• LINE, vector form: $(x, y, z)=(x_{0}, y_{0}, z_{0})+t\,\vec{u}$ . The point comes first, the DIRECTION is the coefficient of $t$ .
• LINE, parametric form: $x=x_{0}+at$ , $y=y_{0}+bt$ , $z=z_{0}+ct$ . One parameter per line, and never the same one for two different lines.
• LINE, symmetric form: $\dfrac{x-x_{0}}{a}=\dfrac{y-y_{0}}{b}=\dfrac{z-z_{0}}{c}$ . The POINT is in the numerators, with its sign changed; the DIRECTION is in the denominators.
• PLANE: $a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$ , expanded into $ax+by+cz+d=0$ . The three coefficients $(a, b, c)$ form the NORMAL vector, never a direction vector.
• A normal to a plane defined by three points is given by $\vec{n}=\vec{AB}\times\vec{AC}$ .

A line is read from a point

A

and a direction vector

\vec{u}

lying ALONG it; a plane from a point

A

and a normal vector

\vec{n}

PERPENDICULAR to it. The same triple never plays both roles.

The reflex that saves you: faced with the equation of a plane, write “ $\vec{n}=(a, b, c)$ ” in the margin right away. Faced with a line, write “ $A(\dots)$ and $\vec{u}=(\dots)$ ”. The rest of the chapter is computed from these two lines.

Relative positions, and the test that decides between them

• LINE AND PLANE: if $\vec{u}\cdot\vec{n}\neq 0$ , the line CROSSES the plane at a single point. If $\vec{u}\cdot\vec{n}=0$ , it is parallel, and it remains to know whether it lies in the plane.
• To decide between strictly parallel and contained: substitute ONE point of the line into the equation of the plane. If it checks, the line is contained in the plane; if not, it is strictly parallel.
• TWO LINES: $\vec{u}$ and $\vec{v}$ parallel give parallel or coincident lines; not parallel, they leave intersecting or SKEW lines.
• The coplanarity test: $\vec{AC}\cdot(\vec{u}\times\vec{v})=0$ with $A$ on the first line and $C$ on the second. Zero, the lines are coplanar, hence intersecting; non-zero, they are skew.
• TWO PLANES: parallel normals give parallel or coincident planes; otherwise the intersection is a LINE, with direction vector $\vec{n_{1}}\times\vec{n_{2}}$ .

The only three positions of a line and a plane. The dot product

\vec{u}\cdot\vec{n}

separates the first from the other two; testing one point of the line separates the last two.

Two planes in space never meet at a single point: their intersection is empty, a line, or the whole plane. An answer “they meet at $(1, 2, 3)$ ” is wrong on sight.

Distances and angles, and what tells them apart

• Point $S$ to the plane $ax+by+cz+d=0$ : $\text{dist}=\dfrac{|ax_{S}+by_{S}+cz_{S}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$ . The equation must first be brought to the form $=0$ .
• Point $S$ to the line through $A$ with direction $\vec{u}$ : $\text{dist}=\dfrac{\|\vec{AS}\times\vec{u}\|}{\|\vec{u}\|}$ .
• Between two SKEW lines: $\text{dist}=\dfrac{|\vec{AC}\cdot(\vec{u}\times\vec{v})|}{\|\vec{u}\times\vec{v}\|}$ , that is, the scalar triple product divided by the norm of the cross product.
• Angle between two planes, or between two lines: $\cos\theta=\dfrac{|\vec{n_{1}}\cdot\vec{n_{2}}|}{\|\vec{n_{1}}\|\,\|\vec{n_{2}}\|}$ , with a COSINE.
• Angle between a line and a plane: $\sin\theta=\dfrac{|\vec{u}\cdot\vec{n}|}{\|\vec{u}\|\,\|\vec{n}\|}$ , with a SINE, because the normal is perpendicular to the plane.

Each distance is the length of a PERPENDICULAR segment, in orange: to the plane, to the line, or to both skew lines (the common perpendicular). Any other segment is longer.

All these formulas carry an absolute value or a norm in the numerator: a distance and an angle between lines are always positive, and a negative result signals an error, never a discovery.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Reading the coefficients of a plane as a direction vector

2 marks, and the rest of the problem, since the vector used is perpendicular to the one that was needed.

What not to write

“The plane $2x-y+3z-5=0$ has direction vector $(2, -1, 3)$ .”

What to write

“ $(2, -1, 3)$ is the NORMAL vector of the plane, so it is perpendicular to it. A direction vector of the plane is any vector orthogonal to $\vec{n}$ , for example $(1, 2, 0)$ , since $2-2+0=0$ .”

The coefficients of the plane

2x-y+3z-5=0

give

\vec{n}

, in orange, PERPENDICULAR to the plane. A direction vector of the plane lies in it, in green:

(1, 2, 0)

, orthogonal to

\vec{n}

Why: The equation of a plane says “the vector joining $A$ to $M$ is orthogonal to $\vec{n}$ ”. The coefficients therefore describe the direction FORBIDDEN to the plane, not a direction of the plane.

2. Swapping point and direction in the symmetric equations

the whole question, since the two objects are swapped from start to finish.

What not to write

“ $\dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z}{4}$ passes through $(2, -1, 4)$ and has direction $(1, -3, 0)$ .”

What to write

“The point is $(1, -3, 0)$ , read off the NUMERATORS with the sign changed, and the direction is $(2, -1, 4)$ , read off the DENOMINATORS.”

Why: The symmetric form is just the parametric form with $t$ isolated: $t=\frac{x-x_{0}}{a}$ . The denominator is therefore the coefficient of $t$ , that is, the direction.

3. Concluding that two lines intersect because their directions are not parallel

3 marks, and an intersection system solved for nothing, since it has no solution.

What not to write

“ $\vec{u}$ and $\vec{v}$ are not parallel, so the two lines intersect.”

What to write

“Not parallel directions only means the lines are NOT PARALLEL. Two cases remain in space: intersecting or SKEW. I compute $\vec{AC}\cdot(\vec{u}\times\vec{v})$ : zero, they are coplanar, hence intersecting; non-zero, they are skew.”

The two drawings are identical up to the projection: only the break in the line, on the left, says that one line passes BEHIND the other. The scalar triple product decides, never the drawing.

Why: In the plane, two non-parallel lines must meet. In space, they can cross without touching, and that is the most common case, not the exception.

4. Using the same parameter for two different lines

the whole question, with the opposite of the true conclusion: the lines are declared skew where they actually intersect.

What not to write

“ $x=1+2t$ for $D_{1}$ and $x=3-t$ for $D_{2}$ : I set them equal and solve for $t$ .”

What to write

“Each line has ITS OWN parameter: $t$ for $D_{1}$ , $s$ for $D_{2}$ . The system to solve is $1+2t=3-s$ , and so on for the three components, in two unknowns.”

The lines meet at

I

, reached for

t=3

D_{1}

and for

s=-1

D_{2}

. With a single parameter, you would look for a common

t

, which does not exist, and draw the wrong conclusion.

Why: The parameter locates a position ON its own line. Two lines can perfectly well meet at a point reached for $t=3$ on one and $s=-1$ on the other.

5. Forgetting the absolute value in a distance to a plane

1 mark, and an answer that cannot exist, which the grader sees at once.

What not to write

“ $d=\dfrac{2\times 1-1\times 4+3\times 0-5}{\sqrt{14}}=\dfrac{-7}{\sqrt{14}}$ , the distance is negative.”

What to write

“The formula has an ABSOLUTE VALUE in the numerator: $d=\dfrac{|-7|}{\sqrt{14}}=\dfrac{7}{\sqrt{14}}=\dfrac{\sqrt{14}}{2}$ . A distance is always positive.”

The distance is the length of the perpendicular

SH

, not that of any segment

SM

: every other segment joining

S

to the plane is strictly longer.

Why: The sign of the numerator tells which SIDE of the plane the point is on, useful information for a half-space, but it has no place in a length.

6. Applying the distance formula without bringing the equation to zero

2 marks, and the error goes unnoticed since the result stays positive and plausible.

What not to write

“For the plane $2x-y+3z=5$ , the distance is $\dfrac{|2x_{S}-y_{S}+3z_{S}+5|}{\sqrt{14}}$ .”

What to write

“I first move everything to one side: $2x-y+3z-5=0$ , so $d=-5$ and the numerator is $|2x_{S}-y_{S}+3z_{S}-5|$ .”

Why: The formula assumes the form $ax+by+cz+d=0$ . The constant term enters with ITS sign, the one it has once moved to the left, not the one it had on the right.

7. Concluding that a line parallel to a plane lies outside it

2 marks, and one of the three cases of the tree forgotten entirely.

What not to write

“ $\vec{u}\cdot\vec{n}=0$ , so the line is parallel to the plane and has no point in common with it.”

What to write

“ $\vec{u}\cdot\vec{n}=0$ leaves TWO cases: strictly parallel, or CONTAINED in the plane. I substitute the point $A(1, 0, 2)$ of the line into the equation of the plane: if it checks, the line lies in the plane.”

In both cases

\vec{u}\cdot\vec{n}=0

. Only the point

A

, tested in the equation of the plane, separates the strictly parallel line (left) from the line contained in the plane (right).

Why: The dot product only speaks about DIRECTIONS. The position of the line relative to the plane also needs the information of a POINT, which the dot product never saw.

8. Computing the angle between a line and a plane with a cosine

2 marks, and an almost right angle where the line is almost lying flat on the plane.

What not to write

“ $\cos\theta=\dfrac{|\vec{u}\cdot\vec{n}|}{\|\vec{u}\|\,\|\vec{n}\|}$ , so the angle between the line and the plane is $84^{\circ}$ .”

What to write

“It is a SINE: $\sin\theta=\dfrac{|\vec{u}\cdot\vec{n}|}{\|\vec{u}\|\,\|\vec{n}\|}$ , so $\theta\approx 6^{\circ}$ . The cosine gives the angle with the NORMAL, and the required angle is its complement.”

The cosine measures the angle with the NORMAL

\vec{n}

(in red); the angle with the plane (in green) is its complement, hence the sine. Angle enlarged: in the trap,

84^{\circ}

versus

6^{\circ}

Why: The only vector available to represent the plane is its normal, which is perpendicular to it. Going from the plane to its normal adds exactly $90^{\circ}$ , hence the switch from cosine to sine.

Which method to choose

Which relative position, from the two tests

First compute a dot product or a cross product on the DIRECTIONS, then, if needed, test ONE point. Never the other way round.

The four positions of two lines: parallel directions (a point decides between parallel and coincident) or not (the scalar triple product decides between intersecting and skew).

If a line and a plane, with $\vec{u}\cdot\vec{n}\neq 0$ → the line CROSSES the plane: substitute the parametric equations into the equation and solve for $t$

Example: $\vec{u}=(2, 1, 3)$ and $\vec{n}=(1, 2, -1)$ give $2+2-3=1\neq 0$
If a line and a plane, with $\vec{u}\cdot\vec{n}=0$ → test a point of the line in the equation: if it checks, the line is CONTAINED in the plane; otherwise, STRICTLY PARALLEL
If two lines with PARALLEL directions → parallel; test a point of one in the other to tell coincident from strictly parallel
If two lines with non-parallel directions, scalar triple product $\vec{AC}\cdot(\vec{u}\times\vec{v})=0$ → coplanar, hence INTERSECTING: solve the system in the two parameters $t$ and $s$
If two lines with non-parallel directions, non-zero scalar triple product → SKEW: do not look for an intersection, compute their distance directly

$\text{dist}=\frac{|\vec{AC}\cdot(\vec{u}\times\vec{v})|}{\|\vec{u}\times\vec{v}\|}$ , and the numerator is already computed.
If two planes with non-parallel normals → intersecting along a LINE with direction vector $\vec{n_{1}}\times\vec{n_{2}}$ ; a common point is found by fixing one variable

No test is done by eye on a drawing: in space, two skew lines project exactly like two intersecting lines. Only the scalar triple product decides.

Which distance formula, from the two objects

Look at the nature of the two objects involved, and the formula follows. Each one has its characteristic numerator.

If a point and a PLANE → $\frac{|ax_{S}+by_{S}+cz_{S}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$ , after bringing the equation to the form $=0$

Example: $S(1, 0, -2)$ and $x+2y-z-6=0$ give $\frac{|-3|}{\sqrt{6}}=\frac{\sqrt{6}}{2}$
If a point and a LINE → $\frac{\|\vec{AS}\times\vec{u}\|}{\|\vec{u}\|}$ , with a CROSS product in the numerator

It is the “area divided by base” reading of the parallelogram built on $\vec{AS}$ and $\vec{u}$ .
If two SKEW lines → $\frac{|\vec{AC}\cdot(\vec{u}\times\vec{v})|}{\|\vec{u}\times\vec{v}\|}$ , with a SCALAR TRIPLE product in the numerator
If two PARALLEL lines, or a line parallel to a plane → reduce to the point-object case by taking ANY point of one of the two

This is the case students handle worst on tests, although it needs no new formula.
If two PARALLEL planes → take a point of one and apply the point-to-plane distance to the other

Universal check: a distance is positive, and it is $0$ exactly when the two objects touch. Finding $0$ between two lines declared skew is therefore a contradiction.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Finding the intersection of a line and a plane

When to use it: The question gives a line and a plane and asks for their intersection, the point where the line pierces the plane, or their relative position.

1 Extract and write down the two basic data: the point $A$ and the direction vector $\vec{u}$ of the line, the normal vector $\vec{n}$ of the plane.
2 Compute $\vec{u}\cdot\vec{n}$ and CONCLUDE on the relative position before any intersection calculation.
3 If the product is non-zero, write the parametric equations of the line, then substitute $x$ , $y$ and $z$ into the equation of the plane.
4 Solve the equation for $t$ , which now has only one unknown, then substitute the value found into the parametric equations to get the point.
5 Check that the point found satisfies the equation of the plane, and write this check on your paper.

Concluding sentence

Since $\vec{u}\cdot\vec{n}=1\neq 0$ , the line crosses the plane at a single point; the substitution gives $t=3$ , so $D\cap P=\{I(7, 3, 7)\}$ .

The trap: Going straight to the substitution without the test $\vec{u}\cdot\vec{n}$ . When the product is zero, the substitution gives either an impossible equation or an identity, and many papers then conclude “no solution” instead of “the line lies in the plane”.

Marking: 0.5 mark for the position test, 1 mark for the substitution, 1 mark for solving, 0.5 mark for the point and its check.

Showing that two lines are skew and computing their distance

When to use it: The question gives two lines and asks for their relative position, or directly for the distance between them.

1 Write the two point-direction pairs: $A$ and $\vec{u}$ for the first line, $C$ and $\vec{v}$ for the second.
2 Check that $\vec{u}$ and $\vec{v}$ are not parallel, by computing $\vec{u}\times\vec{v}$ and noting that it is not zero.
3 Compute $\vec{AC}$ , then the scalar triple product $\vec{AC}\cdot(\vec{u}\times\vec{v})$ , and conclude: non-zero, the lines are skew.
4 Divide the absolute value of the scalar triple product by the norm of $\vec{u}\times\vec{v}$ to get the distance.
5 Conclude in one sentence that gives the relative position AND the distance, with the exact value.

Concluding sentence

The scalar triple product $\vec{AC}\cdot(\vec{u}\times\vec{v})$ equals $12\neq 0$ : the lines are not coplanar, so they are skew, and their distance is $\frac{12}{\|\vec{u}\times\vec{v}\|}=2$ .

The trap: Computing $\vec{u}\times\vec{v}$ a second time for the distance. It is already computed at step 2, and the scalar triple product at step 3: the distance costs only one more division.

Marking: 0.5 mark for the two point-direction pairs, 0.5 mark for the non-parallel directions, 1 mark for the scalar triple product and the conclusion, 1 mark for the distance.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Substitute the point found into both equations

An intersection point must satisfy the equation of the plane AND the equations of the line. Two quick substitutions validate the whole calculation, the parameter included.

$I(7, 3, 7)$ in $x+2y-z-6$ gives $7+6-7-6=0$ : the point is indeed on the plane.
Check the sign of a distance

A distance is always positive or zero. A negative result signals a forgotten absolute value; a zero result between two lines declared skew is a contradiction.

A $-\frac{7}{\sqrt{14}}$ is not a distance: it is the signed numerator, which only says which side of the plane you are on.
Check that a direction vector of the plane is orthogonal to the normal

After building a direction vector of a plane, take its dot product with the normal: it must be $0$ . It is the only check that catches the normal-direction confusion.

$(1, 2, 0)\cdot(2, -1, 3)=2-2+0=0$ : the vector does lie along the plane.
Compare the angle found with the mental picture

An angle between two lines or two planes never exceeds $90^{\circ}$ . An angle between a line and a plane close to $90^{\circ}$ means a line almost perpendicular to the plane, which must show in the data.

$\vec{u}$ and $\vec{n}$ almost orthogonal give a line-plane angle close to $0^{\circ}$ : an answer of $84^{\circ}$ gives away a cosine used instead of the sine.
Test two distinct parameters on an intersection of lines

After solving the system in $t$ and $s$ , substitute $t$ into the first line and $s$ into the second: the two triples must be IDENTICAL. This is the check that catches the single parameter.

$t=3$ gives $(7, 3, 7)$ and $s=-1$ also gives $(7, 3, 7)$ : the lines do meet at this point.

The typical problem, taken apart

A line that crosses a plane: point, distance and angle

The line $D$ passes through $A(1, 0, -2)$ and has direction vector $\vec{u}=(2, 1, 3)$ . The plane $P$ has equation $x+2y-z-6=0$ .

1) Determine the relative position of $D$ and $P$ , then their intersection.

2) Compute the distance from the point $A$ to the plane $P$ .

3) Compute the angle between the line $D$ and the plane $P$ .

The line starts from

A

, below the plane, and crosses it at

I

(hidden part dashed): this is what

\vec{u}\cdot\vec{n}\neq 0

guarantees before any calculation.

Step 1

The plane is already in the form $ax+by+cz+d=0$ , so $\vec{n}=(1, 2, -1)$ and $d=-6$ .

Why

Writing $\vec{n}$ in the margin as soon as you read the question avoids confusing it with a direction vector, and gives at once the data for all three questions.

Step 2

$\vec{u}\cdot\vec{n}=2\times 1+1\times 2+3\times(-1)=2+2-3=1\neq 0$ , so $D$ crosses $P$ at a single point.

Why

This test decides what follows: had it been zero, you would have had to test a point to decide between contained and parallel. Writing it BEFORE the substitution is worth half a mark.

Step 3

Parametric equations of $D$ : $x=1+2t$ , $y=t$ , $z=-2+3t$ .

Why

The parametric form is the only one that substitutes directly into the equation of a plane. The symmetric form would need a useless detour.

Step 4

Substitution: $(1+2t)+2t-(-2+3t)-6=0$ , that is $1+2t+2t+2-3t-6=t-3=0$ .

Why

The terms in $t$ reduce to a single coefficient, which is exactly $\vec{u}\cdot\vec{n}=1$ . A zero coefficient would have confirmed the parallel case: the two calculations check each other.

Step 5

$t=3$ , so $I(1+6, 3, -2+9)=I(7, 3, 7)$ . Check: $7+6-7-6=0$ .

Why

The check is immediate and it tests the parameter along with the three coordinates. Without it, an arithmetic error goes unnoticed.

Step 6

Distance from $A$ to $P$ : $\dfrac{|1+2\times 0-(-2)-6|}{\sqrt{1+4+1}}=\dfrac{|-3|}{\sqrt{6}}=\dfrac{3}{\sqrt{6}}=\dfrac{\sqrt{6}}{2}$ .

Why

The absolute value turns $-3$ into $3$ ; the minus sign only said which side of the plane $A$ is on, information the question does not want.

Step 7

$\|\vec{u}\|=\sqrt{4+1+9}=\sqrt{14}$ and $\|\vec{n}\|=\sqrt{6}$ , so $\sin\theta=\dfrac{|1|}{\sqrt{14}\times\sqrt{6}}=\dfrac{1}{\sqrt{84}}$ .

Why

A SINE, because the angle computed with $\vec{n}$ is the angle with the perpendicular to the plane. The numerator is the dot product already computed at step 2.

Step 8

$\theta=\arcsin\left(\dfrac{1}{\sqrt{84}}\right)\approx 6.3^{\circ}$ , a very small angle.

Why

The order of magnitude confirms the calculation: $\vec{u}\cdot\vec{n}=1$ is small compared with $\sqrt{84}\approx 9.2$ , so the line is almost parallel to the plane, and a result close to $84^{\circ}$ would have given away a cosine.

The conclusion, written out

The line $D$ crosses the plane $P$ at the point $I(7, 3, 7)$ ; the distance from $A$ to $P$ is $\frac{\sqrt{6}}{2}$ units of length, and the angle between $D$ and $P$ is about $6.3^{\circ}$ .

The classic mistake on this problem: Giving $\theta\approx 83.7^{\circ}$ by taking the inverse cosine instead of the inverse sine: that is the angle between the line and the NORMAL, the complement of the one asked for.

Learn by heart

• Line: a POINT and a DIRECTION vector. Plane: a POINT and a NORMAL vector.
• $ax+by+cz+d=0$ : $(a, b, c)$ is the NORMAL, never a direction of the plane.
• Symmetric form: the point in the numerators with the sign changed, the direction in the denominators.
• $\vec{u}\cdot\vec{n}\neq 0$ : the line crosses. $\vec{u}\cdot\vec{n}=0$ : parallel or contained, and a point decides.
• Non-parallel directions do not mean intersecting lines: in space, the SKEW case remains.
• Coplanarity of two lines: $\vec{AC}\cdot(\vec{u}\times\vec{v})=0$ .
• Two distinct parameters, $t$ and $s$ , for two different lines.
• Distance to a plane: absolute value in the numerator, and the equation brought to $=0$ first.
• Angle between two planes or two lines: COSINE. Angle between a line and a plane: SINE.
• Intersection of two non-parallel planes: a LINE, with direction $\vec{n_{1}}\times\vec{n_{2}}$ .

Frequently asked questions

How do you know whether two lines in space are skew?

First check that their direction vectors are not parallel; otherwise the lines are parallel. Then compute the scalar triple product of the vector joining a point of each line with the two direction vectors. If it is zero, the lines are coplanar and therefore intersect; if it is not, they are skew.

What is the difference between the normal vector and a direction vector of a plane?

The normal vector is perpendicular to the plane, and its components are the three coefficients of the scalar equation. A direction vector lies in the plane, so it is orthogonal to the normal. Mixing the two up ruins the whole problem, since the two directions are perpendicular to each other.

How do you find the intersection of a line and a plane?

First test the dot product of the direction vector and the normal vector. If it is not zero, write the parametric equations of the line, substitute the three coordinates into the equation of the plane, and solve the resulting equation, which has only one unknown. The value of the parameter gives back the point.

Why is the angle between a line and a plane computed with a sine?

Because the only vector that represents the plane is its normal, which is perpendicular to it. The usual ratio therefore gives the angle between the line and the normal, and the required angle is its complement. Switching from cosine to sine performs exactly that complement to ninety degrees.

Can a line parallel to a plane have points in common with it?

Yes, if it lies in the plane. The dot product test only speaks about directions, so it cannot tell the two cases apart. To decide, substitute any point of the line into the equation of the plane: if the equality holds, the whole line is contained in the plane.

Can two planes meet at a single point?

No, never. In space, the intersection of two planes is empty if they are strictly parallel, a whole plane if they coincide, and a line in every other case. The direction vector of that line is the cross product of the two normal vectors.

Practise it

Corrected exercises: Lines and planes in space, 201-SN4-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

16 corrected exercises
160 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-sn4-lines-and-planes. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: lines and planes in space (201-SN4-RE)

The essentials

Four ways of writing, and what each one tells you

Relative positions, and the test that decides between them

Distances and angles, and what tells them apart

The mistakes that cost marks

1. Reading the coefficients of a plane as a direction vector

2. Swapping point and direction in the symmetric equations

3. Concluding that two lines intersect because their directions are not parallel

4. Using the same parameter for two different lines

5. Forgetting the absolute value in a distance to a plane

6. Applying the distance formula without bringing the equation to zero

7. Concluding that a line parallel to a plane lies outside it

8. Computing the angle between a line and a plane with a cosine

Which method to choose

Which relative position, from the two tests

Which distance formula, from the two objects

How the answer is expected to be written

Finding the intersection of a line and a plane

Showing that two lines are skew and computing their distance

Check before you hand in

The typical problem, taken apart

A line that crosses a plane: point, distance and angle

Learn by heart

Frequently asked questions

How do you know whether two lines in space are skew?

What is the difference between the normal vector and a direction vector of a plane?

How do you find the intersection of a line and a plane?

Why is the angle between a line and a plane computed with a sine?

Can a line parallel to a plane have points in common with it?

Can two planes meet at a single point?

Corrected exercises: Lines and planes in space, 201-SN4-RE

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