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

Revision sheet: the three products of vectors in space (201-SN4-RE)

This sheet does not repeat the formulas of the three products: they are in your notes and on the formula sheet. It deals with what decides the mark: the KIND of result expected, the factor everyone forgets, and the write-up the 201-SN4-RE marker expects on an area or a volume.

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

Three products, three kinds of result: $\vec{u}\cdot\vec{v}$ is a NUMBER, $\vec{u}\times\vec{v}$ is a VECTOR, $\vec{u}\cdot(\vec{v}\times\vec{w})$ is a NUMBER again. Half of the marks lost come from writing an arrow over a number, or a number where the question expects a vector.

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

The essentials

Three products, three kinds of result

• DOT PRODUCT: $\vec{u}\cdot\vec{v}=a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=\|\vec{u}\|\,\|\vec{v}\|\cos\theta$ . It is a NUMBER, never a vector.
• CROSS PRODUCT: $\vec{u}\times\vec{v}=(b_{1}c_{2}-c_{1}b_{2}, c_{1}a_{2}-a_{1}c_{2}, a_{1}b_{2}-b_{1}a_{2})$ . It is a VECTOR, orthogonal to the other two.
• SCALAR TRIPLE PRODUCT: $\vec{u}\cdot(\vec{v}\times\vec{w})$ . It is a NUMBER, equal to the $3\times 3$ determinant of the three vectors.
• The cross product is NOT commutative: $\vec{v}\times\vec{u}=-(\vec{u}\times\vec{v})$ . The dot product is.
• $\vec{u}\times\vec{u}=\vec{0}$ and $\vec{u}\cdot(\vec{u}\times\vec{v})=0$ : two free equalities that serve as checks.

On the left, the dot product measures how much of

\vec{u}

lies along

\vec{v}

, a number. On the right, the cross product leaves the plane of the two vectors, a vector, and its norm is the area of the parallelogram.

Before writing a line, ask what the question expects: a length, an angle and an area are NUMBERS; a normal and a direction are VECTORS. That is what chooses the product.

What each product MEASURES

• $\vec{u}\cdot\vec{v}=0$ (with $\vec{u}$ and $\vec{v}$ non-zero) means ORTHOGONAL. The sign gives the type of angle: positive acute, negative obtuse.
• $\|\vec{u}\times\vec{v}\|=\|\vec{u}\|\,\|\vec{v}\|\sin\theta$ is the AREA OF THE PARALLELOGRAM built on the two vectors. The area of the triangle is HALF of it.
• $\vec{u}\times\vec{v}=\vec{0}$ means COLLINEAR.
• $|\vec{u}\cdot(\vec{v}\times\vec{w})|$ is the VOLUME OF THE PARALLELEPIPED. The volume of the corresponding tetrahedron is one SIXTH of it.
• $\vec{u}\cdot(\vec{v}\times\vec{w})=0$ means COPLANAR: the parallelepiped is flattened, its volume is zero.

Base: the parallelogram of

\vec{v}

and

\vec{w}

, with area

\|\vec{v}\times\vec{w}\|

; height

h

: the projection of

\vec{u}

onto the normal. Base times height:

|\vec{u}\cdot(\vec{v}\times\vec{w})|

Two factors never to forget, and they are the two most frequent mistakes of the chapter: the $\frac{1}{2}$ of the triangle and the $\frac{1}{6}$ of the tetrahedron.

The orthogonal projection, and its two forms

• VECTOR projection of $\vec{u}$ onto $\vec{v}$ : $\vec{u_{v}}=\dfrac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|^{2}}\,\vec{v}$ , with the norm SQUARED in the denominator.
• SCALAR projection: $\dfrac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|}$ , a number, which can be negative.
• Orthogonal decomposition: $\vec{u}=\vec{u_{v}}+\vec{u_{\perp}}$ , where $\vec{u_{\perp}}=\vec{u}-\vec{u_{v}}$ is orthogonal to $\vec{v}$ . That is the check: their dot product must be $0$ .
• Distance from a point $S$ to the line through $A$ with direction $\vec{u}$ : $\dfrac{\|\vec{AS}\times\vec{u}\|}{\|\vec{u}\|}$ .
• Direction cosines of $\vec{v}=(a, b, c)$ : $\cos\alpha=\frac{a}{\|\vec{v}\|}$ , $\cos\beta=\frac{b}{\|\vec{v}\|}$ , $\cos\gamma=\frac{c}{\|\vec{v}\|}$ , with $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1$ .

In green the projection

\vec{u_{v}}

, in orange the orthogonal component

\vec{u_{\perp}}

. The sign of the dot product tells on which side the foot falls: it is the sign of the scalar projection.

The sum of the squares of the direction cosines is always $1$ : it is the fastest check on a computation of direction angles.

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. Getting the kind of result wrong, number or vector

1 mark per occurrence, and the marker can no longer follow the rest of the reasoning.

What not to write

“ $\vec{u}\cdot\vec{v}=(3, -1, 2)$ ” or “ $\vec{u}\times\vec{v}=7$ ”.

What to write

“ $\vec{u}\cdot\vec{v}=7$ , a number; $\vec{u}\times\vec{v}=(3, -1, 2)$ , a vector.”

Why: This is the founding distinction of the chapter. Every question falls either among those that want a number, angle, area, volume, distance, or among those that want a vector, normal, direction.

2. Giving the area of the triangle without dividing by two

2 marks, and it is the most frequent mistake of the whole chapter.

What not to write

“The area of triangle $ABC$ is $\|\vec{AB}\times\vec{AC}\|=14$ .”

What to write

“ $\|\vec{AB}\times\vec{AC}\|=14$ is the area of the PARALLELOGRAM; the area of the triangle is $\frac{14}{2}=7$ .”

The norm of the cross product measures the whole parallelogram,

14

; the diagonal

BC

cuts it into two equal triangles, and the triangle

ABC

of the question, in green, is only half of it,

7

Why: The cross product measures the parallelogram built on the two vectors, and the triangle is only half of it. The factor is not added, it is taken away.

3. Giving the volume of the tetrahedron without dividing by six

2 marks, and a result six times too large that no order-of-magnitude check rescues.

What not to write

“The volume of tetrahedron $ABCD$ is $|\vec{AB}\cdot(\vec{AC}\times\vec{AD})|=18$ .”

What to write

“ $|\vec{AB}\cdot(\vec{AC}\times\vec{AD})|=18$ is the volume of the PARALLELEPIPED; the volume of the tetrahedron is $\frac{18}{6}=3$ .”

The scalar triple product measures the parallelepiped built on

\vec{AB}

\vec{AC}

\vec{AD}

(in grey),

18

; the tetrahedron

ABCD

(in navy, face

BCD

in green) fills only one sixth of it,

3

Why: The parallelepiped splits into six tetrahedra of equal volume, exactly as the parallelogram splits into two triangles. The factor $\frac{1}{6}$ is not a convention, it comes from that dissection.

4. Treating the cross product as if it were commutative

1 mark on a normal, the whole question on an orientation or a signed scalar triple product.

What not to write

“ $\vec{v}\times\vec{u}=\vec{u}\times\vec{v}$ , the order changes nothing since we will take the norm.”

What to write

“ $\vec{v}\times\vec{u}=-(\vec{u}\times\vec{v})$ . The norms are indeed equal, so the AREA does not change, but the NORMAL reverses, which flips the orientation of the plane and the sign of the scalar triple product.”

Same parallelogram, same norm, so same area; but

\vec{v}\times\vec{u}

(in orange) points to the other side of the plane from

\vec{u}\times\vec{v}

(in green): the normal and the orientation reverse.

Why: Anticommutativity has no effect when only a norm is kept, and it is decisive as soon as the vector is kept. You need to know which of the two cases you are in.

5. Concluding that a vector is zero because a dot product is

the whole proof question, and the mistake comes back with

\vec{u}\times\vec{v}=\vec{0}

, which means collinear.

What not to write

“ $\vec{u}\cdot\vec{v}=0$ , so $\vec{u}=\vec{0}$ or $\vec{v}=\vec{0}$ .”

What to write

“ $\vec{u}\cdot\vec{v}=0$ means that $\vec{u}$ and $\vec{v}$ are ORTHOGONAL, or that one of them is zero. For example $(1, 0, 0)\cdot(0, 1, 0)=0$ although neither is zero.”

\vec{u}=(1, 0, 0)

and

\vec{v}=(0, 1, 0)

are not zero, and yet

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

: a zero dot product says “right angle”, not “zero vector”.

Why: Products of vectors do not have the property of the product of real numbers, where a zero product forces a zero factor. Here, the zero expresses a relative position.

6. Dividing the projection by the norm instead of its square

2 marks, and the orthogonal decomposition that follows becomes wrong.

What not to write

“The vector projection of $\vec{u}$ onto $\vec{v}$ is $\dfrac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|}\,\vec{v}$ .”

What to write

“The VECTOR projection is $\dfrac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|^{2}}\,\vec{v}$ , with the square. Without the square, you get the SCALAR PROJECTION times $\vec{v}$ , a vector $\|\vec{v}\|$ times too long.”

\vec{u}=(3, 2)

\vec{v}=(2, 0)

: the correct projection,

\frac{6}{4}\vec{v}

, stops at the foot of the perpendicular; divided by

\|\vec{v}\|=2

only (in red), it is twice too long.

Why: The square comes from dividing twice by the norm: once to measure the projection, once to turn $\vec{v}$ into a unit vector. The check is immediate: if $\vec{u}=\vec{v}$ , the projection must be exactly $\vec{v}$ .

7. Forgetting the absolute value in an angle between two lines

1.5 marks, and an answer above

90^{\circ}

for an angle between lines, which does not exist.

What not to write

“ $\cos\theta=\dfrac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\,\|\vec{v}\|}=-0.6$ , so the angle between the two lines is $126.9^{\circ}$ .”

What to write

“The angle between two LINES is the smaller of the two, so $\cos\theta=\dfrac{|\vec{u}\cdot\vec{v}|}{\|\vec{u}\|\,\|\vec{v}\|}=0.6$ and $\theta\approx 53.1^{\circ}$ . The angle between two VECTORS keeps the sign.”

Between the VECTORS

\vec{u}

and

\vec{v}

126.9^{\circ}

(in red); the LINES make two supplementary angles and the smaller one is kept,

53.1^{\circ}

(in green).

Why: A line has no sense, only a direction: its two opposite direction vectors give two supplementary angles, and the convention keeps the smaller one.

8. Dropping the parentheses of a vector triple product

the whole question, and the expression without parentheses does not even have a meaning.

What not to write

“ $(\vec{u}\times\vec{v})\times\vec{w}=\vec{u}\times(\vec{v}\times\vec{w})$ , so I write $\vec{u}\times\vec{v}\times\vec{w}$ .”

What to write

“The cross product is NOT associative: with $\vec{u}=\vec{v}=(1, 0, 0)$ and $\vec{w}=(0, 1, 0)$ , $(\vec{u}\times\vec{v})\times\vec{w}=\vec{0}$ while $\vec{u}\times(\vec{v}\times\vec{w})=(0, -1, 0)$ . The parentheses are compulsory.”

\vec{u}=\vec{v}=(1, 0, 0)

\vec{w}=(0, 1, 0)

\vec{u}\times(\vec{v}\times\vec{w})=(0, -1, 0)

in green, while

\vec{u}\times\vec{v}=\vec{0}

cancels the other grouping.

Why: Here $\vec{u}\times\vec{v}=\vec{0}$ , and the cross product of the zero vector with anything stays zero, while the other grouping produces a non-zero vector.

Which method to choose

Which product, depending on what the question asks

Read the word in the question, not the data. Each word points to one product and one only, and the kind of result confirms it.

If “angle”, “orthogonal”, “perpendicular”, “projection” → DOT product, which returns a number

Example: $\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\,\|\vec{v}\|}$
If “normal vector”, “perpendicular to both”, “equation of a plane” → CROSS product, which returns a vector

Example: $\vec{n}=\vec{AB}\times\vec{AC}$ gives the normal of the plane $ABC$
If “area”, “collinear”, “distance from a point to a line” → CROSS product, then take its NORM

Example: area of the triangle $=\frac{1}{2}\|\vec{AB}\times\vec{AC}\|$
If “volume”, “coplanar”, “distance between two skew lines” → SCALAR TRIPLE product, in absolute value

Example: volume of the tetrahedron $=\frac{1}{6}|\vec{AB}\cdot(\vec{AC}\times\vec{AD})|$
If “decompose”, “component along”, “work done by a force” → ORTHOGONAL PROJECTION, vector or scalar depending on what is asked

If the answer must be a vector, the norm is squared in the denominator; if it is a number, it is to the first power.

No branch applies? Look at what the answer must BE. A number comes from a dot or scalar triple product, a vector from a cross product, and there is no other possibility.

Two vectors, or three: what a zero means

Look at which product vanishes, and translate it at once into a relative position.

Three zeros, three positions: zero dot product, right angle; zero cross product, arrows on one line; zero scalar triple product, arrows in one plane (zero volume).

If $\vec{u}\cdot\vec{v}=0$ with $\vec{u},\vec{v}$ non-zero → the two vectors are ORTHOGONAL

Example: $(1, 2, 0)\cdot(2, -1, 5)=0$
If $\vec{u}\times\vec{v}=\vec{0}$ → the two vectors are COLLINEAR, so the area is zero

Example: $(2, 4, 6)$ and $(1, 2, 3)$
If $\vec{u}\cdot(\vec{v}\times\vec{w})=0$ → the three vectors are COPLANAR, so the volume is zero

It is also the criterion that says two lines in space are coplanar, hence intersecting or parallel.
If $\|\vec{u}\times\vec{v}\|=\|\vec{u}\|\,\|\vec{v}\|$ → the two vectors are ORTHOGONAL, since $\sin\theta=1$

The mirror case of the first one, and a handy check.

With three vectors, the scalar triple product is computed as a $3\times 3$ determinant: it is the same object, and two proportional rows make it zero at once.

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.

Computing the area of a triangle in space

When to use it: The question gives three points $A$ , $B$ , $C$ in space and asks for the area of the triangle, or the area of the parallelogram built on two sides.

1 Compute the components of TWO vectors starting from the SAME vertex, $\vec{AB}$ and $\vec{AC}$ , and write them down.
2 Set up the cross product as a symbolic determinant, then give its three components.
3 Compute the norm of this cross product, keeping the exact value under the root.
4 Divide by $2$ for the triangle, and SAY so explicitly: that is where the mark is won or lost.
5 Conclude with the unit of area if the question gives one.

Concluding sentence

$\vec{AB}\times\vec{AC}=(-3, 6, -3)$ , so $\|\vec{AB}\times\vec{AC}\|=\sqrt{54}=3\sqrt{6}$ and the area of triangle $ABC$ is $\frac{3\sqrt{6}}{2}$ square units.

The trap: Crossing the position vectors $\vec{OB}$ and $\vec{OC}$ instead of two sides of the triangle: that measures the triangle $OBC$ , not $ABC$ . Any two sides do the job, even $\vec{AB}$ and $\vec{BC}$ , since $\vec{AB}\times\vec{BC}=\vec{AB}\times\vec{AC}$ ; starting both from the same vertex is simply the habit that never fails.

Marking: 0.5 mark for the two vectors, 1 mark for the cross product, 0.5 mark for the norm, 0.5 mark for the division by two.

Showing that four points are coplanar

When to use it: The question asks you to show that $A$ , $B$ , $C$ , $D$ are coplanar, or that two lines are coplanar.

1 Choose a reference point, usually $A$ , and compute the THREE vectors $\vec{AB}$ , $\vec{AC}$ , $\vec{AD}$ from it.
2 State the criterion: “the four points are coplanar if and only if the scalar triple product of the three vectors is zero”.
3 Compute $\vec{AB}\cdot(\vec{AC}\times\vec{AD})$ , either by the cross product followed by the dot product, or directly as a $3\times 3$ determinant.
4 Conclude from the value found, and interpret it geometrically if the question asks for it.

Concluding sentence

The scalar triple product $\vec{AB}\cdot(\vec{AC}\times\vec{AD})$ equals $0$ , so the parallelepiped built on these three vectors is flat: the points $A$ , $B$ , $C$ and $D$ are coplanar.

The trap: Taking three vectors that are always coplanar, such as $\vec{AB}$ , $\vec{BC}$ and $\vec{AC}$ : since $\vec{AC}=\vec{AB}+\vec{BC}$ , their scalar triple product is zero whatever $D$ is, and the conclusion proves nothing. The safe choice remains three vectors from the same point, $\vec{AB}$ , $\vec{AC}$ and $\vec{AD}$ : reaching $B$ , $C$ and $D$ is not enough, since $\vec{DB}$ , $\vec{BC}$ and $\vec{CD}$ add up to zero and are therefore always coplanar too.

Marking: 0.5 mark for the three vectors from the same point, 0.5 mark for stating the criterion, 1 mark for the computation, 0.5 mark for the geometric conclusion.

Check before you hand in

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

Test the orthogonality of the cross product

Take the dot product of the vector obtained with EACH of the two starting vectors. Both must give exactly $0$ . Two multiplications, no risk.

$(-3, 6, -3)\cdot(1, 2, 3)=-3+12-9=0$ : the cross product is right. A non-zero result points to a sign error in the second component, the most error-prone of the three.
Check the sign of the dot product on the figure

Look at the angle on a quick sketch. If it looks acute, the dot product must be positive; obtuse, negative; right, zero. A disagreement points to a vector taken backwards.

A $\vec{BA}\cdot\vec{BC}$ found positive while the angle at $B$ is clearly obtuse: $\vec{AB}$ was used instead of $\vec{BA}$ .
Check the sum of the squared direction cosines

Add $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma$ : the total must be exactly $1$ . It is an instant check on three results at once.

For $\vec{v}=(2, 3, 6)$ with norm $7$ : $\frac{4}{49}+\frac{9}{49}+\frac{36}{49}=1$ , all good.
Look at the order of magnitude of an area or a volume

The area of a triangle cannot exceed that of the parallelogram, and the volume of a tetrahedron is six times smaller than that of the parallelepiped. Comparing before handing in catches the forgotten factors.

A tetrahedron volume given as $18$ while the scalar triple product is $18$ : the sixth was forgotten, the answer is $3$ .
Run the test $\vec{u}=\vec{v}$ on a projection formula

Replace $\vec{u}$ by $\vec{v}$ in the formula used: the projection of $\vec{v}$ onto itself must give back exactly $\vec{v}$ . If it gives $\|\vec{v}\|\,\vec{v}$ , the square is missing in the denominator.

With $\vec{v}=(3, 0, 4)$ : $\frac{25}{25}\vec{v}=\vec{v}$ , the formula is right; $\frac{25}{5}\vec{v}=5\vec{v}$ gives away the missing square.

The typical problem, taken apart

Distance from a point to a line, by two readings of the same area

In space, the line $L$ passes through $A(1, 0, 2)$ and has direction vector $\vec{u}=(2, 1, 2)$ . Let $S(3, 4, 4)$ .

1) Compute $\vec{AS}$ , then $\vec{AS}\times\vec{u}$ .

2) Deduce the distance from $S$ to the line $L$ .

3) Check the result with the orthogonal projection.

The parallelogram built on

\vec{AS}

and

\vec{u}

has area

\|\vec{AS}\times\vec{u}\|

, and also base times height, that is

\|\vec{u}\|\times d

: the two readings give the formula.

Step 1

$\vec{AS}=(3-1, 4-0, 4-2)=(2, 4, 2)$ .

Why

Always the head minus the tail, in that order. A $\vec{SA}$ written by mistake changes every sign of the cross product.

Step 2

$\vec{AS}\times\vec{u}=(4\times 2-2\times 1, 2\times 2-2\times 2, 2\times 1-4\times 2)=(6, 0, -6)$ .

Why

The second component is the one most often wrong, because its pattern is reversed. Writing it separately avoids the sign error.

Step 3

Check: $(6, 0, -6)\cdot(2, 1, 2)=12+0-12=0$ , and $(6, 0, -6)\cdot(2, 4, 2)=12+0-12=0$ .

Why

The cross product must be orthogonal to both starting vectors. This check costs ten seconds and fully validates the previous step.

Step 4

$\|\vec{AS}\times\vec{u}\|=\sqrt{36+0+36}=\sqrt{72}=6\sqrt{2}$ and $\|\vec{u}\|=\sqrt{4+1+4}=3$ .

Why

Keep the exact form $6\sqrt{2}$ : a decimal value would lose the simplification that comes next, and the marker expects the exact value.

Step 5

$d=\dfrac{\|\vec{AS}\times\vec{u}\|}{\|\vec{u}\|}=\dfrac{6\sqrt{2}}{3}=2\sqrt{2}$ .

Why

The formula is only the “area divided by base” reading of the parallelogram in the figure. Recovering it that way keeps you from confusing it with the distance to a PLANE, which uses a dot product.

Step 6

Check by projection: $\vec{AS}\cdot\vec{u}=4+4+4=12$ , so the projection is $\frac{12}{9}\vec{u}=\left(\frac{8}{3}, \frac{4}{3}, \frac{8}{3}\right)$ .

Why

The squared norm in the denominator, $9$ and not $3$ : this is the check on the projection formula, and here it lands on simple fractions.

Step 7

The orthogonal component is $\vec{AS}-\left(\frac{8}{3}, \frac{4}{3}, \frac{8}{3}\right)=\left(-\frac{2}{3}, \frac{8}{3}, -\frac{2}{3}\right)$ .

Why

It is the vector joining the foot $H$ of the perpendicular to $S$ : its norm IS the distance sought, and it also gives the coordinates of $H$ .

Step 8

Its norm is $\sqrt{\frac{4}{9}+\frac{64}{9}+\frac{4}{9}}=\sqrt{8}=2\sqrt{2}$ , the same value as in step 5.

Why

Two independent methods giving the same number: it is the best possible guarantee before handing in, and it answers question 3 with no extra work.

The conclusion, written out

The distance from the point $S$ to the line $L$ is $2\sqrt{2}$ units of length, a value confirmed by the cross product and by the orthogonal decomposition.

The classic mistake on this problem: Dividing $\|\vec{AS}\times\vec{u}\|$ by $\|\vec{u}\|^{2}$ instead of $\|\vec{u}\|$ , copying the projection formula. The result would then be $\frac{6\sqrt{2}}{9}$ , a length three times too small.

Learn by heart

• $\vec{u}\cdot\vec{v}$ is a NUMBER, $\vec{u}\times\vec{v}$ is a VECTOR, $\vec{u}\cdot(\vec{v}\times\vec{w})$ is a NUMBER.
• $\vec{u}\cdot\vec{v}=0$ : orthogonal. $\vec{u}\times\vec{v}=\vec{0}$ : collinear. Zero scalar triple product: coplanar.
• Area of the parallelogram $=\|\vec{u}\times\vec{v}\|$ ; area of the triangle: HALF.
• Volume of the parallelepiped $=|\vec{u}\cdot(\vec{v}\times\vec{w})|$ ; volume of the tetrahedron: one SIXTH.
• $\vec{v}\times\vec{u}=-(\vec{u}\times\vec{v})$ , and the cross product is not associative: the parentheses matter.
• Vector projection: $\frac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|^{2}}\vec{v}$ , with the SQUARE. Scalar projection: $\frac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|}$ .
• Distance from a point to a line $=\frac{\|\vec{AS}\times\vec{u}\|}{\|\vec{u}\|}$ .
• Angle between two LINES: absolute value in the numerator, so the answer is at most $90^{\circ}$ .
• $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1$ for the direction cosines.

Frequently asked questions

What is the difference between the dot product and the cross product?

The dot product gives a number, which measures the angle between the two vectors and equals zero when they are perpendicular. The cross product gives a vector, perpendicular to both, whose length is the area of the parallelogram they form, and which is the zero vector when they are collinear.

How do you find the area of a triangle in space?

Take two vectors starting from the same vertex, for example from A to B and from A to C, compute their cross product, then the length of that vector. This number is the area of the parallelogram. The area of the triangle is half of it, and that division by two is what students forget most often.

What is the scalar triple product of three vectors used for?

Its absolute value is the volume of the parallelepiped built on the three vectors, and the volume of the corresponding tetrahedron is one sixth of it. It is used above all as a coplanarity test: if the scalar triple product is zero, the solid is flat and the three vectors lie in one plane.

Why does the cross product change sign when you swap the order?

Because it defines an orientation. The resulting vector is perpendicular to the plane of the two vectors, but it can point to one side or the other, and the order of the factors chooses the side. The length, and so the area, does not change, but the direction of the normal vector is reversed.

How do you find the distance from a point to a line in space?

Take a point A on the line and its direction vector, compute the cross product of the vector from A to the point with the direction vector, then divide its length by the length of the direction vector. This amounts to reading the area of a parallelogram as base times height, the height being the distance you want.

Can the dot product be negative?

Yes, and its sign has a precise meaning. Positive, the angle between the two vectors is acute; zero, they are perpendicular; negative, the angle is obtuse. That is why the angle between two lines is computed with an absolute value in the numerator: a line has no sense, so the smaller of the two angles is kept.

Practise it

Corrected exercises: Vectors in space: the three products, 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.

17 corrected exercises
170 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.

Next sheet Lines and planes in space

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

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Revision sheet: the three products of vectors in space (201-SN4-RE)

The essentials

Three products, three kinds of result

What each product MEASURES

The orthogonal projection, and its two forms

The mistakes that cost marks

1. Getting the kind of result wrong, number or vector

2. Giving the area of the triangle without dividing by two

3. Giving the volume of the tetrahedron without dividing by six

4. Treating the cross product as if it were commutative

5. Concluding that a vector is zero because a dot product is

6. Dividing the projection by the norm instead of its square

7. Forgetting the absolute value in an angle between two lines

8. Dropping the parentheses of a vector triple product

Which method to choose

Which product, depending on what the question asks

Two vectors, or three: what a zero means

How the answer is expected to be written

Computing the area of a triangle in space

Showing that four points are coplanar

Check before you hand in

The typical problem, taken apart

Distance from a point to a line, by two readings of the same area

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Frequently asked questions

What is the difference between the dot product and the cross product?

How do you find the area of a triangle in space?

What is the scalar triple product of three vectors used for?

Why does the cross product change sign when you swap the order?

How do you find the distance from a point to a line in space?

Can the dot product be negative?

Corrected exercises: Vectors in space: the three products, 201-SN4-RE

See also

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