Week 6Stewart §12.1, 12.2, 12.3, 12.591 min full read
7 concepts21 worked examples28 exercises4 exam-level7 figures
What are you here for?
06Vectors, lines, and planes in space
You drag a loaded sled across flat ground with a rope, and because the rope runs up to your shoulder the pull leaves at about thirty degrees above the floor. The sled slides forward, but you can feel that not all of your effort is going into moving it — some of the pull is trying to lift the sled instead. Two questions hide in that feeling: how much of a pull aimed one way actually acts along another, and how would you even write down a push that has both a size and a direction?
By the end you can pin any point in a room with three numbers and measure the distance between two of them; write a push, a pull or a displacement as a single object you can add and scale; find the angle between two directions and decide in one line whether they are perpendicular; work out exactly how much of one pull acts along another; and write the equation of any line or plane in space.
In 60 seconds
Everything here is three numbers and one product: a point is three coordinates, a vector is a length with a direction stored as three components, and the dot product turns two vectors into the single number that carries the angle between them.
you are describing a line (a point and a direction) or a plane (a point and a normal)
Three most common mistakes
Treating where a vector is drawn as part of what it is. The vector from $A$ to $B$ is $B-A$, the same arrow no matter where you start it — and it is $B-A$, not $A-B$.
Answering a dot product with an arrow. $\mathbf a\cdot\mathbf b$ is a number; the product that returns a vector is the cross product, and that is next section, not this one.
Reading the coefficients in $ax+by+cz=d$ as a direction lying in the plane. They are the normal $\langle a,b,c\rangle$, the direction perpendicular to the plane.
On the syllabus the visible weights attached to this material are quizzes at ten percent and homework at ten percent. The heavier reason to get it solid is structural rather than numerical: every later part of Calculus II reads positions, lengths and angles off vectors, so an unpaid gap here is charged again in every partial derivative, gradient and surface integral that follows.
How much time do you have?
10 minutes
You leave able to find a distance, a and a dot product, and to test two vectors for perpendicularity. Enough for a short computation; not enough for a line-and-plane word problem.
The 60 second card, Formula card, The dot product and the angle between two vectors, Mistake ledger
45 minutes
You add the two ideas the formulas cannot supply on their own: how to read a picture into components, and how a point plus a direction becomes a line while a point plus a normal becomes a plane.
The 60 second card, Vectors: displacement with a length and a direction, The dot product and the angle between two vectors, Projection: the shadow of one vector on another, Equations of a line in space, Equations of a plane, and distance to it, Fading ladder, Practice B (computation)
Full reading
The whole build, in order: the missing third number that forces coordinates, vectors and their algebra, the dot product and everything it measures, and finally lines and planes written from a point and a direction.
Hook, Recall first, Try it yourself first, All seven concept blocks, Method boxes, Contrast pairs, Fading ladder, Full exam-style question, Practice A to D, Mistake ledger, Check yourself
By the end of this section
Locate a point in space with an ordered triple, and compute the distance between two points and the equation of a .
Represent a displacement as a vector in , find the vector from one point to another, and compute its magnitude.
Add and scale vectors, and build the unit vector in a given direction using the standard basis.
Compute the dot product, use it to find the angle between two vectors, and test two vectors for orthogonality.
Find the scalar and of one vector onto another, and use the dot product to compute work.
Write the vector, parametric and of a line through a point in a given direction, or through two points.
Write the equation of a plane from a point and a normal, find the angle between two planes, and the distance from a point to a plane.
Syllabus coverage
12.1
Three-dimensional coordinate systems
The distance formula and the equation of a sphere both come straight from the plane distance formula with a third square added under the root.
covered
12.2
Vectors
Split over two blocks: what a vector is and how to find its length first, then the algebra — addition, scaling and unit vectors — in the block after it.
covered
12.3
The dot product
Also split over two blocks, because projection and work are a distinct skill from computing the angle, and each fails for its own reasons.
covered
12.5
Equations of lines and planes
Lines first, then planes. Planes are built from a normal that is given to you; constructing a normal from two directions in the plane needs the cross product.
covered
12.4
The cross product
The syllabus line for this week jumps from 12.3 to 12.5, and 12.4 opens the following week. It is not covered here, so wherever a plane needs a normal we take the normal as given rather than build it from two directions.
deferred
Recall first
Distance in the plane
The distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
The distance in space is this same formula with one more squared difference added under the root; nothing new is invented, a third leg is just attached.
The Pythagorean theorem
In a right triangle the square of the hypotenuse equals the sum of the squares of the two legs.
It is the reason every length in this section is a square root of a sum of squares, in two dimensions and in three.
The dot product delivers $\cos\theta$; reading the angle back off it needs these reference values, and the sign of the cosine is the sign of the dot product.
The law of cosines
In a triangle with sides $a$, $b$ and included angle $\theta$ between them, the opposite side $c$ satisfies $c^{2}=a^{2}+b^{2}-2ab\cos\theta$.
It is where the geometric formula $\mathbf a\cdot\mathbf b=\vert\mathbf a\vert\vert\mathbf b\vert\cos\theta$ comes from, so it is recalled once inside the dot product block and not needed again.
Solving a linear equation for one letter
From $x=1+4t$ we get $t=\dfrac{x-1}{4}$.
Eliminating the parameter $t$ from the three of a line is exactly this, done three times and set equal.
Try it yourself first (3 questions)
1§06.0 — distance in the plane●○○○○
No vectors yet. This is the plane distance you have used since precalculus, and the point is to check it is automatic before a third coordinate joins in.
Given
The points $(1,2)$ and $(4,6)$ in the plane.
Find
(a) Find the distance between them.
Hint 1/4
You are not being asked for a formula to memorise, only to build the right triangle whose hypotenuse joins the two points.
Hint 2/4
Distance in the plane is $\sqrt{(\Delta x)^2+(\Delta y)^2}$.
Hint 3/4
Here $\Delta x=4-1=3$ and $\Delta y=6-2=4$, so the triangle has legs $3$ and $4$.
Hint 4/4
The distance is $\sqrt{9+16}=5$.
Show solutionDifferences then root
$$\Delta x=4-1=3,\qquad \Delta y=6-2=4$$
the two legs of the right triangle whose hypotenuse is the distance
$$d=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5$$
Pythagoras on those legs
Answer $$\boxed{d=5}$$
Check
The 3-4-5 triangle is the standard check; the answer is a whole number, as intended.
2§06.0 — do lengths simply add●●○○○
One trap, chosen because almost everyone falls for it once. Trust a picture over the words.
Given
Two vectors $\mathbf a$ and $\mathbf b$.
The claim: $\vert\mathbf a+\mathbf b\vert=\vert\mathbf a\vert+\vert\mathbf b\vert$ always.
Find
(a) True or false: the length of a sum is always the sum of the lengths?
Hint 1/4
Do not reach for a rule. Draw $\mathbf a$ across and $\mathbf b$ straight up, then draw the single arrow from start to finish.
Hint 2/4
Compare the diagonal of a right triangle with the sum of its two legs.
Hint 3/4
Take $\mathbf a=\langle 3,0\rangle$ and $\mathbf b=\langle 0,4\rangle$: the sum is $\langle 3,4\rangle$ with length $5$, while the lengths are $3$ and $4$.
the straight crossing is shorter than going along one leg then the other
Answer $$\boxed{\text{False}}$$
Check
The triangle inequality says $\vert\mathbf a+\mathbf b\vert\le\vert\mathbf a\vert+\vert\mathbf b\vert$, with equality only for same-; the $3,4,5$ case shows the strict gap.
Magnitude is not additive. This single fact is behind a large share of the wrong answers in this section.
3§06.0 — reading a cosine back to an angle●●○○○
Last warm-up. The dot product will keep handing you a cosine; you need to turn a cosine into an angle without a calculator for the standard values.
Given
$\cos\theta=-\tfrac12$, with $\theta$ between $0$ and $\pi$.
Find
(a) Find $\theta$.
Hint 1/4
Split the job in two: first find the angle whose cosine is one half in size, then fix the quadrant from the sign.
Hint 2/4
On $[0,\pi]$, a positive cosine gives an acute angle and a negative cosine an obtuse one.
Hint 3/4
The reference angle is $\tfrac{\pi}{3}$; since the cosine is negative, take $\pi-\tfrac{\pi}{3}$.
Hint 4/4
So $\theta=\tfrac{2\pi}{3}$.
Show solutionUse the reference value and the sign
$$\cos\tfrac{\pi}{3}=\tfrac12$$
the reference angle whose cosine has size one half
a negative cosine puts the angle in the second quadrant of the allowed range
$$\theta=\pi-\tfrac{\pi}{3}=\tfrac{2\pi}{3}$$
reflecting the reference angle across the right angle
Answer $$\boxed{\theta=\tfrac{2\pi}{3}}$$
Check
Check: $\cos\tfrac{2\pi}{3}=-\tfrac12$, and $\tfrac{2\pi}{3}$ does lie in $[0,\pi]$.
A negative cosine always means an obtuse angle here; that link is used on every dot-product angle you meet.
Notation
symbol
reads as
means
watch out
$\mathbf a,\ \vec a$
the vector a
a quantity with both a size and a direction; bold in print, an arrow by hand.
A vector is not a point. $\mathbf a=\langle 2,5\rangle$ is a displacement; $(2,5)$ is a place.
$\langle a_1,a_2,a_3\rangle$
the vector with components a-one, a-two, a-three
the vector that moves $a_1$ in the $x$ direction, $a_2$ in $y$, $a_3$ in $z$.
Angle brackets flag a vector; round brackets $(a_1,a_2,a_3)$ flag a point. Keeping them apart on paper prevents most sign slips.
$\vert\mathbf a\vert$
the magnitude of a
the length of the vector, $\sqrt{a_1^{2}+a_2^{2}+a_3^{2}}$.
Always a number, never a vector, and never negative.
$\mathbf i,\ \mathbf j,\ \mathbf k$
i, j, k
the unit vectors one step along $x$, $y$ and $z$: $\langle 1,0,0\rangle$, $\langle 0,1,0\rangle$, $\langle 0,0,1\rangle$.
$\langle 2,-1,3\rangle$ and $2\mathbf i-\mathbf j+3\mathbf k$ are two spellings of one vector.
$\mathbf a\cdot\mathbf b$
a dot b
the dot product, the number $a_1b_1+a_2b_2+a_3b_3$.
A number, not a vector. The centred dot is not ordinary multiplication and not the cross product.
$\hat{\mathbf a}$
a hat
the unit vector in the direction of $\mathbf a$, namely $\mathbf a/\vert\mathbf a\vert$.
Defined only when $\mathbf a\ne\mathbf 0$; you cannot point in the direction of a zero-length arrow.
$\operatorname{proj}_{\mathbf a}\mathbf b$
the projection of b onto a
the part of $\mathbf b$ that lies along $\mathbf a$, a vector parallel to $\mathbf a$.
Projecting onto $\mathbf a$ and onto $\mathbf b$ gives different vectors; the order matters.
$\mathbf n$
n, the normal
a vector perpendicular to a plane; it fixes the plane's tilt.
The normal points out of the plane, not along it. Any nonzero multiple of $\mathbf n$ is an equally good normal.
Conventions used here
A vector is written bold, and its components in angle brackets
In print a vector is bold, $\mathbf a$; by hand you write an arrow, $\vec a$ — they mean the same thing. Its components go in angle brackets, $\mathbf a=\langle a_1,a_2,a_3\rangle$, to keep them visually apart from a point $(a_1,a_2,a_3)$, which is a location rather than a displacement.
Vertical bars mean length
$\vert\mathbf a\vert$ is the magnitude of $\mathbf a$, a length, so it is a number and it is never negative. It is zero only for the zero vector.
The dot product is a number
$\mathbf a\cdot\mathbf b$ is a scalar. If a calculation hands you a vector for a dot product, an error has crept in. The product that returns a vector, the cross product, belongs to the next section.
The angle between two vectors is taken between 0 and pi
The angle $\theta$ between two vectors is the non-reflex angle, so $0\le\theta\le\pi$. That is exactly the range on which $\cos$ is one-to-one, which is why $\theta=\arccos(\cdots)$ has a single answer.
Keep exact values; round only when asked
A magnitude such as $\sqrt{29}$ or an angle such as $\arccos\!\left(-\tfrac13\right)$ is a finished answer. Convert to a decimal only when the question asks for one, and say how many places you kept.
6.1Locating a point in space, and measuring distance
Pin any point in a room with three numbers, and measure the straight-line distance between two of them.
On paper two numbers place any point, because a page is flat. A room is not flat, and that is the whole problem this block fixes.
Solvable with what we have
Place a point on a map with an across value and an up value.
Measure the straight-line distance between two points in the plane.
Not solvable yet
Say where a light fitting sits relative to a corner of the room, because two of your numbers are the floor position and the fitting is on the ceiling.
Tell apart a point on the floor from the point directly above it: both have the same across-and-up reading on the floor plan.
Keep using $(x,y)$ and hope height sorts itself out. Then the floor point $(2,3)$ and the ceiling point above it are both $(2,3)$, and the distance between them comes out as $0$.
Why it fails
A flat pair of numbers has no slot for height. The fix is one more axis, perpendicular to both of the first two, carrying a third number $z$.
TheoremThe distance formula in space
Conditions
The three axes are mutually perpendicular (a right-handed system).
The distance between two points is the square root of the sum of the squares of the three coordinate differences. It is the flat distance formula with a third leg attached under the same root.
The point $\textcolor{#1f6feb}{P=(2,3,4)}$ is reached by going $2$ along $x$, $3$ along $y$, then $4$ straight up. The $\textcolor{#128a5a}{\text{dashed box}}$ records those three moves; its far corner is $P$.
Looks like this, but is not
The two points $(2,3,0)$ and $(2,3,7)$ have the same $x$ and $y$, so they might be counted as the same location.
They sit one directly above the other, $7$ units apart. The third coordinate is the only thing telling them apart, and the distance formula returns $\sqrt{0+0+49}=7$, not $0$.
Distance between (1, -3, 2) and (5, 0, 14)
A deliberately clean case: the three differences are $4$, $3$ and $12$, a Pythagorean quadruple, so the answer is a whole number.
head minus tail in each coordinate; the order does not matter because each is squared
Square, add, root
$$d=\sqrt{4^{2}+3^{2}+12^{2}}=\sqrt{16+9+144}$$
the distance formula
$$=\sqrt{169}=13$$
the sum is a perfect square by design
Answer $$\boxed{\vert PQ\vert=13}$$
Check
Order check: swapping $P$ and $Q$ flips every difference to its negative, and squaring erases the sign, so the distance is the same either way.
Distance formula applied once.
The three coordinate differences are the edges of a box; the distance is its space diagonal. Every length in this section is a root of a sum of squares for exactly this reason.
Centre and radius of a sphere by completing the square
A sphere is the set of points a fixed distance from a centre, so its equation is a squared distance set equal to $r^{2}$. When the equation arrives expanded, three completions of the square put it back into that shape.
Given
$x^{2}+y^{2}+z^{2}-6x+4y-2z=11$
Find
the centre and radius
SolutionGroup each variable and complete the square
$$(x^{2}-6x)+(y^{2}+4y)+(z^{2}-2z)=11$$
collect the terms of each variable together
$$(x-3)^{2}-9+(y+2)^{2}-4+(z-1)^{2}-1=11$$
each bracket adds a constant that must be subtracted back out
Move the constants across
$$(x-3)^{2}+(y+2)^{2}+(z-1)^{2}=11+9+4+1=25$$
collect the three subtracted constants on the right
$$\text{centre }(3,-2,1),\qquad r=\sqrt{25}=5$$
read the centre off the signs inside the brackets and the radius as the root of the right side
Answer $$\boxed{\text{centre }(3,-2,1),\ \ r=5}$$
Check
Check one point: $(3,-2,6)$ is directly above the centre and should be on the sphere; its distance from the centre is $\sqrt{0+0+25}=5$, matching the radius.
Completing the square three times.
The sign flips: $(x-3)^{2}$ means the centre's $x$ is $+3$, and $(y+2)^{2}$ means the centre's $y$ is $-2$. The bracket hides the sign you want.
Checkpoint
§06.1 — inside, on, or outside a sphere●○○○○
Thirty seconds. No formula beyond one squared distance.
Given
The sphere $(x-2)^{2}+(y+1)^{2}+z^{2}=20$.
The point $(2,-5,1)$.
Find
(a) Is the point inside, on, or outside the sphere?
Hint 1/4
You do not need the actual distance, only whether it is more or less than the radius, so keep everything squared.
Hint 2/4
Compare $(x-h)^{2}+(y-k)^{2}+(z-l)^{2}$ with $r^{2}$: less means inside, equal means on, more means outside.
Hint 3/4
Here the centre is $(2,-1,0)$ and $r^{2}=20$; the squared distance is $0+16+1=17$.
Hint 4/4
Since $17<20$, the point is inside.
Show solutionSquared distance to the centre
$$(2-2)^{2}+(-5-(-1))^{2}+(1-0)^{2}=0+16+1=17$$
no need to take the root; compare squares directly
$$17<20$$
the squared distance is below the squared radius, so the point sits within
Answer $$\boxed{\text{inside}}$$
Check
Comparing squares avoids a root; the inequality $17<20$ is the whole decision.
⚠ Only two of the three squares
the plane formula is the one in muscle memory, and the $z$ term is easy to leave off when the numbers look familiar
The vector from $P$ to $Q$ is head minus tail, one subtraction per coordinate. Its magnitude is the root of the sum of the squares of its components, which is just the distance from its tail to its head.
The same components $\langle 3,2\rangle$ drawn from two different starting points give the $\textcolor{#1f6feb}{\text{same}}$ $\textcolor{#d1690a}{\text{vector}}$: three across and two up, wherever it begins.
Looks like this, but is not
$\langle 3,2\rangle$ started at the origin and $\langle 3,2\rangle$ started at $(5,1)$ look like two different arrows in two different places.
A vector stores only how far and which way, not where it starts. Both arrows move three across and two up, so both are the vector $\langle 3,2\rangle$. Position belongs to points, not to vectors.
The vector from A(2, -1, 3) to B(5, 1, -1), and its length
The most common single computation in the section: turn two points into the arrow between them, then measure it.
Given
$A=(2,-1,3)$
$B=(5,1,-1)$
Find
$\overrightarrow{AB}$ and $\vert\overrightarrow{AB}\vert$
Both radicands are perfect squares, so a non-integer here would signal an arithmetic error in the squares.
Two magnitudes.
$\langle 1,2,2\rangle$ with length $3$ and $\langle 2,3,6\rangle$ with length $7$ come back repeatedly; knowing their lengths on sight speeds up every unit-vector step later.
Checkpoint
§06.2 — from two points to an arrow●○○○○
Thirty seconds, in the plane so the arithmetic is instant.
Given
$A=(4,1)$
$B=(1,5)$
Find
(a) Find $\overrightarrow{AB}$ and its length.
Hint 1/4
Point the arrow from $A$ to $B$ first, then measure it; do not mix the two steps.
Hint 2/4
The vector is $B-A$ component by component, and its length is the root of the sum of the squares.
Hint 3/4
With $A=(4,1)$ and $B=(1,5)$ the components are $\langle 1-4,5-1\rangle=\langle -3,4\rangle$.
Add and stretch vectors component by component, and strip any vector down to a pure direction.
One arrow is a single push. Two pushes at once, or one push doubled, need a way to combine and stretch arrows. All of it happens component by component.
RuleVector algebra and the unit vector
Conditions
The unit vector step needs $\mathbf a\ne\mathbf 0$.
Add two vectors by adding matching components. Scale a vector by multiplying every component by the same number. To get the unit vector, divide the vector by its own length, which leaves the direction and sets the size to one.
Laying $\textcolor{#d1690a}{\mathbf b}$ tip-to-tail after $\textcolor{#1f6feb}{\mathbf a}$ leads from the start of $\mathbf a$ to the tip of $\mathbf b$: that single arrow is $\textcolor{#128a5a}{\mathbf a+\mathbf b}$.
Looks like this, but is not
To add $\langle 3,0\rangle$ and $\langle 0,4\rangle$ you might add their lengths, $3+4=7$.
Addition is componentwise, giving $\langle 3,4\rangle$, whose length is $5$. Lengths add only when the two vectors point the same way; here they are perpendicular.
c
c u
length |c u|
-2
$\langle -2,-4,-4\rangle$
6
-1
$\langle -1,-2,-2\rangle$
3
0
$\langle 0,0,0\rangle$
0
1
$\langle 1,2,2\rangle$
3
3
$\langle 3,6,6\rangle$
9
The length column is always $3\vert c\vert$: scaling stretches the length by the size of the factor, and the sign only turns the arrow around.
A combination: 2a - b
Scaling and adding at once, which is the shape of almost every vector expression you will simplify.
6.4The dot product and the angle between two vectors
One product of two vectors returns the single number that carries the angle between them.
Adding and scaling never produced a single number that compared two directions. The sled question needs exactly that: how aligned are two arrows? The dot product answers it.
TheoremThe dot product, two faces
Conditions
Both faces agree for all vectors; the angle face needs $\mathbf a,\mathbf b\ne\mathbf 0$.
Multiply matching components and add them; the result is a single number. That same number equals the product of the two lengths times the cosine of the angle between them, which is why one product carries the angle.
The sign of $\textcolor{#1f6feb}{\mathbf a}\cdot\textcolor{#d1690a}{\mathbf b}$ tracks the $\textcolor{#128a5a}{\text{angle}}$: positive when the arrows lean together, zero at a right angle, negative when they lean apart.
Looks like this, but is not
$\langle 1,2\rangle\cdot\langle 3,4\rangle$ might be written as $\langle 3,8\rangle$, multiplying the slots and keeping a vector.
The dot product adds those products into one number: $3+8=11$. A vector answer here means the multiply-and-add was left half done.
The angle between (1, 2, 2) and (3, 0, -4)
Both faces of the dot product are used here: the component face computes the number, the angle face reads the angle out of it.
a negative product in one slot is added as if positive, most often in the last term
wrong$$(2)(1)+(2)(-2)+(-1)(-2)=2-4-2=-4$$
right$$(2)(1)+(2)(-2)+(-1)(-2)=2-4+2=0$$
6.5Projection: the shadow of one vector on another
Find how much of one vector runs along another, and compute the work a force does.
The dot product measured the angle. The sled question wanted a length: how much of the pull actually runs along the ground. That length is a projection, and the work done is the dot product itself.
RuleScalar and vector projection, and work
Conditions
The vector being projected onto, $\mathbf a$, must be nonzero.
The is the dot product divided by the length of $\mathbf a$: it is how far $\mathbf b$ reaches along $\mathbf a$, as a single number. The vector projection multiplies that by the unit vector of $\mathbf a$, giving the actual arrow. Work is force dotted with displacement.
Dropping a perpendicular from the tip of $\textcolor{#d1690a}{\mathbf b}$ onto the line of $\textcolor{#1f6feb}{\mathbf a}$ marks off $\textcolor{#128a5a}{\operatorname{proj}_{\mathbf a}\mathbf b}$, the part of $\mathbf b$ that lies along $\mathbf a$.
Looks like this, but is not
Projecting $\mathbf b$ onto $\mathbf a$ might be taken to give the same vector as projecting $\mathbf a$ onto $\mathbf b$.
They point along different lines, so they cannot be equal in general: $\operatorname{proj}_{\mathbf a}\mathbf b$ lies along $\mathbf a$, while $\operatorname{proj}_{\mathbf b}\mathbf a$ lies along $\mathbf b$. Order matters, and the two denominators differ as well.
Projecting (1, 4, 8) onto (2, 2, 1)
The base vector $\mathbf a$ has a clean length, so both projections come out whole. Watch which formula divides by the length and which by its square.
Given
$\mathbf a=\langle 2,2,1\rangle$
$\mathbf b=\langle 1,4,8\rangle$
Find
the scalar and the vector projection of $\mathbf b$ onto $\mathbf a$
The leftover $\mathbf b-\operatorname{proj}_{\mathbf a}\mathbf b=\langle -3,0,6\rangle$ should be perpendicular to $\mathbf a$, and indeed $\langle -3,0,6\rangle\cdot\langle 2,2,1\rangle=-6+0+6=0$.
One dot product, one length, two divisions.
The vector projection is the scalar projection times the unit vector of $\mathbf a$; splitting it that way is why one formula has $\vert\mathbf a\vert$ and the other $\vert\mathbf a\vert^{2}$.
The work done dragging the sled
This is the opening scene. The pull is $50$ newtons at thirty degrees above the ground, and the sled slides twenty metres along the ground.
Given
Pull of magnitude $50$ N at $30^{\circ}$ above the horizontal.
Displacement $\mathbf d=\langle 20,0\rangle$ metres along the ground.
The upward component $25$ N multiplied the displacement's zero vertical component, contributing nothing; the feeling that part of the pull was wasted is exactly that zero.
One dot product, after resolving the force.
This closes the opening question: the work is the projection of the pull onto the motion, times the distance. Effort aimed across the motion is not lost to friction, it is simply never work.
Checkpoint
§06.5 — projecting onto an axis●○○○○
Thirty seconds. Projecting onto a coordinate axis just reads off one component.
Given
$\mathbf b=\langle 3,4,12\rangle$
$\mathbf a=\langle 1,0,0\rangle$
Find
(a) Find the scalar projection of $\mathbf b$ onto $\mathbf a$.
Hint 1/4
Notice what $\mathbf a$ is before computing anything: it is the unit vector along one axis.
Hint 2/4
The scalar projection is $\dfrac{\mathbf a\cdot\mathbf b}{\vert\mathbf a\vert}$.
Hint 3/4
With $\mathbf a=\langle 1,0,0\rangle$ the dot product is $3$ and $\vert\mathbf a\vert=1$.
Sweep out a line as a starting point plus every multiple of one direction.
In the plane a line was a slope and an intercept. In space slope alone cannot say which way a line tilts, so we go back to what a line really is: a starting point and one direction, walked forwards and backwards.
RuleLine: vector, parametric, symmetric forms
Conditions
The direction vector $\mathbf v=\langle a,b,c\rangle$ is nonzero.
Start at the point $\mathbf r_0$ and add every multiple of the direction $\mathbf v$; as $t$ runs over all numbers, the tip sweeps the whole line. In coordinates that is three parametric equations, and solving each for $t$ and setting them equal gives the symmetric form.
Every point on the line is the base point plus $t$ copies of the $\textcolor{#128a5a}{\text{direction}}$: the $\textcolor{#1f6feb}{\text{dots}}$ at $t=0,1,2$ are evenly spaced because equal steps of $t$ move equal distances.
Looks like this, but is not
In $\dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z}{4}$ the numbers $2,-1,4$ under the fractions might be read as a point on the line.
Those are the direction components, the denominators. A point on the line is read from the numbers being subtracted in the numerators: $(1,-3,0)$. Direction lives downstairs, position upstairs.
Line through (1, 2, 3) in the direction (4, -1, 2)
A point and a direction are everything a line needs; the three written forms are three spellings of the same set of points.
Given
point $(1,2,3)$
direction $\mathbf v=\langle 4,-1,2\rangle$
Find
the parametric and symmetric equations
SolutionParametric: base plus t times direction
$$x=1+4t,\qquad y=2-t,\qquad z=3+2t$$
each coordinate starts at the point and grows at the rate set by its direction component
Symmetric: eliminate t
$$t=\frac{x-1}{4}=\frac{y-2}{-1}=\frac{z-3}{2}$$
solve each parametric equation for $t$ and set the three equal
At $t=0$ every parametric equation returns the point $(1,2,3)$, confirming the line passes through it.
One direction read into three equations.
The symmetric form is just the parameter eliminated; if a direction component were zero you could not divide by it, and that coordinate would be written separately as a constant.
Line through two points, and where it meets the xy-plane
Two points give a direction by subtraction; after that it is the previous example, plus one substitution to find a crossing.
Given
$A=(2,4,-3)$
$B=(3,-1,1)$
Find
the parametric equations, and the point where the line crosses the plane $z=0$
Answer $$\boxed{x=2+t,\ y=4-5t,\ z=-3+4t;\quad \text{crosses }z=0\text{ at }\left(\tfrac{11}{4},\tfrac14,0\right)}$$
Check
The crossing point has $z=0$ as required, and its parameter $t=\tfrac34$ lies between $A$ at $t=0$ and $B$ at $t=1$, so the plane is crossed between the two given points.
One subtraction for the direction, one equation solved for the crossing.
Starting from $B$ instead of $A$, or using $\overrightarrow{BA}$, gives a different pair of $(\mathbf r_0,\mathbf v)$ but the very same line.
Checkpoint
§06.6 — reading a line off its symmetric form●○○○○
Thirty seconds. Direction downstairs, point upstairs.
Given
The line $\dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z}{4}$.
Find
(a) Give a direction vector and one point on the line.
Hint 1/4
Separate the two things the symmetric form stores: which way the line goes, and one place it passes through.
Hint 2/4
The denominators are the direction; the point is where each numerator equals zero.
Hint 3/4
Denominators $2,-1,4$ give the direction; numerators $x-1$, $y+3$, $z$ vanish at $(1,-3,0)$.
Hint 4/4
So a direction is $\langle 2,-1,4\rangle$ and a point is $(1,-3,0)$.
Show solutionDenominators are the direction
$$\mathbf v=\langle 2,-1,4\rangle$$
the three denominators are the direction components
Numerators give the point
$$x-1,\ y+3,\ z\ \Rightarrow\ (1,-3,0)$$
each numerator is zero at the point, so the point is where each bracket vanishes
Fix a plane with a point and a perpendicular normal, and measure distances to it.
A line needed a point and a direction to follow. A plane needs a point and the one direction it must avoid: the normal, perpendicular to everything lying in the plane.
RulePlane from a point and a normal, and distance
Conditions
The normal $\mathbf n=\langle a,b,c\rangle$ is nonzero.
A point $\mathbf r$ is on the plane exactly when the arrow from $\mathbf r_0$ to it is perpendicular to the normal, so their dot product is zero. Multiplying out, the coefficients of $x,y,z$ are the normal. The distance from any point to the plane is the plane's equation evaluated at the point, in size, divided by the length of the normal.
A plane is pinned by one $\textcolor{#1f6feb}{\text{point on it}}$ and its $\textcolor{#128a5a}{\text{normal}}$; the $\textcolor{#d1690a}{\text{distance}}$ from an outside point is measured straight along that normal, not slantwise.
Looks like this, but is not
The planes $2x-y+3z=5$ and $2x-y+3z=9$ have different right-hand sides, so they might be expected to cross somewhere.
They share the normal $\langle 2,-1,3\rangle$, so they are parallel and never meet; the different constants only slide the plane along its normal. Same normal means parallel, whatever the constant.
Plane through (1, 2, -1) with normal (3, -2, 1)
The normal is handed to us, so this is one point-normal line written out and tidied. Building a normal from two directions in the plane is the cross product, and that is next section.
Given
point $(1,2,-1)$
normal $\mathbf n=\langle 3,-2,1\rangle$
Find
the equation of the plane
SolutionPoint-normal form
$$3(x-1)-2(y-2)+1(z-(-1))=0$$
each coefficient is a component of the normal; each subtraction uses the point
Expand and collect
$$3x-3-2y+4+z+1=0$$
multiply out the three brackets
$$3x-2y+z+2=0$$
collect the constants; the $x,y,z$ coefficients are still the normal
Answer $$\boxed{3x-2y+z+2=0}$$
Check
Check the given point: $3(1)-2(2)+(-1)+2=3-4-1+2=0$, so $(1,2,-1)$ lies on the plane.
One point-normal expansion.
Read backwards, the coefficients $3,-2,1$ in the final equation are the normal again: in $ax+by+cz+d=0$ the normal is always $\langle a,b,c\rangle$.
Distance from (1, 1, 1) to the plane 2x - 2y + z = 6
The distance formula is the plane's own expression evaluated at the point, made positive, over the length of the normal.
Given
point $(1,1,1)$
plane $2x-2y+z=6$, i.e. $2x-2y+z-6=0$
Find
the distance from the point to the plane
SolutionEvaluate the plane expression at the point
$$2(1)-2(1)+(1)-6=2-2+1-6=-5$$
put the point into $ax+by+cz+d$; the sign tells which side, its size feeds the distance
the normal is $\langle 2,-2,1\rangle$, a clean length
$$D=\frac{\vert-5\vert}{3}=\frac{5}{3}$$
the size of the evaluation over the length of the normal
Answer $$\boxed{D=\tfrac{5}{3}}$$
Check
Order-of-magnitude check: the point is close to the plane, and $\tfrac53\approx1.7$ is a small distance, consistent with the point sitting just off it.
One evaluation, one magnitude, one division.
The sign of the evaluation, here negative, says the point is on the opposite side of the plane from the normal; the distance itself only uses the size.
Angle between two planes
The angle between planes is the angle between their normals, so a plane problem becomes a dot-product problem.
you have two points and want a distance, or an expanded equation you suspect is a sphere.
Subtract coordinate by coordinate
Form $\Delta x$, $\Delta y$, $\Delta z$. The order of subtraction does not matter; each is about to be squared.
Root the sum of the three squares
The distance is $\sqrt{\Delta x^{2}+\Delta y^{2}+\Delta z^{2}}$, never with a term missing.
For a sphere, complete the square once per variable
Group $x$ terms, $y$ terms, $z$ terms, complete each square and carry the subtracted constants to the right; the right side becomes $r^{2}$.
Where it goes wrong
Leaving off the $z$ term and using the plane formula.
Reading the right-hand number as $r$ instead of $r^{2}$.
From two points to a vector and its length
a problem gives points but the method needs a direction or a displacement.
Head minus tail
$\overrightarrow{PQ}=Q-P$, component by component. Decide first which point is the head.
Length is the distance
$\vert\overrightarrow{PQ}\vert=\sqrt{\text{sum of squared components}}$, the same number as the distance from $P$ to $Q$.
Where it goes wrong
Computing $P-Q$ and getting the arrow backwards.
Taking the magnitude componentwise instead of rooting the sum of squares.
Building a unit vector, or a vector of a chosen length
you need a pure direction, or a vector of a specified length in a known direction.
Find the length
Compute $\vert\mathbf a\vert$ once, for the whole vector.
Divide the vector by its length
$\hat{\mathbf a}=\mathbf a/\vert\mathbf a\vert$ is the unit vector; expect fractions.
Scale to the length you want
A vector of length $L$ in that direction is $L\hat{\mathbf a}$.
Where it goes wrong
Skipping the division and calling $\mathbf a$ its own unit vector.
Dividing each component by a different number.
The angle between two vectors
an angle is asked for, or a perpendicularity test is needed.
Dot product from components
Multiply matching slots, add to one number. If it is zero you are done: the angle is a right angle.
Divide by the two lengths
$\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{\vert\mathbf a\vert\,\vert\mathbf b\vert}$, a product of lengths in the denominator, not a sum.
Take the arccosine and read the sign
A negative cosine is obtuse, a positive one acute; report the angle in $[0,\pi]$.
Where it goes wrong
Using $\mathbf a\cdot\mathbf b$ itself as the cosine.
Dividing by the sum of the lengths rather than the product.
Projection and work
you need the part of one vector along another, or the work done by a force.
Dot product and length of the base vector
Compute $\mathbf a\cdot\mathbf b$ and $\vert\mathbf a\vert$; you are projecting onto $\mathbf a$.
Scalar projection divides by the length
$\operatorname{comp}_{\mathbf a}\mathbf b=\dfrac{\mathbf a\cdot\mathbf b}{\vert\mathbf a\vert}$, a number.
Vector projection divides by the square
$\operatorname{proj}_{\mathbf a}\mathbf b=\dfrac{\mathbf a\cdot\mathbf b}{\vert\mathbf a\vert^{2}}\mathbf a$, an arrow along $\mathbf a$. Work is simply $\mathbf F\cdot\mathbf d$.
Where it goes wrong
Using $\vert\mathbf a\vert$ where the vector projection needs $\vert\mathbf a\vert^{2}$.
Returning a vector for the scalar projection.
Writing a line
you have a point and a direction, or two points, and need the equations of the line.
Secure a point and a direction
From two points, the direction is their difference.
Parametric: base plus t times direction
$x=x_0+at$, and likewise for $y$ and $z$.
Symmetric: solve each for t and equate
$\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$; if a direction component is zero, write that coordinate as a constant instead of dividing.
Where it goes wrong
Reading the denominators as a point.
Dividing by a zero direction component in the symmetric form.
Writing a plane and measuring distance to it
you have a point and a normal, or need an angle between planes, or a point-to-plane distance.
Point-normal form
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$, then expand; the coefficients are the normal.
Angle between planes is the angle between normals
Apply the dot-product angle formula to $\mathbf n_1$ and $\mathbf n_2$, and take the acute answer.
Distance from a point
$D=\dfrac{\vert ax_1+by_1+cz_1+d\vert}{\sqrt{a^{2}+b^{2}+c^{2}}}$; keep the $+d$ before taking the size.
Where it goes wrong
Treating the coefficients as a direction lying in the plane.
Forgetting the constant $d$ in the distance numerator.
Neither vector is a multiple of the other, so they are certainly not parallel; the zero dot product places them at a right angle.
A parallel pair
Now the ratios are the test.
Given
$\mathbf u=\langle 2,-1,2\rangle$
$\mathbf w=\langle 6,-3,6\rangle$
Find
whether the vectors are parallel
SolutionLook for a common factor
$$\langle 6,-3,6\rangle=3\langle 2,-1,2\rangle$$
every component of $\mathbf w$ is three times that of $\mathbf u$
$$\Rightarrow\ \mathbf w=3\mathbf u$$
one vector is a scalar multiple of the other
Answer $$\boxed{\text{parallel}}$$
Check
Their dot product is $12+3+12=27$, far from zero, confirming they are not perpendicular but pointing the same way.
Perpendicular and parallel are opposite extremes: a zero dot product versus a constant ratio between the components.
How to tell them apart
Test the dot product first: zero means perpendicular. If it is not zero, check whether one vector is a scalar multiple of the other; if so they are parallel, otherwise neither.
Scaffolding comes off
The common skeleton
Write both vectors in components, lined up slot by slot.
Compute the dot product: multiply matching slots and add to a single number.
Compute the two magnitudes, each the root of a sum of squares.
Divide: $\cos\theta$ is the dot product over the product of the magnitudes.
Take the arccosine, and check the sign of the cosine against the angle you get.
1 · fully worked
Fully worked: the angle between (1, 0, 1) and (1, 1, 0)
Every line carries its reason. Read the reasons, because they are what the next rung takes away.
the standard cosine one half, an acute angle as the positive dot product promised
Answer $$\boxed{\theta=60^{\circ}}$$
Check
The positive dot product and the acute $60^{\circ}$ agree, which is the sign check doing its job.
One dot product, two magnitudes, one arccosine.
The denominator is a product of magnitudes; that single fact is the one most often lost on the rungs below.
2 · you write the reasoning
Easier this time: the vectors are perpendicular, so the arccosine is trivial once you see it. The algebra is done for you. Your job is the missing column: write why each line is allowed, in your own words, before opening the model reasons. Writing the reason is what transfers; copying the algebra is not.
Line 2: the dot product is matching slots multiplied and added; the middle term is $(-1)(2)=-2$, and the sum lands on $0$.
$\cos\theta=0$
reasoning
Line 3: a zero dot product forces $\cos\theta=0$ regardless of the magnitudes, because the numerator of the cosine is the dot product. No lengths are needed.
$\theta=90^{\circ}$
reasoning
Line 4: on $[0,\pi]$ the only angle with cosine zero is a right angle, so $\theta=90^{\circ}$. This is why a zero dot product is read as perpendicularity on sight.
3 · find the buried error
Harder than the rung above, and nobody is checking now. The work below finds the angle between $\mathbf a=\langle 2,2,1\rangle$ and $\mathbf b=\langle -1,2,2\rangle$ and contains exactly two errors. Each step looks locally reasonable; the errors do not announce themselves as jumps.
Sign check: the dot product is positive, so the angle is acute, and $68^{\circ}$ is indeed below $90^{\circ}$.
Both magnitudes were whole numbers, which is the only easy part; the denominator is still their product, never their sum.
Full exam-style question
A full triangle in space: length, angle, projection, lineexam format
The kind of multi-part question a quiz builds from three points: one figure, four skills. Each part reuses the answer before it.
Given
$A=(1,0,1)$
$B=(3,2,1)$
$C=(4,1,2)$
Find
(a) $\overrightarrow{AB}$ and its length; (b) the angle at $A$; (c) the scalar projection of $\overrightarrow{AC}$ onto $\overrightarrow{AB}$; (d) the line through $A$ and $B$
The scalar projection $2\sqrt2$ equals $\vert\overrightarrow{AB}\vert$ exactly, which reads as: $\overrightarrow{AC}$ reaches as far along $\overrightarrow{AB}$ as $B$ itself does. Because the direction has $z=0$, the whole line sits in the plane $z=1$, consistent with $A$ and $B$ both having $z=1$.
Four parts, but only one new dot product after part (a).
Notice the reuse: the dot product $8$ from part (b) is the numerator again in part (c), and $\overrightarrow{AB}$ from part (a) is the direction in part (d). Exam questions are built to reward doing part (a) carefully.
Practice
A · concept 4 questions
1§06.4 — perpendicular is not opposite●●○○○
A wording trap. Perpendicular and opposite are different relationships, and the dot product only tests one of them.
Given
Two nonzero vectors with $\mathbf a\cdot\mathbf b=0$.
Find
(a) True or false: a zero dot product means the two vectors point in opposite directions?
Hint 1/4
Ask what angle a zero cosine corresponds to, and compare it with the angle for opposite directions.
Hint 2/4
$\mathbf a\cdot\mathbf b=\vert\mathbf a\vert\vert\mathbf b\vert\cos\theta$, so a zero dot product forces $\cos\theta=0$.
Hint 3/4
That is $\theta=90^{\circ}$; opposite directions would be $\theta=180^{\circ}$ with a negative dot product.
that coordinate is constant along the line, written separately
Answer $$\boxed{\text{False}}$$
Check
A line running parallel to a coordinate plane always has a zero component and needs the constant form.
B · computation 6 questions
1§06.1 — a distance and a sphere●●○○○
Two standard computations of the first block in one question.
Given
$P=(2,3,6)$ and the origin $O$.
The surface $x^{2}+y^{2}+z^{2}+2x-4y-6z=11$.
Find
(a) Find $\vert OP\vert$.
(b) Find the centre and radius of the surface.
Hint 1/4
Treat the two parts separately; the first is one magnitude, the second is three completions of the square.
Hint 2/4
Distance to the origin is $\sqrt{x^{2}+y^{2}+z^{2}}$; a sphere in standard form reads its centre off the brackets and its radius as the root of the right side.
Hint 3/4
Here $\sqrt{4+9+36}=7$, and grouping gives $(x+1)^{2}+(y-2)^{2}+(z-3)^{2}=25$.
Hint 4/4
So $\vert OP\vert=7$, centre $(-1,2,3)$, radius $5$.
Since $\mathbf b$ lies along the $x$ axis, $\theta$ is the angle $\mathbf a$ makes with the $x$ axis, whose cosine is the $x$ component over the length, $3/5$, agreeing exactly.
5§06.5 — a projection that comes out clean●●●○○
Scalar and vector projection with a deliberately whole result.
Given
$\mathbf a=\langle 1,2,2\rangle$
$\mathbf b=\langle 5,2,0\rangle$
Find
(a) Find $\operatorname{comp}_{\mathbf a}\mathbf b$ and $\operatorname{proj}_{\mathbf a}\mathbf b$.
Hint 1/4
Compute the two shared ingredients first, then feed them into the two different formulas.
Hint 2/4
Scalar projection is $\dfrac{\mathbf a\cdot\mathbf b}{\vert\mathbf a\vert}$; vector projection is $\dfrac{\mathbf a\cdot\mathbf b}{\vert\mathbf a\vert^{2}}\mathbf a$.
Hint 3/4
Here $\mathbf a\cdot\mathbf b=9$ and $\vert\mathbf a\vert=3$, so $\vert\mathbf a\vert^{2}=9$.
Hint 4/4
So the scalar projection is $3$ and the vector projection is $\langle 1,2,2\rangle$.
The leftover $\mathbf b-\operatorname{proj}_{\mathbf a}\mathbf b=\langle 4,0,-2\rangle$ is perpendicular to $\mathbf a$: $\langle 4,0,-2\rangle\cdot\langle 1,2,2\rangle=4+0-4=0$.
6§06.6 — a line through two points●●●○○
Two points to a direction, then all three forms and one crossing.
Given
$A=(2,-1,4)$
$B=(5,3,-2)$
Find
(a) Write the parametric equations of the line through $A$ and $B$.
(b) Find where it crosses the plane $z=0$.
Hint 1/4
Get the direction from the two points before writing any equation.
Hint 2/4
The direction is $B-A$; parametric equations are each coordinate's start plus $t$ times its direction component.
Hint 3/4
Here $\mathbf v=\langle 3,4,-6\rangle$, and $z=4-6t=0$ gives $t=\tfrac23$.
Hint 4/4
So the line is $x=2+3t,\ y=-1+4t,\ z=4-6t$, crossing at $\left(4,\tfrac53,0\right)$.
evaluate the plane expression at the origin, over the length of the normal
Answer $$\boxed{2x-2y+z+3=0,\quad D=1}$$
Check
Check the given point on the plane: $2(1)-2(2)+(-1)+3=2-4-1+3=0$; and the numerator at the origin is the constant $3$, giving the clean distance $1$.
4§06.7 — the angle between two planes●●●●○
The angle between planes is the angle between normals, with one catch about which of the two angles to report.
Given
Plane $2x-y+2z=3$ with normal $\langle 2,-1,2\rangle$.
Plane $x+2y-2z=1$ with normal $\langle 1,2,-2\rangle$.
Find
(a) What is the (acute) angle between the two planes?
Hint 1/4
Reduce the plane question to a vector one: the angle between planes is the angle between their normals.
Hint 2/4
Compute $\cos\alpha=\dfrac{\mathbf n_1\cdot\mathbf n_2}{\vert\mathbf n_1\vert\vert\mathbf n_2\vert}$, then take the acute angle by using the size of the cosine.
Hint 3/4
Here $\mathbf n_1\cdot\mathbf n_2=-4$ and both lengths are $3$, so $\vert\cos\theta\vert=\tfrac49$.
Hint 4/4
So the acute angle is $\arccos\tfrac49\approx 63.6^{\circ}$.
Because a normal can point either way, the sign of the cosine is not meaningful for planes; taking the size $\tfrac49$ gives the acute answer directly.
D · interleaved 3 questions
1§06.1 — a sequence of points settling down●●●○○
A sequence question wearing a vector coat: it uses limits from the series unit and distance from this one, and you have to spot which tool each part wants.
Given
The points $P_n=\left(\dfrac1n,\ 1-\dfrac1n,\ \dfrac{(-1)^{n}}{n}\right)$ for $n=1,2,3,\dots$
Find
(a) Do the points $P_n$ approach a single point as $n\to\infty$, and if so which?
(b) How far is that limiting point from the origin?
Hint 1/4
Decide which unit each part belongs to: part (a) is a limit of a sequence, part (b) is a distance.
Hint 2/4
A sequence of points converges to the point of coordinatewise limits, and distance to the origin is $\sqrt{x^2+y^2+z^2}$.
Hint 3/4
Here $\tfrac1n\to 0$, $1-\tfrac1n\to 1$, and $\tfrac{(-1)^n}{n}\to 0$, so the limit is $(0,1,0)$.
The third coordinate alternates in sign but its size $1/n$ still tends to zero, so the alternation does not stop convergence; the limit sits one unit up the $y$ axis.
2§06.3 — a geometric series of one vector●●●○○
A geometric series from the earlier unit, but the terms are a fixed vector scaled down; the section-6 skills are the scaling and the magnitude.
Given
The sum $\displaystyle\sum_{n=1}^{\infty}\left(\tfrac13\right)^{n}\mathbf v$ with $\mathbf v=\langle 2,-2,1\rangle$.
Find
(a) Find the sum as a single vector.
(b) Find its magnitude.
Hint 1/4
Separate the two units: summing $(1/3)^n$ is the series skill, scaling and measuring the vector is this section's.
Hint 2/4
A constant vector comes out of the sum, leaving a geometric series $\sum (1/3)^n=\dfrac{1/3}{1-1/3}$.
Hint 3/4
That sum is $\tfrac12$, so the vector sum is $\tfrac12\langle 2,-2,1\rangle$, and $\vert\mathbf v\vert=3$.
Hint 4/4
So the sum is $\left\langle 1,-1,\tfrac12\right\rangle$ with length $\tfrac32$.
Show solution(a) Pull the constant vector out of the sum
Directly, $\vert\langle 1,-1,\tfrac12\rangle\vert=\sqrt{1+1+\tfrac14}=\sqrt{9/4}=\tfrac32$, matching the shortcut through $\tfrac12\vert\mathbf v\vert$.
3§06.4 — an angle that shrinks to zero●●●●○
An angle from this section evaluated along a sequence from the earlier one; you decide it is a dot-product problem, then take a limit.
Given
For each $n$, the angle $\theta_n$ between $\langle 1,0\rangle$ and $\left\langle 1,\tfrac1n\right\rangle$.
As $n$ grows, $\left\langle 1,\tfrac1n\right\rangle$ approaches $\langle 1,0\rangle$, so the angle between them must close to zero, which the cosine tending to one confirms.
Mistake ledger (16 entries)
⚠ Only two of the three squares in a distance
the plane formula is the reflex, and the $z$ term is easy to drop
Close the page. On one sheet, from memory: the distance formula in space; how to turn two points into a vector and find its length; the two faces of the dot product and the angle formula that comes from them; the scalar and vector projection, and which one divides by the length and which by its square; and how a point plus a direction gives a line while a point plus a normal gives a plane. Then open the formula card and mark what was missing. What you missed is your revision list; the rest of this section is not.
Find the distance between two points in space and the centre and radius of a sphere given in expanded form?
c-space
Turn two points into the vector between them, in the right order, and compute its magnitude?
c-vectors
Add and scale vectors, and build the unit vector in a given direction without leaving out the division?
c-vector-algebra
Compute a dot product, turn it into an angle with the lengths in the denominator, and test for perpendicularity in one line?
c-dot-product
Find both projections of one vector onto another, keeping the length and its square straight, and compute work as a dot product?
c-projection
Write a line from a point and a direction or from two points, and read a point and a direction back off the symmetric form?
c-lines
Write a plane from a point and a normal, find the angle between two planes, and the distance from a point to a plane?
c-planes
Glossary (14 terms)
konum vektörü
The vector from the origin to a point; it carries the same three numbers as the point but is an arrow rather than a place.
component formbileşen biçimi
A vector written as $\langle a_1,a_2,a_3\rangle$, listing how far it moves along each axis.
magnitudebüyüklük
The length of a vector, $\sqrt{a_1^{2}+a_2^{2}+a_3^{2}}$; a non-negative number.
unit vectorbirim vektör
A vector of length one, obtained by dividing a nonzero vector by its own magnitude.
birim temel vektörler
The vectors $\mathbf i,\mathbf j,\mathbf k$ of length one along the $x$, $y$ and $z$ axes.
dot productskaler çarpım
The number $a_1b_1+a_2b_2+a_3b_3$, equal to $\vert\mathbf a\vert\vert\mathbf b\vert\cos\theta$, which carries the angle between two vectors.
dik
Perpendicular; two nonzero vectors are orthogonal exactly when their dot product is zero.
scalar projectionskaler izdüşüm
The signed length of the part of one vector that lies along another, $\dfrac{\mathbf a\cdot\mathbf b}{\vert\mathbf a\vert}$.
vector projectiondik izdüşüm
The actual vector part of $\mathbf b$ lying along $\mathbf a$, a multiple of $\mathbf a$.
normal vektör
A vector perpendicular to a plane; its components are the coefficients in the plane's equation.
direction vectordoğrultu vektörü
A vector pointing along a line; any nonzero multiple of it serves equally well.
parametric equationsparametrik denklemler
Equations giving each coordinate of a line as a base value plus $t$ times a direction component.
symmetric equationssimetrik denklemler
The form of a line obtained by eliminating the parameter, three fractions set equal.
sphereküre
The set of points a fixed distance $r$ from a centre, with equation a squared distance equal to $r^{2}$.
What comes next
§07 · The cross product, vector functions, and partial derivatives
This section built a plane only when a normal was handed to us. Next comes the tool that makes a normal out of two directions lying in the plane, the cross product, along with the calculus of vector-valued functions and the first partial derivatives.
Sources
James Stewart, Calculus, Ninth Edition — sections 12.1, 12.2, 12.3 and 12.5 Section numbers are quoted only as they appear on the week 6 line of the syllabus. Section 12.4, the cross product, is on the next week's line and is not covered here.
Course syllabus: assessment weights The weights the syllabus attaches to this material are quizzes at 10% and homework at 10%.
Conventions used on this page Vectors are bold with components in angle brackets; magnitudes are non-negative; the dot product is a scalar; the angle between vectors is taken in $[0,\pi]$; exact values are kept unless a decimal is asked for.