MATH 133 dot product and planes: revision sheet

Q: What is the difference between the projection and the scalar component?

The projection of u on v is a vector: it points along v and has the length of the shadow of u on that line. The scalar component is a single signed number, the dot product divided by the length of v. Its absolute value is the length of the projection, and its sign says whether the shadow points with v or against it.

Q: How do I find the distance from a point to a plane in MATH 133?

Write the plane as ax plus by plus cz equals d, so that its normal is (a, b, c). Substitute the point into the left side, subtract d, take the absolute value, and divide by the length of the normal. The justification asked in exams is that this is the length of the projection, on the normal, of any vector joining a point of the plane to the given point.

Q: Why do we use the normal vectors to find the angle between two planes?

Because a plane contains infinitely many directions but has only one normal direction. Each normal stands at a right angle on its plane, so turning one plane onto the other turns the normals by the same angle. Take the absolute value of the dot product of the normals so that the answer is the acute angle, since the sign of a normal is arbitrary.

Q: How can I tell if a line is parallel to a plane or lies in it?

Compute the dot product of the direction of the line with the normal of the plane. If it is not zero, the line crosses the plane in exactly one point. If it is zero, the line is parallel to the plane, and you substitute one point of the line into the equation: if the point satisfies it the whole line lies in the plane, otherwise the line never meets it.

The thread of the chapter

Every length and every angle in this chapter is a projection onto one well-chosen direction. For a line you project on its direction $\mathbf{d}$ and keep what is LEFT; for a plane you project on its normal $\mathbf{n}$ and keep the PROJECTION. A plane is known by its normal, never by a vector lying in it.

Topics covered

Dot product
Angle between two vectors
Orthogonal vectors
Orthogonal projection
Scalar component
Normal vector
Scalar equation of a plane
Line and plane intersection
Point to line distance
Point to plane distance
Parallel planes
Angle between two planes

The essentials

The dot product, the projection, and the two halves

• $\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 = \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta$ , a NUMBER, with $\theta \in [0, \pi]$ . Its sign gives the type of angle: positive acute, zero right, negative obtuse.
• $\mathbf{u} \cdot \mathbf{v} = 0$ with both vectors non-zero means ORTHOGONAL. It never means that one of them is zero.
• Projection on $\mathbf{v}$ : $\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \dfrac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v}$ , a VECTOR along $\mathbf{v}$ . Scalar component: $\dfrac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|}$ , a signed number.
• The leftover $\mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u}$ is orthogonal to $\mathbf{v}$ , and $\|\mathbf{u}\|^2 = \|\operatorname{proj}_{\mathbf{v}}\mathbf{u}\|^2 + \|\mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u}\|^2$ .
• Distance from $Q$ to a LINE through $P$ with direction $\mathbf{d}$ : project $\overrightarrow{PQ}$ on $\mathbf{d}$ and take the length of the LEFTOVER. Distance from $Q$ to a PLANE through $P$ with normal $\mathbf{n}$ : project $\overrightarrow{PQ}$ on $\mathbf{n}$ and take the length of the PROJECTION.

Same

P

and

Q

, two objects. For the line, the part of

\overrightarrow{PQ}

along

\mathbf{d}

is thrown away and the leftover is the distance; for the plane, the part along

\mathbf{n}

IS the distance.

The point $P$ used to describe the line or the plane is arbitrary, so a correct distance never depends on it. That is the test for which half to keep: the leftover for a line, the projection on $\mathbf{n}$ for a plane, are the two quantities that do not move when $P$ moves.

The sign of the dot product reads the angle

• $\mathbf{u} \cdot \mathbf{v} > 0$ : acute angle. $(2, 1, -2) \cdot (1, 1, 0) = 3$ , and $\cos\theta = \frac{3}{3\sqrt{2}}$ gives $\theta = 45^{\circ}$ .
• $\mathbf{u} \cdot \mathbf{v} = 0$ : right angle. $(1, 2, 2) \cdot (2, -1, 0) = 2 - 2 + 0 = 0$ .
• $\mathbf{u} \cdot \mathbf{v} < 0$ : obtuse angle. $(1, 0, 1) \cdot (-1, 1, 0) = -1$ , and $\cos\theta = \frac{-1}{\sqrt{2}\sqrt{2}} = -\frac{1}{2}$ gives $\theta = 120^{\circ}$ .
• The sign alone classifies the angle; its size also depends on the two lengths, so $3$ is not a bigger angle than $-1$ in any sense.

Same vector

\mathbf{v}

, three positions of

\mathbf{u}

: the dot product is positive while

\mathbf{u}

leans forward, zero at the right angle, negative once it leans back.

A plane is a point and a normal

• Plane through $P$ with normal $\mathbf{n} = (a, b, c)$ : $\mathbf{n} \cdot (X - P) = 0$ , that is $ax + by + cz = d$ with $d = \mathbf{n} \cdot P$ . The coefficients of $x$ , $y$ , $z$ ARE the normal, once everything is on the left.
• Plane perpendicular to a line: normal $=$ direction of the line. Plane parallel to a given plane: SAME normal, new $d$ .
• Two planes: proportional normals means parallel or identical; otherwise the angle between the planes is $\cos\theta = \dfrac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{\|\mathbf{n}_1\|\,\|\mathbf{n}_2\|}$ , acute. Perpendicular planes: $\mathbf{n}_1 \cdot \mathbf{n}_2 = 0$ .
• Line $P + t\mathbf{d}$ and plane: $\mathbf{d} \cdot \mathbf{n} \neq 0$ gives exactly one point; $\mathbf{d} \cdot \mathbf{n} = 0$ gives a line PARALLEL to the plane, inside it or not depending on whether $P$ satisfies the equation.
• Distance from $(x_0, y_0, z_0)$ to $ax + by + cz = d$ : $\dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$ . Between parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ with the SAME $(a, b, c)$ : $\dfrac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$ .

Every vector

X - P

that stays in the plane is orthogonal to

\mathbf{n}

: that single right angle is the equation

\mathbf{n} \cdot (X - P) = 0

No three-point plane and no line of intersection of two planes here: both need the cross product, which is the next chapter.

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. Dividing by the length instead of its square in a projection

2 marks, and every distance computed from it

What not to write

“ $\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \dfrac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|}\,\mathbf{v} = \dfrac{8}{3}(1, 2, 2)$ for $\mathbf{u} = (4, 0, 2)$ , $\mathbf{v} = (1, 2, 2)$ .”

What to write

“ $\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \dfrac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v} = \dfrac{8}{9}(1, 2, 2)$ .”

In green the projection

\frac{8}{9}\mathbf{v}

, which stops at the foot of the perpendicular from

\mathbf{u}

. In red

\frac{8}{3}\mathbf{v}

, three times too long: it overshoots the tip of

\mathbf{u}

itself.

Why: The wrong version changes when $\mathbf{v}$ is doubled, and a shadow on a line cannot depend on which arrow names the line. With $\|\mathbf{v}\|^2$ the scale factor appears twice on top and twice below and cancels. The check: $\mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u}$ must dot to zero with $\mathbf{v}$ .

2. Taking the length of the projection for the distance to a line

the whole part

What not to write

“The line through the origin with direction $(2, 1, 2)$ and $Q = (5, 4, 2)$ : $\operatorname{proj}_{\mathbf{d}}\overrightarrow{OQ} = (4, 2, 4)$ , so the distance is $6$ .”

What to write

“The leftover is $(5, 4, 2) - (4, 2, 4) = (1, 2, -2)$ , orthogonal to $\mathbf{d}$ , so the distance is $3$ and the closest point is $(4, 2, 4)$ .”

The projection stops at

(4, 2, 4)

6

units ALONG the line from

O

. The distance is the green side,

3

, and the right triangle checks it:

45 = 36 + 9

Why: The projection tells you WHERE to stop on the line; it measures a length ALONG the line, which depends on the chosen point. The distance is the other side of the right triangle, and Pythagoras checks it: $45 = 36 + 9$ .

3. Forgetting the denominator in the distance to a plane

2 marks

What not to write

“Distance from $(1, 2, 3)$ to $2x - y + 2z = 1$ : $|2 - 2 + 6 - 1| = 5$ .”

What to write

“ $\dfrac{|2 - 2 + 6 - 1|}{\sqrt{4 + 1 + 4}} = \dfrac{5}{3}$ .”

Why: The numerator is the scalar component multiplied by $\|\mathbf{n}\|$ . Multiply the equation by $2$ and the wrong answer doubles while the plane has not moved: dividing by $\|\mathbf{n}\|$ is what makes the distance independent of how the equation is written.

4. Subtracting right-hand sides of parallel planes whose normals differ

the whole part

What not to write

“ $x - 2y + 2z = 1$ and $3x - 6y + 6z = 21$ are $\dfrac{|21 - 1|}{3} = \dfrac{20}{3}$ apart.”

What to write

“Dividing the second by $3$ : $x - 2y + 2z = 7$ . Same normal, so the distance is $\dfrac{|7 - 1|}{3} = 2$ .”

Why: The formula $\dfrac{|d_1 - d_2|}{\|\mathbf{n}\|}$ holds only when both equations use the SAME $(a, b, c)$ . The safe route needs no rescaling: take one point of the first plane, here $(1, 0, 0)$ , and use the point-to-plane formula on the second, $\dfrac{|3 - 21|}{9} = 2$ .

5. Reading a zero dot product between a line and a normal as perpendicular

the whole part, and every part built on it

What not to write

“ $(1, 1, 0) \cdot (1, -1, 5) = 0$ , so the line $(0, 0, 1) + t(1, 1, 0)$ is perpendicular to the plane $x - y + 5z = 2$ .”

What to write

“ $\mathbf{d} \cdot \mathbf{n} = 0$ , so the line is PARALLEL to the plane. Its point $(0, 0, 1)$ gives $5 \neq 2$ , so it does not meet the plane.”

\mathbf{d}

is orthogonal to

\mathbf{n}

, so the line runs alongside the plane, here above it since

5 > 2

. A line perpendicular to the plane would follow

\mathbf{n}

, the red dashes.

Why: $\mathbf{d}$ is orthogonal to the NORMAL, so it lies along the plane. Perpendicular to the plane would mean $\mathbf{d}$ is a multiple of $\mathbf{n}$ . And parallel is not the end of the answer: one substitution decides between no point and the whole line.

6. Giving the obtuse angle between two planes

1 mark

What not to write

“ $\mathbf{n}_1 = (1, 0, 1)$ , $\mathbf{n}_2 = (-1, 1, 0)$ , $\cos\theta = -\frac{1}{2}$ , so the planes make $120^{\circ}$ .”

What to write

“ $\cos\theta = \dfrac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{\|\mathbf{n}_1\|\,\|\mathbf{n}_2\|} = \dfrac{1}{2}$ , so the planes make $60^{\circ}$ .”

Cut across both planes: each appears as a line, each normal as an arrow. The normals make

120^{\circ}

, in red; the planes make the acute angle

60^{\circ}

, in green.

Why: The sign of a normal is arbitrary: writing the second plane with $(1, -1, 0)$ describes the same plane and gives $+\frac{1}{2}$ . The angle between two vectors lives in $[0, \pi]$ ; the angle between two planes, like the angle between two lines, is the acute one.

7. Taking a vertical gap for the distance to a slanted plane

2 marks

What not to write

“The roof $3y + 4z = 24$ is at $z = 5.25$ above $Q = (2, 1, 4)$ , so $Q$ is $1.25$ m from the roof.”

What to write

“ $\dfrac{|3 + 16 - 24|}{5} = 1$ m. The vertical gap, $1.25$ m, is longer, by the factor $\dfrac{1}{\cos\theta} = \dfrac{5}{4}$ .”

From

Q

, the dashed vertical segment meets the roof after

1.25

m, but the solid segment along the normal reaches it after only

1

m, at a right angle.

Why: The shortest segment to a plane runs along the NORMAL, not along the $z$ -axis. The two agree only for a horizontal plane, where $\mathbf{n}$ is vertical, which is exactly the one case where no formula was needed.

8. Reading the normal off an equation that is not in standard form

1 mark, plus every angle and distance computed with it

What not to write

“The plane $z = 3x - 2y + 1$ has normal $(3, -2, 1)$ .”

What to write

“Rewritten as $3x - 2y - z = -1$ , the plane has normal $(3, -2, -1)$ .”

Why: The normal is read only once $x$ , $y$ and $z$ are all on the left and the constant is alone on the right. The $1$ in the faux line is the constant, not a coordinate of anything, and the coefficient of $z$ is $-1$ , not missing.

Which method to choose

Which projection the question is asking for

Read what the statement gives, a line, a plane or two vectors, before choosing a formula

Left:

\mathbf{d} \cdot \mathbf{n} \neq 0

, the line pierces the plane once. Middle and right:

\mathbf{d} \cdot \mathbf{n} = 0

in both, and only a point of the line tells none from all.

If two vectors and the word angle → $\cos\theta = \dfrac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\,\|\mathbf{v}\|}$ ; the sign alone if only acute or obtuse is asked

Example: $(2, 1, -2) \cdot (1, 1, 0) = 3$ over $3\sqrt{2}$ gives $\frac{\pi}{4}$
If a point and a LINE $P + t\mathbf{d}$ → project $\overrightarrow{PQ}$ on $\mathbf{d}$ : the projection gives the closest point, the LEFTOVER gives the distance

Example: $(5, 4, 2)$ and $\mathbf{d} = (2, 1, 2)$ : closest point $(4, 2, 4)$ , distance $3$
If a point and a PLANE $ax + by + cz = d$ → project on $\mathbf{n}$ : distance $\dfrac{|ax_0 + by_0 + cz_0 - d|}{\|\mathbf{n}\|}$ , foot $Q - \dfrac{\mathbf{n} \cdot Q - d}{\|\mathbf{n}\|^2}\mathbf{n}$

Example: $(2, 4, 5)$ and $x + 2y + 2z = 2$ : distance $6$ , foot $(0, 0, 1)$
If a line and a plane → compute $\mathbf{d} \cdot \mathbf{n}$ FIRST; non-zero, substitute and solve for $t$ ; zero, test one point

Example: $\mathbf{d} \cdot \mathbf{n} = 4$ , then $4t + 2 = 6$ gives $t = 1$
If two planes → compare the normals: proportional, rescale and compute a distance; otherwise the acute angle from $|\mathbf{n}_1 \cdot \mathbf{n}_2|$

Example: $(1, 0, -1)$ and $(0, 1, -1)$ : $\cos\theta = \frac{1}{2}$ , so $60^{\circ}$

If a question seems to need a vector lying IN a plane, or two planes meeting in a line, it belongs to the cross product chapter: nothing on this sheet produces a normal from two directions.

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.

Distance from a point to a plane, justified

When to use it: Any question that says derive, justify or show, or that gives the plane by a point and a normal

1 Read the normal $\mathbf{n}$ from the equation written as $ax + by + cz = d$ , and compute $\|\mathbf{n}\|$ .
2 Name a point $P$ of the plane and CHECK it in the equation, in one line.
3 Compute $\overrightarrow{PQ}$ and $\overrightarrow{PQ} \cdot \mathbf{n}$ .
4 State the reason: the distance is the length of the projection of $\overrightarrow{PQ}$ on $\mathbf{n}$ , because the normal is the direction that leaves the plane fastest.
5 Conclude with $\dfrac{|\overrightarrow{PQ} \cdot \mathbf{n}|}{\|\mathbf{n}\|}$ and, if asked, the foot $F = Q - \dfrac{\overrightarrow{PQ} \cdot \mathbf{n}}{\|\mathbf{n}\|^2}\,\mathbf{n}$ , checked in the equation.

Concluding sentence

“The distance from $Q$ to the plane is the length of the projection of $\overrightarrow{PQ}$ on the normal: $\dfrac{|\overrightarrow{PQ} \cdot \mathbf{n}|}{\|\mathbf{n}\|} = \dfrac{9}{3} = 3$ .”

The trap: Writing the distance as $\|\overrightarrow{PQ}\|$ . The point $P$ was arbitrary, so its distance to $Q$ means nothing; only the component along $\mathbf{n}$ does.

Marking: Typically 1 mark for the normal and a checked point, 1 for the projection argument in words, 1 for the computation, and 1 for the foot checked in the equation. A bare formula with the right number earns the computation mark only.

Check before you hand in

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

The foot goes back into the equation

Any closest point on a plane, or intersection with a line, must satisfy the plane's equation. Substitute it: two seconds, and it catches a projection divided by $\|\mathbf{n}\|$ instead of $\|\mathbf{n}\|^2$ .

$(0, 0, 1)$ in $x + 2y + 2z = 2$ gives $2$ : correct. The foot $(-4, -8, -7)$ gives $-34$ : wrong, before any other check.
The leftover dots to zero

After a projection on $\mathbf{v}$ , compute $(\mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u}) \cdot \mathbf{v}$ . Anything but zero means the projection is wrong.

$(1, 2, -2) \cdot (2, 1, 2) = 2 + 2 - 4 = 0$ : the distance $3$ to the line stands.
The rescaling test for a distance

Multiply the plane's equation by $2$ in your head and redo the formula. A correct distance does not move; a distance that doubles has lost its denominator.

$Q = (2, 4, 5)$ and $2x + 4y + 4z = 4$ : $\frac{|4 + 16 + 20 - 4|}{6} = 6$ , the same $6$ as with $x + 2y + 2z = 2$ .
Cosines between minus one and one, plane angles below ninety

A cosine outside $[-1, 1]$ means a length was not square-rooted. A negative cosine for two PLANES means the absolute value was forgotten.

$\frac{3}{3\sqrt{2}} = 0.71$ is plausible; $\frac{3}{9 \times 2}$ would signal lengths left squared.

The typical problem, taken apart

The point of a plane closest to a given point

Find the distance from $Q = (2, 4, 5)$ to the plane $\Pi: x + 2y + 2z = 2$ , the point $F$ of $\Pi$ closest to $Q$ , and the equation of the plane through $Q$ parallel to $\Pi$ .

The plane

\Pi

, its normal

\mathbf{n}

and the point

Q

on the side

\mathbf{n}

points to: the shortest path from

Q

\Pi

runs along

\mathbf{n}

, not along an axis.

Step 1

Normal $\mathbf{n} = (1, 2, 2)$ , $\|\mathbf{n}\| = 3$ . A point of $\Pi$ : $P = (2, 0, 0)$ , since $2 + 0 + 0 = 2$ .

Why

Reading $\mathbf{n}$ first fixes the direction of every later move. Checking $P$ in the equation costs one line and protects the whole computation from a point that is not on the plane.

Step 2

$\overrightarrow{PQ} = (0, 4, 5)$ and $\overrightarrow{PQ} \cdot \mathbf{n} = 0 + 8 + 10 = 18$ .

Why

This one number carries everything: its sign says which side $Q$ is on, its size over $\|\mathbf{n}\|$ gives the distance, and its size over $\|\mathbf{n}\|^2$ gives the step back to the foot.

Step 3

Distance $= \dfrac{18}{3} = 6$ . The formula agrees: $\dfrac{|2 + 8 + 10 - 2|}{3} = 6$ .

Why

Writing the projection argument and then the formula shows the marker both the reason and the result; the agreement of the two is the first verification.

Step 4

$F = Q - \dfrac{18}{9}\,\mathbf{n} = (2, 4, 5) - (2, 4, 4) = (0, 0, 1)$ . Check: $0 + 0 + 2 = 2$ , and $\overrightarrow{FQ} = (2, 4, 4) = 2\mathbf{n}$ , of length $6$ .

Why

The step back uses $\|\mathbf{n}\|^2 = 9$ , not $\|\mathbf{n}\| = 3$ , because it multiplies the vector $\mathbf{n}$ , which already has length $3$ . The substitution in $\Pi$ is what proves it.

Step 5

Parallel plane through $Q$ : same normal, $x + 2y + 2z = 2 + 8 + 10 = 20$ . Distance between the two planes: $\dfrac{|20 - 2|}{3} = 6$ .

Why

This is a free cross-check: the plane through $Q$ parallel to $\Pi$ is at exactly the distance from $Q$ to $\Pi$ , and the two numbers must agree.

The conclusion, written out

“ $Q$ is at distance $6$ from $\Pi$ , the closest point is $F = (0, 0, 1)$ , and the plane through $Q$ parallel to $\Pi$ is $x + 2y + 2z = 20$ .”

The classic mistake on this problem: Stepping back by $\frac{18}{3}\mathbf{n} = (6, 12, 12)$ gives $F = (-4, -8, -7)$ , which gives $-4 - 16 - 14 = -34 \neq 2$ in the equation: the foot is off the plane, and the substitution says so at once.

Learn by heart

• $\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta$ ; zero means ORTHOGONAL, never that a vector is zero.
• $\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \dfrac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v}$ : the SQUARE of the length below.
• Distance to a line: project on $\mathbf{d}$ , keep the LEFTOVER. Distance to a plane: project on $\mathbf{n}$ , keep the PROJECTION.
• $ax + by + cz = d$ has normal $(a, b, c)$ , read only when everything is on the left.
• Line and plane: $\mathbf{d} \cdot \mathbf{n} \neq 0$ , one point; $\mathbf{d} \cdot \mathbf{n} = 0$ , PARALLEL, then test one point.
• Distance to a plane $= \dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$ ; parallel planes only after making the normals EQUAL.
• Angle between planes $=$ angle between normals, with $|\mathbf{n}_1 \cdot \mathbf{n}_2|$ : always acute.

Frequently asked questions

What is the difference between the projection and the scalar component?

The projection of u on v is a vector: it points along v and has the length of the shadow of u on that line. The scalar component is a single signed number, the dot product divided by the length of v. Its absolute value is the length of the projection, and its sign says whether the shadow points with v or against it.

How do I find the distance from a point to a plane in MATH 133?

Write the plane as ax plus by plus cz equals d, so that its normal is (a, b, c). Substitute the point into the left side, subtract d, take the absolute value, and divide by the length of the normal. The justification asked in exams is that this is the length of the projection, on the normal, of any vector joining a point of the plane to the given point.

Why do we use the normal vectors to find the angle between two planes?

Because a plane contains infinitely many directions but has only one normal direction. Each normal stands at a right angle on its plane, so turning one plane onto the other turns the normals by the same angle. Take the absolute value of the dot product of the normals so that the answer is the acute angle, since the sign of a normal is arbitrary.

How can I tell if a line is parallel to a plane or lies in it?

Compute the dot product of the direction of the line with the normal of the plane. If it is not zero, the line crosses the plane in exactly one point. If it is zero, the line is parallel to the plane, and you substitute one point of the line into the equation: if the point satisfies it the whole line lies in the plane, otherwise the line never meets it.

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Corrected exercises: Dot product, projections and planes, MATH 133 at McGill

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Revision sheet: dot product, projections and planes (MATH 133)

The essentials

The dot product, the projection, and the two halves

The sign of the dot product reads the angle

A plane is a point and a normal

The mistakes that cost marks

1. Dividing by the length instead of its square in a projection

2. Taking the length of the projection for the distance to a line

3. Forgetting the denominator in the distance to a plane

4. Subtracting right-hand sides of parallel planes whose normals differ

5. Reading a zero dot product between a line and a normal as perpendicular

6. Giving the obtuse angle between two planes

7. Taking a vertical gap for the distance to a slanted plane

8. Reading the normal off an equation that is not in standard form

Which method to choose

Which projection the question is asking for

How the answer is expected to be written

Distance from a point to a plane, justified

Check before you hand in

The typical problem, taken apart

The point of a plane closest to a given point

Learn by heart

Frequently asked questions

What is the difference between the projection and the scalar component?

How do I find the distance from a point to a plane in MATH 133?

Why do we use the normal vectors to find the angle between two planes?

How can I tell if a line is parallel to a plane or lies in it?

Corrected exercises: Dot product, projections and planes, MATH 133 at McGill

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