DSE Maths | Vectors and the dot product lesson plan (for tutors) | One dot product gives the angle and the projection

2026-11-04

Tutor lesson-prep guides · Secondary Maths

Students can find an angle with the cosine rule but treat a dot product as a formula to memorise, not a number that carries both length and angle. This lesson starts from the single value 14 and connects angle, perpendicularity, projection and plane equations to it.

Objectives — Students can: (1) compute a dot b = 14 for a = (3, 4) and b = (-2, 5), with |a| = 5, |b| = 5.385, giving an angle of 58.7 degrees; (2) solve for an unknown component from a zero dot product, for example (2, k) dot (3, -4) = 0 gives k = 1.5; (3) write the projection of a on b as (-0.97, 2.41) with scalar projection 2.60; (4) write the plane through (1, 2, -1) with normal (2, -1, 3) as 2x – y + 3z + 3 = 0.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Trigonometric ratios	In a right triangle, which two sides give cos theta?	Adjacent over hypotenuse
Cosine rule	With three sides known, how do you get an angle?	cos C equals (a squared plus b squared minus c squared) over 2ab
Line equations	What does a point plus a direction determine?	A straight line

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Opener	Two drones leave the same point along (3, 4) and (-2, 5). What is the angle between them?	Most students want to draw and measure
8-20 min	Two meanings of the dot product	Write a dot b = 3 times -2 plus 4 times 5 = 14 beside a dot b = \|a\|\|b\| cos theta	Substitute one into the other
20-30 min	Find the angle	cos theta = 14 over (5 times 5.385) = 0.5199	theta is about 58.7 degrees
30-40 min	Perpendicularity and unknowns	A zero dot product means 90 degrees, so solve for k	(2, k) dot (3, -4) gives k = 1.5
40-52 min	Projection and planes	Projection is (a dot b over \|b\| squared) times b; in 3D the normal does the same job	(-0.97, 2.41); 2x – y + 3z + 3 = 0
52-60 min	Exit check	Three items	Two or fewer correct means redo act two

03 Teaching script (read it out)

Scene 1: The dot product is one number answering two questions

First compute a dot b = 3 times -2 plus 4 times 5 = 14. It is a scalar with no direction. Then write a dot b = |a||b| cos theta with |a| = 5 and |b| = root 29 about 5.385, so cos theta = 14 over 26.926, about 0.5199, giving 58.7 degrees. Stress the sign: negative means the angle exceeds 90 degrees, zero means exactly perpendicular.

Board: a dot b = 14 beside a dot b = |a||b| cos theta

Scene 2: A projection is a shadow built straight from the dot product

Ask how much of a points along b. The answer is |a| cos theta, which equals (a dot b) over |b|, here 14 over 5.385, about 2.60. For a vector, multiply by the unit vector of b: (14 over 29) times (-2, 5) gives (-0.97, 2.41). The denominator is |b| squared, not |b|, because the unit vector already divided by |b| once.

Draw the projection, marking |a| cos theta = 2.60 and (-0.97, 2.41)

Scene 3: Three dimensions only add a component, and the normal is the key

Take a point P(1, 2, -1) and a normal n = (2, -1, 3). Every point X on the plane satisfies n dot (X – P) = 0, which expands to 2(x-1) – (y-2) + 3(z+1) = 0, that is 2x – y + 3z + 3 = 0. Remind them that the normal is perpendicular to every vector in the plane, so “perpendicular” in solid geometry almost always means “dot product is zero”.

Board: expand n dot (X – P) = 0 in four steps

04 Question types and worked steps

Type	Example	Steps	Answer
Calculate	Dot product and angle from components	Dot first, divide by the product of lengths, then inverse cosine	14 and 58.7 degrees
Unknown	Find component k given perpendicularity	Set the dot product to zero and solve the linear equation	k = 1.5
Projection	Vector projection of a on b	Compute a dot b over \|b\| squared, then multiply by b	(-0.97, 2.41)
Solid geometry	Plane equation from a normal and a point	Expand n dot (X – P) = 0	2x – y + 3z + 3 = 0

Sanity check (teach this): Check: cos theta = 0.5199 must lie between -1 and 1, so anything outside means a slip in the dot product or a length. Verify backwards: 5 times 5.385 times 0.5199 is 14, matching the direct dot product. The projection length must equal 14 over 5.385, about 2.60, and the squared components 0.97 and 2.41 give about 6.76, whose root is 2.60. Substituting P(1, 2, -1) into the plane gives 2 – 2 – 3 + 3 = 0.

05 Independent practice (3 questions)

Question	Answer	What to watch
a = (1, 2), b = (3, -1): find the angle	About 81.9 degrees	Dot is 1, product of lengths is root 50, about 7.071
(2, k) and (3, -4) are perpendicular: find k	k = 1.5	6 minus 4k equals 0
Projection of (3, 4) on (-2, 5)	(-0.97, 2.41)	Scalar projection 2.60

06 Common mistakes and what to say

Mistake	Student writes	Say this
Mixing up dot and cross products	“a dot b should be a vector”	“Dot gives a number, cross gives a vector; angles always use the dot product.”
Squaring dropped from the projection formula	“Just divide by \|b\|”	“The unit vector already divides by \|b\| once, so multiplying by b divides by \|b\| squared overall.”
Calculator left in radian mode	“Inverse cos gives 1.02”	“1.02 is radians; if the question wants degrees, switch to DEG first.”
Sign slip when expanding a plane equation	“2x – y + 3z + 3 becomes minus 3”	“Substitute the known point back; it must give zero if the plane passes through it.”

07 Homework (paper, section and question numbers)

Maths chapter index — All chapters
Practice paper — Questions 7 to 10: dot product, projection and planes

08 Exit check (last 5 minutes)

Question	Answer	Decision
What is (3, 4) dot (-2, 5)?	14	Wrong: redo act one
What is the dot product of two perpendicular vectors?	Zero	One wrong: redo before next lesson
Plane with normal (2, -1, 3) through (1, 2, -1)	2x – y + 3z + 3 = 0	Two wrong: restart act three

09 Tutor FAQ

Is cosine always needed for the dot product?

Only for angles; perpendicularity, projection and plane equations need only the component form.

Why divide by |b| squared for a projection?

Because the projection is the scalar projection times the unit vector of b, and the unit vector already divides by |b|.

Is the cross product needed in 3D?

It can produce a normal, but DSE questions usually give the normal or let students build the equation with a dot product.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.