1. “Teach A Level Maths” Vol. 2: A2 Core Modules
54: Applications of the Scalar Product
© Christine Crisp

2. Module C4
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3. Finding Angles between Vectors
The scalar product can be rearranged to find the angle between the vectors.
Notice how careful we must be with the lines under the vectors.
The r.h.s. is
the product of 2 vectors divided by
the product of the 2 magnitudes of the vectors

4. e.g. Find the angle between and
Solution:
Tip: If at this stage you get zero, STOP.
The vectors are perpendicular.
( 3s.f. )

5. Exercise
1. Find the angle between the following pairs of vectors.
(a)
and
(b)
and
(c)
and

6. Solutions:
(a)
and
( 3s.f. )

7. Solutions:
(b)
and
( 3s.f. )

8. Solutions:
(c)
and
The vectors are perpendicular.

9. When solving problems, we have to be careful to use the correct vectors.
e.g. The triangle ABC is given by
Find the cosine of angle ABC.
Solution:
We always sketch and label a triangle A B C
( any shape triangle will do )
Use BUT
the a and b of the formula are not the a and b of the question.
We need the vectors and

10. When solving problems, we have to be careful to use the correct vectors.
e.g. The triangle ABC is given by
Find the cosine of angle ABC.
Solution:
We always sketch and label a triangle B
( any shape triangle will do )
Use BUT C
the a and b of the formula are not the a and b of the question. A
We need the vectors and

13. We use the 2 direction vectors only since these define the angle.
Finding Angles between Lines
e.g. a
With lines instead of vectors, we have 2 possible angles. We usually give the acute angle.
( If the obtuse angle is found, subtract from )
We use the 2 direction vectors only since these define the angle.

14. e.g. Find the acute angle, a, between the lines
Solution:
and
where

15. e.g. Find the acute angle, a, between the lines
and
Solution:
where
and

16. e.g. Find the acute angle, a, between the lines
and
Solution:
where
and

17. e.g. Find the acute angle, a, between the lines
and
Solution:
where
and
(nearest degree)

18. Diagram

19. We can find the angle between 2 lines even if they are skew lines.O F A C B D E G

20. We can find the angle between 2 lines even if they are skew lines.
e.g. The line through C and A and the line through O and F D E G F O
To define the angle we just draw a line parallel to one line meeting the other. A C B

21. We can find the angle between 2 lines even if they are skew lines.
e.g. The line through C and A and the line through O and F D E G F O
To define the angle we just draw a line parallel to one line meeting the other. A C B
The direction vector of the new line is the same as the direction vector of one of the original lines so we don’t need to know whether or not the lines intersect.

22. SUMMARY
To find the angle between 2 vectors
Form the scalar product.
Find the magnitude of both vectors.
Rearrange to
and substitute.
To find the angle between 2 lines
Use the direction vectors only and apply the method above.
If the angle found is obtuse, subtract from .

23. Exercise
1. Find the acute angle between the following pairs of lines. Give your answers to the nearest degree.
(a)
and
(b)
and

24. Solutions:
(a)
where
and
(nearest whole degree)

25. (a)
where
and
(nearest whole degree)

26. M Q Another Application of the Scalar Product
Suppose we have a line,
and a point not on the line.
If we draw a perpendicular from the point to the line . . . M
we can find the coordinates of M, the foot of the perpendicular. x Q
The scalar product of QM . . .

27. M Q Another Application of the Scalar Product
Suppose we have a line,
and a point not on the line.
If we draw a perpendicular from the point to the line . . . M
we can find the coordinates of M, the foot of the perpendicular. x Q
The scalar product of QM
and the direction vector of the line . . .

28. M Q Another Application of the Scalar Product
Suppose we have a line,
and a point not on the line.
If we draw a perpendicular from the point to the line . . . M
we can find the coordinates of M, the foot of the perpendicular. x Q
The scalar product of QM
and the direction vector of the line
equals zero ( since the vectors are perpendicular )

29. M x Q
M is a point on the line so its position vector is given by one particular value of the parameter s.
So,
where
We can therefore substitute into and solve for s.

30. e.g. Find the coordinates of the foot of the perpendicular from the point to the line
Q (1, 2, 2)

31. Solution:
Q (1, 2, 2) x M

32. Q (1, 2, 2)
Solution: x M

33. Q (1, 2, 2)
Solution: x M

35. Finally we can find m by substituting for s in the equation of the line.
The coordinates of M are

36. M is the foot of the perpendicular and is a value of r so .
SUMMARY
To find the coordinates of the foot of the perpendicular from a point to a line:
Sketch and label the line and point Q
Substitute into , where
This is because it is so easy to substitute the wrong vectors into the equation.
M is the foot of the perpendicular and is a value of r so
is the direction vector of the line
Solve for the parameter, s
Substitute for s into the equation of the line
Change the vector m into coordinates.

37. Exercise
1. Find the coordinates of the foot of the perpendicular from the points given to the lines given:
(a)
and
(b)
and

38. (a)
Q (-1, 2, -8)
Solution: x M

39. Point is

40. (b)
Q (1, 1, -4)
Solution: x M

41. Point is

43. The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.
For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

44. SUMMARY
To find the angle between 2 vectors
Use the direction vectors only and apply the method above.
Form the scalar product.
Find the magnitude of both vectors.
Rearrange to
and substitute.
To find the angle between 2 lines
If the angle found is obtuse, subtract from .

45. e.g. Find the angle between and
Solution:
e.g. Find the angle between
and
( 3s.f. )
Tip: If at this stage you get zero, STOP.
The vectors are perpendicular.

46. When solving problems, we have to be careful to use the correct vectors.
e.g. The triangle ABC is given by
Find the cosine of angle ABC.
Solution:
( any shape triangle will do )
We always sketch and label a triangle A B C
Use BUT
the a and b of the formula are not the a and b of the question.
We need the vectors and

47. ( 3 s.f. )

48. We use the 2 direction vectors only since these define the angle.
Finding Angles between Lines
e.g.
With lines instead of vectors, we have 2 possible angles. We usually give the acute angle. a
We use the 2 direction vectors only since these define the angle.
( If the obtuse angle is found, subtract from )

49. e.g. Find the acute angle, a, between the lines
and
Solution:
where
(nearest degree)

50. M Q Another Application of the Scalar Product Suppose we have a line,x Q M
If we draw a perpendicular from the point to the line
and the direction vector of the line
equals zero ( since the vectors are perpendicular )
Suppose we have a line,
and a point not on the line.
we can find the coordinates of M, the foot of the perpendicular.
The scalar product of QM

51. x Q M
M is a point on the line so its position vector is given by one particular value of the parameter s.
So,
where
We can therefore substitute into and solve for s.

52. M is the foot of the perpendicular and is a value of r so .
SUMMARY
To find the coordinates of the foot of the perpendicular from a point to a line:
Substitute into , where
is the direction vector of the line
Solve for the parameter, s
Substitute for s into the equation of the line
Change the vector m into coordinates.
Sketch and label the line and point Q
M is the foot of the perpendicular and is a value of r so

53. e.g. Find the coordinates of the foot of the perpendicular from the point to the line
Q (1, 2, 2)
Solution: x M

55. Finally we can find m by substituting for s in the equation of the line.
The coordinates of M are