1. Revision videos

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

3. e.g. Find the acute angle, , between the lines and Solution: whereand (nearest degree)

4. O F A CB D E G 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 To define the angle we just draw a line parallel to one line meeting the other. 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.

5. 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.

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

7. Solutions: (a) whereand (nearest whole degree)

8. (b) whereand (nearest whole degree)

22. Vectors consolidation Find the coordinates of the foot of the perpendicular from the point given to the line given Recap and secure all work on vectors

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

24. Another Application of the Scalar Product x Q M If we draw a perpendicular from the point to the line... 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 and the direction vector of the line...

25. Another Application of the Scalar Product 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

26. 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.

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

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

29. Solution: x M

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

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

33. 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 This is because it is so easy to substitute the wrong vectors into the equation.  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.

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

36. (a) Q (  1, 2,  8) Solution: x M

37. Point is

38. (b) Q (1, 1,  4) Solution: x M

39. Point is

46. Revision videos