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

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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

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

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

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

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

Solutions: (c) and The vectors are perpendicular.

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

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

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.

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

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

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

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

Diagram

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

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.

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 .

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

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

(a) where and (nearest whole degree)

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 . . .

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 . . .

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 )

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.

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

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

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

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

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

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.

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

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

Point is

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

Point is

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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 .

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.

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

( 3 s.f. )

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 . )

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

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

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.

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 .

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

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