Summary: Vectors. This theorem is used to find any points (or position vectors) on a given line (direction vector). Two ways RT can be applied:

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Summary: Vectors

This theorem is used to find any points (or position vectors) on a given line (direction vector). Two ways RT can be applied:

Case 1: If the point lies BETWEEN two known position vectors

For example, if a point X lies along AB (ie position vectors of A and B are known), position vectors of X can be found if the ratio of AX: XB is known.

For AX: XB = λ_{: µ as shown in the diagram below,}

µ λ λ µ + + = OA OB OX using RT.

Case 2: If the point lies OUTSIDE two known position vectors

This applies to the scenario where the point X lies on the line AB produced as shown in the figure below: O A X B λ µ O A B X λ µ

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Summary: Vectors

µ λ λ µ + + = OA OX

OB . Do note that the subject (i.e. OB) is always the point in the middle. The point X can be found by making OX as the subject, i.e.

λ µ µ λ OB OA OX =( + ) − . Note:

(1) The use of RT is very common in A Levels question. (2) If λ_{= µ, RT becomes the mid-point theorem}

(3) The vectors used in the RT are always POSITION VECTORS! 2) Definition of scalar and vector product

The formula (or definition) of Scalar ‘dot’ product is a•b= abcosθ. The formula (or definition) of Vector ‘cross’ product is a×b= absinθ. LHS of both formulas can be determined using vector operations.

For example, if           = 3 2 1 a a a a and           = 3 2 1 b b b b , 3 3 2 2 1 1 3 2 1 3 2 1 b a b a b a b b b a a a b a = + +           •           = • &           − − − − =           ×           = × 1 2 2 1 1 3 3 1 2 3 3 2 3 2 1 3 2 1 ) ( b a b a b a b a b a b a b b b a a a b a Note:

(1) Both vectors a & b must point in the same direction. Refer to lecture notes to check what I mean!

(2) a x b will produce a vector that is mutually perpendicular to a and b. This is very useful in application problems involving planes.

3) Vector line equation

A vector line equation can be obtained in two ways using: (1) Two known position vectors OR

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Summary: Vectors

(1) Parametric form i.e. l:r =a+λb, where r is any point on the line, a is a given point on the line, b is a direction vector parallel to the line and λ_{is the parameter.}

Note: the vector line equation resembles the straight line equation i.e. y = mx + c from A Math. (2) Cartesian form i.e. by letting r =

          z y x

and rearranging the parametric form by making λ_the subject, the Cartesian equation of the line can be obtained.

For example, If           +           =           ⇒           +           = 1 2 3 3 2 1 1 2 3 3 2 1 : λ λ z y x r l 3 2 / ) 2 ( 3 / ) 1 ( 3 2 2 3 1 − = − = − = ⇒ + = + = + = z y x z y x λ λ λ λ λ λ

Hence, the Cartesian form is 3

2 2 3 1 − = − = − z y x

Note: Students are advised to know how to convert between the two forms of the vector line equations.

4) Vector plane equation

A vector plane equation can be obtained in three ways using: (1) Three known position vectors OR

(2) Two known position vectors and one vector (or direction vector) parallel to the plane OR (3) One known position vector and two vectors (or direction vectors) parallel to the plane

A vector line equation can be expressed in three forms:

(1) Parametric form i.e. Π:r =a+λb+µc, where r is any point on the plane, a is a given point on the plane, b and c are direction vectors parallel to the plane and λ_{& µ are the} parameters.

(2) Cartesian form e.g. ax + by + cz = d, where a, b, c and d are constants.

(3) Scalar Dot Product form i.e. r•n=a•n, where r is any point on the plane, a is a given point on the plane, n is the vector perpendicular to the plane (i.e. also called the normal

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Summary: Vectors

vector). This is the most useful and easiest to use of the three forms. Most applications require the plane to be in this form.

Special Note:

It is very useful to know how to change the forms of the vector plane equation, especially when solving applications problems involving vectors. Students can convert the vector plane equations in the following manner:

The figure above can be explained as follows:

a) Parametric form can be converted to scalar product form, then to Cartesian form, in this sequence.

b) Scalar product form can be converted to Cartesian form only. c) Cartesian form can be converted to scalar product form. d) Cartesian form can be converted to parametric form.

Note: It also means that there is no way to convert scalar product form to parametric form! Conversions (a) to (c) are relatively easy, which should be documented in most lecture notes. Here, I will present the conversion of Cartesian form directly to parametric form.

For example, consider the Cartesian equation of a plane in the form 2x + 3y + 4z = 5. Now, by letting x as the subject, equation becomes

2 5 4 3 − + − = y z x -- (*)

Parametric form Scalar dot product

form

Cartesian form

Parametric form Scalar dot product

form

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Summary: Vectors

Note: Students can also make y or z as the subject.

Now, we can let y = λ_&_{z = µ}_{which will give us the two parameters in the parametric form.} Equation (*) thus becomes,

2 5 4 3 − + − = λ µ x Since

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              − − =             − − + =           ⇒           = µ λ µ λ µ λ µ λ ₂ 2 3 2 5 2 5 4 3 z y x z y x r          − +               − +             = Π ⇒          − +               − +               =           ∴ 1 0 2 0 12 3 0 0 2 5 : 0 2 0 2 3 0 0 2 5 µ λ µ µ λ λ r z y x Hence,             = 0 0 2 5

a which is a point on the plane. The two vectors parallel to the plane are

         −               − 1 0 2 & 0 1 2 3 . 5) Applications of vectors a) Finding angles

The key to finding angles is to use the direction vectors of the line and the plane (i.e. normal vector). To determine the angles, students should use the scalar dot product formula i.e. a•b= a bcosθ

b a b a• = ⇒cosθ

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Students should recall that if cos is positive, the angle must lie in the first quadrant (from A Math θ trigonometry) i.e. acute angle. To find the angle, simply replace vectors a and b with the appropriate directional and/or normal vectors.

i) Finding angle between two lines

Vectors a and b are the direction vectors of each line.

ii) Finding angle between a plane and a line

Vectors a and b can be the normal vector of the plane and direction vector of the line. The angle between them is thus 90° - θ_._{See figure below}_:

iii) Finding angle between two planes

Vectors a and b are the normal vectors of the planes.

b) Finding point of intersection i) Between two lines

The point of intersection can be found by equating the two vector line equations and solving for the values of the two parameters (e.g. λ_{and µ).}

Note: Three equations can be formed, but only two unknowns are solved. The third equation is to

check for consistency of the values of λ_{and µ}_{found. If the third equation is not satisfied, this}

Line Normal vector of plane θ Angle required

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Summary: Vectors

will imply that there is no intersection between the two lines. Lecture notes should contain this example (please refer).

ii) Between a line and a plane

The point of intersect will lie on both the line and the plane. Since the point lie on the line, it must take the vector equation of the line.

For example, if the intersect lie on

          +           = 1 2 3 3 2 1 :r λ

l , the position vector of the point of intersect

must be           + + + λ λ λ 3 2 2 3 1 .

To find the point of intersect, students are advised to change the vector plane equation (if given in the cartesian or parametric form) to the scalar dot product form i.e. r•n=a•n.

For example, to find the point of intersection between the line and plane below, 2x + 3y + 4z = 5 &           +           = 1 2 3 3 2 1 :r λ l

Here, the position vector of the intersection must be

          + + + λ λ λ 3 2 2 3 1 , as explained earlier. By first changing the vector plane equation to the scalar dot product form

i.e. 2x + 3y + 4z = 5 5 4 3 2 =           • ⇒r

Next, r can be replaced by

          + + + λ λ λ 3 2 2 3 1

since the point will also lie on the plane.

Note: r is any point on the plane

Thus, 5 4 3 2 3 2 2 3 1 =           •           + + + λ λ λ

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Summary: Vectors

iii) Between two planes

When two planes intersect, they will form a vector line. To find the line, the plane equations should be in

• Scalar dot product form & • Parametric form

For example, if we have 7

1 3 2 : 1 =−           − • Π r &           − − +           − +           = Π 9 1 6 1 0 2 8 2 0 : 2 r λ µ By changing Π2 to           + − − − = µ λ µ µ λ 9 8 2 6 2

r and substituting this into Π1, we will get

7 1 3 2 9 8 2 6 2 − =           − •           + − − − µ λ µ µ λ

Here, we will get a relationship between λ_{and µ. By either making}λ_{or µ as the subject and} substituting it back into equation of Π₂(i.e. parametric form), the line of intersection can be found

Note: This method reduces the equation of Π₂ to a single parameter, which now defines a vector line equation.

Alternatively, students can change the plane equations into the Cartesian form and use GC applets to define the equation of the intersection.

iv) Between three planes

When 3 planes intersect, there are three possible outcomes • They form a point (i.e. position vector)

• They form a line (i.e. vector line equation) • There is no intersect

Students can determine the intersection between three planes using GC applets.

Note: This section is related to system of linear equations. Under that chapter, there is usually a unique solution (i.e. they intersect to form a point), but under vectors, either one of the 3 outcomes is possible.

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Summary: Vectors

c) Finding foot of perpendicular

When finding foot of perpendicular, students should recall that if two vectors are perpendicular, their dot product is zero (i.e. a•b=0).

i) Between a point and a line

To find the foot of perpendicular, we first must construct a direction vector between the point (i.e. position vector) and the foot of perpendicular (i.e. red point). Since the foot of perpendicular lies on the line, it must take the equation of the line.

For example, if the line equation is

          +           = 1 2 3 3 2 1 :r λ

l and the given point (i.e. OA) is           5 5 5 .

Let N be the foot of perpendicular

          + + + = ⇒ λ λ λ 3 2 2 3 1 ON

The direction vector is

          + − + − + − =           −           + + + = − = λ λ λ λ λ λ 2 2 3 3 4 5 5 5 3 2 2 3 1 OA ON AN

Since AN is perpendicular to the line 0 1 2 3 =           •

⇒ AN , which allows you to solve for λ_and subsequently solve for ON.

Foot of perpendicular Given point

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is the direction vector of the line. Once the foot of perpendicular is found, the shortest distance (i.e. perpendicular distance) can also be found, by taking the modulus of the direction vector.

ii) Between a point and a plane

To find the foot of perpendicular (ON) to the plane, we can first construct a line containing the given point and parallel to the normal vector of the plane. Since the foot of perpendicular is also a point on the line, it must take the equation of the line.

The same point (i.e. foot of perpendicular) is also a point on the plane, hence we can now substitute the equation into the equation of the plane, which should be in the scalar dot product form.

For example, if equation of plane is 5

4 3 2 : =           •

Π r and the point is           5 5 5 .

To find foot of perpendicular, first construct the line containing           5 5 5 and parallel to           4 3 2 i.e.           +           = 4 3 2 5 5 5 :r λ l Normal vector Foot of perpendicular, ON Given point

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Summary: Vectors

          + + + = λ λ λ 4 5 3 5 2 5

ON . Also, since ON also lies on the plane, thus

5 4 3 2 4 5 3 5 2 5 =           •           + + + ⇒ λ λ λ

. The position vector of the foot of perpendicular can be found once λ_is solved.

Note: The vector line is intentionally introduced to determine the foot of perpendicular by means of intersecting a line with a plane.

d) Finding length of projection

Most JC lecture notes contain the formula to determine the length of projection of a direction vector onto another direction vector. The formula is

Length of projection =a•bˆ

Students are advised to refer to lecture notes for relevant examples! Miscellaneous

These are other points which may also appear in exam, but not as popular as those mentioned above: i) Finding distance of a plane from the origin

By modifying the scalar dot product form of the plane, the distance between the plane and the origin can be found

e.g. Π:r•n=a•n

Dividing both sides by |n|,

n n a n n r • = •

⇒ or r•nˆ=a•nˆ, where nˆ is the unit vector of normal vector of the plane.

The distance between the plane and the origin is thus a•nˆ. If the distance is negative, it implies that the plane is below the origin.

Note: This approach can also be used to determine the perpendicular distance between two planes i.e. by knowing the distance of each plane from the origin, the perpendicular distance is simply the difference between the two distances. Can you visualize?

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Summary: Vectors

Area of triangle = a b a xb 2 1 sin ) )( ( 2 1 = θ iii) Finding reflection

The mid-point theorem can be used to find the position vector of the reflected point. Refer to diagram. Two position vectors must be provided in order to solve. This method also works if the point is reflected about the plane.

O Reflected point Axis or point of reflection Point to be reflected