161 solution

8/8/2019 161 Solution

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2010 SAJC JC2 H2 Mathematics

Tutorial 16.1 : Vectors I

DIY Questions

1. Objective/skill: To find position vectors and to test for understanding of collinearity.

The points A, B, C have position vectors p + q, 2p – 2q and 6p + λ q respectively relative to

an origin. Find AB

and AC

. If A, B and C are collinear, find the value of λ .

( ) (2 2

3

AB OB OA= −

= − − +

= −

p q p q

p q

)

( ) (

( )

6

5 1

AC OC OA

λ

λ

= −

= + − +

= + −

p q p q

p q

)

A dIf ,B an C are collinear,

( )

( )

3 5 1

3 5 1

AB AC μ

μ λ

μ μ λ

=

⎡ ⎤− = + −⎣ ⎦

− = + −

p q p q

p q p q

μ 51 = (1)

( )13 −=− λ (2)

From (1),5

1=μ

( )

14

15

13

−=

−=−

λ

λ

2. Objective/skill: To find vector equation and Cartesian equation.

A, B, C have position vectors i + 3 j – k, 2i – 2 j – k, – i + 2 j + 5k respectively. Find the vector

equation and the Cartesian equation of the line AM where M is the midpoint of BC .

By midpoint theorem

( )1

2

2 11

2 221 5

12

0

2

OM OB OC = +

⎡ ⎤−⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟

= − +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥−⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

112

0 3

2 1

12

3

3

11

62

6

AM OM OA= −

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= − ⎜ ⎟⎜ ⎟ ⎜ ⎟

−⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞−⎜ ⎟⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟

= − ⎜ ⎟⎜ ⎟−⎝ ⎠

Equation of line AM in vector

equation form

1 1

3 6

1 6

λ

⎛ ⎞ ⎛ ⎜ ⎟ ⎜

= +⎜ ⎟ ⎜⎜ ⎟ ⎜

⎞⎟⎟⎟

− −⎝ ⎠ ⎝

r

⎠

λ

λ

λ

61

63

1

−−=

+=

+=

z

y

x

6

1

6

3

1

1 −−=

−=

−=

z y xλ

Equation of line AM in Cartesian

form

6

1

6

3

1

1 −−=−=− z y x

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3. Objective/skill: To find vector equation and points of intersection.

Write the equations z y

x 242

312 −=

−=− in vector equation form. Hence, find the

coordinates of the point where the line cuts (a) the plane z = 0 (b) the xz-plane.

Let λ =−=−

=− z y

x 242

312

λ

λ

+=

=−

2

2

x

x

λ

λ

3

2

3

1

2

31

−=

=−

y

y

4 2

12

2

z

z

λ

λ

− =

= −

Equation of line in vector equation form

2 2 11 2 1 2

3 33 3

2 11 222

x

y

z

λ

λ λ

λ

⎛ ⎞⎛ ⎞⎜ ⎟+ ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟= = − = + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎜ ⎟⎝ ⎠ −⎝ ⎠⎜ ⎟−⎝ ⎠

r

(a) When line cuts the plane 0= z ,

( )

4

024

=

=−

λ

λ

( )

2 4 61 2 74

33 300

+⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟= − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

r

(b) When the line cuts the xz-plane, 0= y

( )

2

1

2

031

=

=−

λ

λ

12 5

2 2

0 0

71 12 4

2 2

⎛ ⎞+ ⎛ ⎞⎜ ⎟

⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟= =⎜ ⎟⎜ ⎟

⎛ ⎞ ⎜ ⎟⎜ ⎟− ⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

r

Practice Questions

4. Objective/skill: To apply ratio theorem and prove a point lies on a line.

In a parallelogram OABC , P is the midpoint of AB. Q is divides OP in the ratio 2 : 1.Show that Q lies on AC and find the ratio of AQ : QC

By midpoint theorem ( )1

2OP = +a b

.

O A

BC

PQ

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( )

( )

2

3

2 1

3 2

13

OQ OP=

= × +

= +

a b

a b

If we are able to show

AQ QC , then we can

conclude that Q lies on

AC .

( )1

3

2 1

3 3

AQ OQ OA= −

= + −

= − +

a b a

a b

( )1

3

QC OC OQ= −

= − +c a b

Furthermore, we observe

that = +b c a

We want to express QC

in

terms of a and b, then

compare with AQ

.

( ) ( )13

2 4

3 3

2 12

3 3

2

QC

AQ

= − − +

= −

⎡ ⎤= − +⎢ ⎥⎣ ⎦

=

b a a b

b a

a b

Since 2QC AQ=

and Q is common, points A,C,Q are collinear. The ratio of AQ : QC is 1 : 2.

Alternative

(1

3OQ = +a b

) (see above for detailed workings)

We also observe that = +b c a

If Q were to lie on line AC , then OQ

would satisfy ratio theorem, i.e.

( ) ( )1 1 2 1 2

3 3 3 3OQ

3

+= + = + + = + =

a ca b a c a a c

Since 131

32 =+ , from ratio theorem, Q divides AC internally in the ratio 1 : 2. Therefore

points A, Q, C are collinear and AQ : QC is 1 : 2.

5. Objective/skill: To find points of intersection and angle between two lines.

N2002/P2/4

a) Show that the lines given by

( ) ( )k jik jir +++++= 3425 λ and ( ) ( )k jik jir 5743 +++++=

intersects, and find their point of intersection.

b) Calculate the acute angle between the lines.

(a) To check whether the lines intersect, we equate both lines together, i.e.

5 3 4

2 3 1 7

4 1 5

λ μ

λ μ

λ μ

+ +⎛ ⎞ ⎛ ⎜ ⎟ ⎜

+ = +⎜ ⎟ ⎜⎜ ⎟ ⎜+ +⎝ ⎠ ⎝

⎞⎟⎟⎟

⎠

Consider the first two equations 5 3 4λ μ + = + and 2 3 1 7λ μ + = + . Solving them

simultaneously, we obtain 2λ = , 1μ = .

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Now we check for the consistency in the third equation. Substitute 2λ = , 1μ = into the

equation 4 1 5λ μ + = + , we obtain L.H.S. 4 2 6 1 5 R.H.S.= + = = + = Hence the lines

intersects.

Their point of intersection is

( )7,8,6 .

(b) Let θ be the angle between the two lines. By scalar-dot product rule, we have

1 4 1 4

3 7 3 7 cos

1 5 1 5

θ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⋅ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Simplifying, we get30 10

cos1111 90

θ = = , or 17.5θ = ° .

6. Objective/skill: To find modulus, unit vectors, line of projection and length of

perpendicular.

N2001/P2/15

a) The points P and Q have position vectors k ji +−3 and k ji 279 −− . Show that

9PQ = .

Find the unit vector in the direction of PQ

, and also a Cartesian equation for the

line PQ.

The line l , which passes through P, has equation

2

1

1

1

2

3 −=+=−− z y x

Find(i) The length of the projection of PQ onto l .

(ii) The length of the perpendicular from Q to l .

(a) The first part is straight-forward.

9 3 6

7 1

2 1 3

PQ OQ OP

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

6= − = − − − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠

.

Hence ( ) ( )2 226 6 3PQ PQ= = + − + − =

9 .

The unit vector in the direction of PQ

is

6 21 1 1

6 29 3

3 1

PQPQ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

.

The equation of line PQ is

3 2

1

1 1

λ

⎛ ⎞ ⎛ ⎜ ⎟ ⎜

= − + −⎜ ⎟ ⎜⎜ ⎟ ⎜

2

⎞⎟⎟⎟

−⎝ ⎠ ⎝

r

⎠

. In other words,

3 2

1 2

1 1

x

y

z

λ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

= − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

.

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We write the vector form into three separate equations, and make λ the subject in order to

obtain the Cartesian form1

1

2

1

2

3 +−=

−−=

− z y x.

(i) Let m denote the direction vector for line l. Then the length of projection of PQ onto l is

given by ˆPQ ⋅m

In order to find m, we rewrite the equation of line l into vector form, i.e. let

3 1

2 1 2

1 x y zμ

− + −= = =

−

Rearranging, we get

3 2

1

1 2

μ

μ

μ

−⎛ ⎞⎜ ⎟

= − +⎜ ⎟⎜ ⎟+⎝ ⎠

r , or

3 2

1 1

1 2

μ

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

r . Hence

2

1

2

−⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟⎝ ⎠

m .

Q

Hence length of projection of PQ onto l is given by

( )( )

2 2 2

6 21 1

6 1 12 6 6 8 units32 1 2 3 2

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

− ⋅ = − − − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− + + −⎝ ⎠ ⎝ ⎠

(ii) By Pythagoras’ Theorem, the length of the perpendicular from Q to the line l is simply

the length QR, which is given by2 29 8 17− = .

7. Objective/skill: To find perpendicular distance and angle between two lines.

The equation of a line l is r = λ i – (λ + 4) j + k. Given that OA is perpendicular to l, find,

with respect to origin O, the position vector of the point A on l. Hence find the

perpendicular distance of the line l from O. Find also the acute angle between l and

another line with equation5

3

1

1

4

+=

−=

−

z y x.

Equation of line l Equation of line 2l

λ =+

=−

=

− 5

3

1

1

4

z y x

λ 4−= x

P

l

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −

2

1

2

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−

1

2

2

R θ

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0 1

4 4

1 1

λ

λ λ

⎛ ⎞ ⎛ ⎞ ⎛ ⎜ ⎟ ⎜ ⎟ ⎜

= − − = − + −⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎝

r 1

0

⎞⎟⎟⎟

⎠

4

1

t

OA t ⎛ ⎞⎜= − −⎜⎜ ⎟⎝ ⎠

⎟⎟ for some t ∈

Since OA

perpendicular to l

1

4 1 0

1 0

4 0

t

t

t t t

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

− − ⋅ − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+ + = ⇒ = −2

Hence

2 2

4 2 2

1 1

OA

− −⎛ ⎞ ⎛ ⎜ ⎟ ⎜

= − + = −⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎟⎟

⎠

Shortest distance between l and origin

2

2 3

1

OA

−⎛ ⎞⎜ ⎟

= = − =⎜ ⎟⎜ ⎟⎝ ⎠

11

1

y yλ λ

−= ⇒ = +

33 5

5

z zλ λ

+= ⇒ = − +

0 41

3 5

0 4

1 1

3 5

x y

z

λ λ

λ

λ

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− +⎝ ⎠ ⎝ ⎠

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

r

r

The acute angle between l and 2l

1 1

1 4

1 10 5 5

cos cos 56.91 4 84

1 1

0 5

− −

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− ⋅⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= = =

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

°

8. Objective/skill: Prove two lines are perpendicular and area of triangle using cross product.

N2007/1/6

Referred to the origin O , the position vectors of points A and B are given

i – j +2k and 2i +4 j + k respectively.

(i) Show that OA is perpendicular to OB.

(ii) Find the position vector of the point M on the line segment AB such that AM : MB = 1:2.

(iii) The point C has position vector -4i +2 j + 2k . Use a vector product to find theexact area of triangle OAC .

(i) Given that

1

1

2

OA

⎛ ⎞⎜ ⎟

= −⎜ ⎟⎜ ⎟⎝ ⎠

,

2

4

1

OB

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟⎝ ⎠

. Since

1 2

1 4 2 4 2 0

2 1

OA OB

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⋅ = − ⋅ = − + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

, OA

and OB are perpendicular.

(ii) By ratio theorem,

42 1

23 3

5

OA OBOM

⎛ ⎞+ ⎜ ⎟

= = ⎜ ⎟⎜ ⎟

⎝ ⎠

.

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(iii) Area of triangle OAC is given by

1 4 61 1 1

1 2 10 352 2 2

2 2 2

OA OC

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

× = − × = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

.

9. Objective/skill: To use ratio theorem, prove two lines are perpendicular, understandwhat are dot and cross products.

N2009/P2/2

Relative to the origin O, two points A and B have position vectors given by

14 14 14= + +a i j k and 11 13 2=b i - j + k respectively.

(i) The point P divides the line AB in the ratio 2 : 1. Find the coordinates of P.

(ii) Show that AB and OP are perpendicular.

(iii) The vectors c is a unit vector in the direction of OP

. Write c as a column vector,

and give the geometrical meaning of .a c .

(iv) Find ×a p , where p is the vector OP

, and give the geometrical meaning of ×a p .

Hence write down the area of triangle OAP.

14 11 122 1

14 2 13 42 1 3

14 2 6

OA OBOP

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟

= = + − =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

−

⎞

⎟⎟⎟ ⎠

.

The coordinates of is .P ( )12, 4, 6−

ii)

12 14 3

4 14 27

6 14 12

AB OB OA

−⎛ ⎞ ⎛ ⎞ ⎛

⎜ ⎟ ⎜ ⎟ ⎜= − = − − = −⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟ ⎜ ⎟ ⎜ −⎝ ⎠ ⎝ ⎠ ⎝

.

3 12

27 4 36 108 72 0

12 6

AB OP

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⋅ = − ⋅ − = − + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

Hence and are perpendicular. AB OP

iii)

A BP2 1

O

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( )22 2

12 61 1

4

12 4

27

36 6

OP

OP

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠

+ − +⎝ ⎠

c

⋅a c is the length of projection of OAonto the line with the direction

OP

.

iv)

( ) ( ) ( )( )( )( ) ( )( )( )

( )( ) ( )( )

14 12 1 6 1 3 1 2 5

14 4 28 1 2 28 1 3 1 6 28 3

14 6 1 3 81 2 1 6

⎡ ⎤⎛ ⎞⎡ ⎤ − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

− = − = − − =⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ −− − −⎝ ⎠ ⎝

× =

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎢ ⎥⎝ ⎠⎣ ⎦

× ×pa

×a p is the area of parallelogram with two sides OA and .OP

Area of triangle ( )22 25 3 8 98 2 uni

51 1

2 t8 3 142 2

8

sOAP⎛ ⎞⎜ ⎟

= = =⎜ ⎟⎜ ⎟−

× + + − =

⎝ ⎠

pa2

Mastery Questions

10. Objective/skill: To find equation of line, points of intersection and angle between

(J94/1/14) In the diagram, O is centre of the square base ABCD of a right pyramid,

vertex V . Perpendicular unit vectors i, j, k are parallel to AB, AD, OV respectively. The

length of AB is 4 units and the length of OV is 2h units. P, Q, M and N are the mid-

points of AB, BC , CV and VA respectively. The point O is taken as the origin for position vectors.

(a) Show that the equation of the line PM may be expressed as

0 1

2

0

t

h

⎛ ⎞ ⎛ ⎜ ⎟ ⎜

= − +⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ ⎠ ⎝

r 3

⎞⎟⎟⎟ ⎠

,

where t is a parameter.

(b) Find an equation for the line QN .

(c) Show that the lines PM and QN intersect,

and that the position vector OX

of their

point of intersection is⎟⎟⎟

⎠

⎞

⎜⎜ .⎜

⎝

⎛ −

h21

21

21

(d) Given that OX is perpendicular to VB,

find the value of h and calculate the acute angle between PM and QN , giving your

answer correct to the nearest 0.1°.

D

i

N

V

k j

P

C

M

A B

O

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(a) By ratio theorem, ( ) ( )1 1

2 2 22 2

OM OV OC h h= + = + + = + +i j k i j k

.

Hence ( )2 3PM OM OP h h= − = + + − − = + +i j k j i j k

.

The equation of the line PM is

0 1

2 3

0

OP tPM t

h

⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟= + = − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

r

.

(b) Similarly, ( ) ( )1 1

2 2 22 2

ON OA OV h h= + = − − + = − − +i j k i j k

.

Hence 2 3QN ON OQ h h= − = − − + − = − − +i j k i i j k

.

The equation of the line QN is

2 3

0 1

0

OQ sQN s

h

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= + = + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

r

.

(c) To show that both lines, we equate them, i.e.

2 3

2 3

t s

t s

ht hs

−⎛ ⎞ ⎛ ⎜ ⎟ ⎜

− + = −⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎟⎟ ⎠

. Clearly, by

observation, for a solution set that satisfy all three equations, we require1

2t s= = . Hence

12

12

12

OX

h

⎛ ⎞⎜ ⎟

= −⎜ ⎟

⎜ ⎟⎝ ⎠

.

(d)

2

2

2

VB

h

⎛ ⎞⎜ ⎟

= −⎜ ⎟⎜ ⎟−⎝ ⎠

.

12

12

12

2

0 2 0

2

OX VB OX VB h

h h

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⊥ ⇒ ⋅ = ⇒ − ⋅ − = ⇒ = ±⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠

2 . Clearly from the

diagram, 2h = .

From scalar-dot product rule,

1 3 1 3

3 1 3 1 cos

h h h h

θ

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⋅ − = −

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

. Substituting 2h = and solving,

we get 1 1cos 70.5

3θ −= = ° .

Vectors 9

161 solution

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