A lvl H2 Maths: Vectors

Question from http://www.sgforums.com/forums/2297/topics/407847

Relative to an origin O, the position vectors of points A and B are a and b respectively. Givent that angle AOB is 90deg, show that the position vector of the foot of the perpendicular from O to AB is

a + (|a|² / (|a|² + |b|²)) (b - a)

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Answer:

Using the concept of vector projection, (i.e. AF = |AF| * unit vector AB),

OF = OA + AF = a + |a| cos θ * (b - a) / |b - a|, where θ = angle between OA and BA
= a + |a| [ {a.(a - b)} / {|a| |a - b| }] * (b - a) / |b - a|
= a + {a.a - a.b} / (|a|² + |b|²) * (b - a)
= a + |a|² / (|a|² + |b|²) * (b - a) since a.a = |a|² and a.b = 0

(shown)

A lvl H2 Maths: Vectors

TJC 2007 P1 Q4

Relative to an origin O, position vectors of A, B and C are a, b and c respectively, where a, b and c are non-parallel vectors. M is the mid-point of AC and P is on the line AB produced such that AB:BP = 2:3. The line PM meets the line BC at a point S. Show that the position vector of S is ⅛ (5b + 3c) .

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Answer:

By repeated use of Ratio theorem,

OM = ½(a + c) and OP = ½(5b - 3a)

OS = μOP + (1-μ) OM
= ½( μ(5b - 3a)+(1-μ)(a + c) )
= (½ - 2μ)a + (5μ/2)b + ((1-μ)/2) c


Also, by ratio theorem again (seeing that B, S and C are collinear),
OS = λc + (1-λ)b

Since a, b and c are non-parallel, we can compare coefficients of the vectors

½ - 2μ = 0 ====> μ = ¼
λ = ((1-μ)/2) = ¾/2 = ⅜

Hence, from OS = λc + (1-λ)b
OS = ⅜ c + (1-⅜) b
OS = ⅛ (5b + 3c) (shown)

A lvl H2 Maths: Vectors

TPJC 2000 P2 Q14

Referred to the origin O, the points A, B and C have position vectors given respectively by


Find
(i) the cartesian equations of the line l passing through A and B;
(ii) the length of projection of BC on l;
(iii) the image of C in the line l;
(iv) the points where l meets the plane z = 0;
(v) the 2 points P on l such that cos ∡POB =

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Answer:

(i)



Hence, line l :



(ii)



Thus, length of projection





(iii) The graphic is as shown

So, point M is point B, moved along downwards along line l

Unit vector along line l =


Hence,




(iv) Let the point be


Meeting the plane z = 0 means 3 + 2λ = 0
λ = -1.5

Hence, substituting λ = -1.5, we get the point of intersection as




v) Let
for some k since P is a point on l.

Note: So we need to find out what is the value of k.



Since cos ∡POB = , and
cos ∡POB =






Square both sides.

169k² + 26k + 1 = 112k² -16k + 16
57k² + 42k - 15 = 0
19k² + 14k - 5 = 0
(19k - 5)(k + 1) = 0
k = -1 or 5/19

Hence, the two points P are
(-1, 4, -1) or (5/19, 4/19, 29/19)

A lvl H2 Maths: Vectors

HCJC 1999 P1 Q19

Points A, B, C D have position vectors -3i + 4j, 3i + pj, 2i + 3j and i - 2j respectively.
(i) Given that AB·AD = 0, find the value of p.
(ii) Show that ∡BAC = ∡DAC.
(iii) Show that the points B, C, D are collinear and write down the vector equation of the line l through these three points.
(iv) Find the position vector of N, the foot of the perpendicular from the point E (5, 18, -4) to the line l. Find also the shortest distance from E to the line l.

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Answer:

(i) AB = OB - OA = 6i + (p-4)j
AD = OD - OA = 4i - 6j

AB·AD =0
24 - (p-4)*6 = 0
24 - 6p + 24 = 0
p = 8


(ii) AC = OC - OA = 5i - j
AB = 6i + 4j
AD = 4i - 6j









Since cos^-1 ∡BAC = cos^-1 ∡DAC
=> ∡BAC = ∡DAC (shown)

(iii) BC = OC - OB = -i - 5j
BD = OD - OB = -2i - 10j = 2 BC

Thus, B, C and D are collinear (shown)

Line l for λ ε R


(iv) Since N is on l, let N =

Hence,


Since is perpendicular to the line l,

2 - λ + 50 - 25λ = 0
26λ = 52
λ = 2

Hence,

A lvl H2 Maths: Vectors

HCJC Year 2000 Paper 1 Q11

Given that a = 3i + j + k, b = i - k, c = 4i - 3j + 2k are position vectors of the points A, B and C respectively, find the position vectors of the points and Q which divide BC internally and externally in the ratio 1:2, respectively.

Show that AP is perpendicular to BC.

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Answer:


By Ratio Theorem,
OP = ⅓(OC + 2OB)
OP = ⅓(4i - 3j + 2k + 2i - 2k)
OP = 2i - j









By Ratio Theorem,
OB = ½(OQ + OC)
OQ = 2OB - OC
OQ = 2i - 2k - (4j - 3j + 2k)
OQ = -2i + 3j -4k




AP = OP - OA = -i - 2j - k
BC = OC - OB = 3i - 3j + 3k

AP · BC = -3 + 6 - 3 = 0
Hence, AP is perpendicular to BC (shown)

A lvl H2 Maths: Vectors

H2 Maths 2008 P1 Q11

The equations of three planes, p1, p2, p3 are

2x – 5y + 3z = 3,
3x + 2y – 5z = -5,
5x + λy + 17z = μ,

respectively, where λ and μ are constants. When λ = -20.9 and μ =16.6, find the coordinates of the point at which these planes meet. [2]

The planes p1 and p2 intersect in a line l.

(i) Find a vector equation of l. [4]
(ii) Given that all three planes meet in the line l, find λ and μ. [3]
(iii) Given instead that the three planes have no point in common, what can be said about the values of λ and μ? [2]
(iv) Find the Cartesian equation of the plane which contains l and the point (1, -1, 3). [4]

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Answer:

Use Graphic calculator to solve for 3 unknowns
x = -4/11, y=-4/11, z = 6/11

(i) 2x – 5y + 3z = 3 ---- (1)
3x + 2y – 5z = -5 ---- (2)

line parallel to both planes = (2, -5, 3) x (3, 2, -5)
= (19,19,19) = 19 (1,1,1)

Let z = 0
2x - 5y = 3
3x + 2y = -5
Solving, x = -1, y = -1
(-1,-1,0) is a point on both planes

So, equation of vector l = (-1,-1,0) + t (1,1,1)
Note: Alternatively, use the GC to solve.

(ii) (5,λ,17) dot (1,1,1) = 0
5 + λ + 17 = 0
λ = -22

5x + -22y + 17z = μ
Sub (0,0,1)
μ = 17


(iii) Three planes have no points in common, means p3 does not intersect vector l
This means p3 must be parallel to vector l, but have no points on vector l
Thus, λ will still be -22, and μ can be anything except 17


(iv) Plane must also contain points (0,0,1) and (1,-1,3)
and be perpendicular to (1,1,1)

Vector linking points (0,0,1) and (1,-1,3) is (1,-1,2).

So plane must be perpendicular to both (1,1,1) and (1,-1,2)
Using cross product,
(1,1,1) x (1,-1,2) = (3,-1,-2)

so, cartesian equation of plane is 3x -y -2z = D
sub in (0,0,1)
D = -2

So final answer: 3x -y -2z = -2