Vectors (3)

Question 1

The points W, X and Y have position vectors (1,3), (-2,1) and (5,4) respectively

Find the position vector of Z such that WX = YZ

Answer 1

WX = OX - OW = (-2,1) - (1,3) = (-3,-2)
YZ = OZ - OY = (a, b) - (5, 4) = (a-5, b-4)
WX = YZ => (a-5, b-4) = (-3, -2) => a=2, b=2
OZ = (2, 2)

Question 2

The points W, X and Y have position vectors (1,3), (-2,1) and (5,4) respectively

The points A and B have position vectors -2i + 4j and 5i + j respectively. Point P on the line AB such that AP:PB = 1:3. Determine the position vector of P

Answer 2

AB = (5i + j) - (-2i + 4j) = (5-(-2))i + (1-4)j = 7i - 3j

P is (1/4) of the way along AB, so AP = (1/4)AB = (7/4)i - (3/4)j

OP = OA + AP = ((-2) + (7/4))i + (4 - (3/4))j
= -(1/4)i + (13/4)j

Question 3

The following sketch shows a triangle ABC
M is the midpoint of line BC
Given that AB = -5i + 2j and AC = -2i + 4j, find |AM|

Answer 3

BC = AC - AB = -2i + 4j - (-5i + 2j) = 3i + 2j
BM = (1/2)BC = (3/2)i + j
AM = AB + BM = -5i + 2j + (3/2)i + j = (-7/2)i + 3j
|AM| = ¬((-7/2)^2 + 3^2) = (¬85)/2

Question 4

The movement of a particle is modelled by vector p. The particle is travelling at a speed of 7m/s with direction of 15degrees above the horizontal. Write P in component form

The particle strikes another particle. Its movement is now modelled by vector q=(2¬2)(i+j). Find the amount by which the particle’s speed has decreased and state the particle’s new direction

Answer 4

p = 7 cos 15degrees i + 7 sin 15degrees j

The speed is given by the magnitude of q
|q| = ¬((-7/2)^2 + (2¬2)^2) = ¬16 = 4 m/s
Particles speed has decreased by 3 m/s. The direction of q is 45degrees

Question 5

Points S, T, U and V make a parallelogram STUV. The position vectors of S, T and V are i+j, 8i+j and -i+5j respectively. Find the distance between S and U. Leave answer in exact surd form

Answer 5

TU is parallel to SV
So TU is the same length as SV

TU = SV = -i + 5j - (i + j) = -2i + 4j
SU = ST + TU = (8i + j) - (i + j) + (-2i + 4j) = 5i + 4j
|SU| = ¬(5^2 + 4^2) = ¬41

Question 6

Points P, Q and R have position vectors (2,4,-2), (-2,-1,-1) and (2,7,-5) respectively. Show that OP is parallel to QR

Answer 6

(2,7,-5) - (-2,-1,-1) - (2,4,-2) = 2(2,4,-2)

So QR = 2OP, which means they’re parallel

Question 7

A skateboard ramp has vertices modelled with position vectors O=(0i + 0j + 0k)m. X=(3i - j)m, Y=(5j)m and Z=(4k)m

Calculate the distance from the midpoint of OZ to the midpoint of XY

Answer 7

Midpoint of OZ has position vector (1/2)OZ = 2k m
Midpoint of XY has position vector (1/2)(OX + OY)
= (1/2)[5i + (3i - j)] = (1/2)(3i + 4j) = (1.5i + 2j)m

So the distance between the two midpoints is |(1.5i + 2j) - (2k)| = ¬((1.5)^2 + 2^2 + (-2)^2) = ¬10.25 = 3.20m (3 s.f.)

Question 8

The points A and B have position vectors 2i + 3j + 4k and -3i + j - 3k respectively. Show that AB = -5i - 2j - 7k

The point M divides the line AB in the ratio 2:1. Calculate the distance of M from the origin

Answer 8

AB = OB - OA = -3i + j - 3k - (2i + 3j + 4k) => -5i - 2j - 7k

M is two-thirds of the way along AB, so AM = (2/3)AB
OM = OA + AM = 2i + 3j + 4k + (2/3)AB
= (2+(-10/3))i + (3 + (-4/3))j + (4 + (-14/3))k
=-(4/3)i + (5/3)j - (2/3)k
|OM| = ¬(-(4/3)^2 + (5/3)^2 + (-14/3)^2) = ¬5