The Student Room Group

P3 Vectors

1) State a vector equation of the line passing through the points A and B whose position vectors are i-j+3k and i+2j+2k respectively. Determine the position vector of the point C which divides the line segment AB internally such that AC=2CB.

2) The point P lies on the line which is parallel to the vector 2i+j-k and which passes through the point with position vector i+j+2k. The point Q lies on another line which is parallel to the vector i+j-2k and which passes through the point with position vector i+j+4k. The line PQ is perpendicular to both these lines. Find a vector equation of the line PQ and the coordinates of the mid-point of PQ.

3) The position vectors of three points A, B and C with respect to a fixed origin O are

(2i-2j+k), (4i+2j+k) and (i+j+3k)

respectively.

a) Write down unit vectors in the direction of the lines CA and CB and calculate the size of ACB in degrees, to 1 decimal place.

The mid-point of AB is M.

b) Find a vector equation of the straight line passing through C and M.

c) Show that AB and CM are perpendicular.

d) Find the position vector of the point N on the line CM such that ONC=90 degrees.

3) The lines L1 and L2 intersect at the point B. Given that the equations of the lines are

L1: r = (2+3t)i + (3+4t)j + (4+2t)k
L2: r = (1+2s)i + (1+3s)j + (4s-2)k

where t and s are scalar parameters, find

a) the coordinates of B.

b) the acute angle, in degrees to 1 decimal place, between the lines.

4) Referred to an origin O, the points A, B, C and D are (3,1,-1), (6,7,8), (2,5,0) and (0,7,-2) respectively.

a) Find the vector equations for AB and CD.

b) Show that AB and CD intersect at the point (4,3,2).

c) Calculate the size of ACD to the nearest degree.

Answers:

1) r = 25i/11 + 25j/11 + 16k/11 + @(-i+3j+k).
26i/11 + 22j/11 + 15k/11.

2a) (1/rt14)(i-3j-2k), (1/rt14)(3i+j-2k), 73.4 degrees.
b) r = i+j+3k + @(2i-j-2k).
d) (19i/9 + 4j/9 + 17k/9).

3a) (5,7,6).
b) 26.3 degrees.

4a) Equation of AB is r=3i+j-k + @(i+2j+3k)
Equation of CD is r=2i+5j+ lambda(-i+j-k).
c) 123 degrees.

Cheers :wink:
Reply 1
Are you sure about those answers ??

1)
A(1,-1,3)
B(1,2,2)
AB = (0,3,-1)

l: r = OA + t.AB
r = (1, -1, 3) + t(0,3,-1)
===================

c(x,y,z)
C is 2/3 of theway along AB
C - A = (2/3)(B - A) = (2/3)BA

x - 1 = (2/3)(0) --> x = 0
y + 1 = (2/3)(3) --> y = 1
z + 3 = (2/3)(-1) --> z = -7/3

C = (1, 1, -7/3)
============
Reply 2
Reply 3
3.
a)
Solve the simultaneous equations:
2+3t = 1+2s
3+4t = 1+3s

To verify that these lines intersect, plug in the values you got for t and s in:
4+2t = 4s-2

Now substitute these values in either L1 or L2.

b)
Find the direction vectors of both lines:
L1: r = (2+3t)i + (3+4t)j + (4+2t)k = (2, 3, 4) + t(3, 4, 2) -- direction vector = a = (3, 4, 2).
L2: r = (1+2s)i + (1+3s)j + (4s-2)k = (1, 1, -2) + s(2, 3, 4) -- direction vector = b = (2, 3, 4).

Now:
a . b = |a| |b| cosθ
(3, 4, 2) . (2, 3, 4) = (3² + + 2²)0.5 (2² + + 4²)0.5 cosθ
6 + 12 + 8 = 29 cosθ
cosθ = 26/29
θ = 26.3

4.
a)
AB: r = A + t(B-A)
CD: r = C + s(D-C)

b) & c) are the same as question 3 above.
Reply 4
3/ ( the other one)
A(2,-2,1), B(4,2,1), C(1,1,3)
a)
CA = (1,-3,-2), |CA| = √(1 + 9 + 4) = √14
CB = (3,1,-2), |CB| = √(9 + 1 + 4) = √14

unit vectors are:
(1/√14)(i-3j-2k),
(1/√14)(3i+j-2k),
=============

CA.CB = (1,-3,-2).(3,1,-2) = 3 - 3 + 4 = 4
cosθ = CA.CB / |CA||CB| = 4 / 14 = 2/7
θ = 73.4 deg
======

b)
M = ½(B + A)
M = (3,0,1)
C = (1,1,3)
CM = M - C = (2,-1,-2)
r = C + t.CM
r = (1,1,3) + t.(2,-1,-2)
==================

c)
AB = B - A = (4-2,2+2,1-1) = (2,4,-2)
AB.CM = (2,4,0).(2,-1,-2) = 4 - 4 + 0 = 0
:. AB and CM are perpindicular
=======================

d)
N lies on CM, so
N = (1+2t, 1-t, 3-2t) for some value of t.

ON.r = 0
(1+2t, 1-t, 3-2t).(2,-1,-2)
= 2 + 4t - 1 + t - 6 + 4t
= -5 + 9t = 0
t = 5/9
=====

N = (1/9)(19, 4, 17)
===============