P3 vectors

Referred to an origin O, the points A, B and C have position vectors (9i – 2j + k), (6i + 2j + 6k) and (3i + pj + qk) respectively, where p and q are constants.

(a) Find, in vector form, an equation of the line l which passes through A and B. (2)

Given that C lies on l,
(b) find the value of p and the value of q, (2)

(c) calculate, in degrees, the acute angle between OC and AB.(3)

The point D lies on AB and is such that OD is perpendicular to AB.
(d) Find the position vector of D. (6)

i got the first 3 answers

a)l = (9i-2j+k)- µ (-3i +4k+5k)

b)µ=-2 p=6 q=11

c) angle= 39.8 degree

bt im really stuck on d dont know wot to do at all

i thought id use
a.b=0 (coz a is perpendicular to b)
bt still cant do it!
thanks 2 any1 who can do it!

Reply 1

qqqwwweee

2

tammypotato

Referred to an origin O, the points A, B and C have position vectors (9i – 2j + k), (6i + 2j + 6k) and (3i + pj + qk) respectively, where p and q are constants.

(a) Find, in vector form, an equation of the line l which passes through A and B. (2)

Given that C lies on l,
(b) find the value of p and the value of q, (2)

(c) calculate, in degrees, the acute angle between OC and AB.(3)

The point D lies on AB and is such that OD is perpendicular to AB.
(d) Find the position vector of D. (6)

i got the first 3 answers

a)l = (9i-2j+k)- µ (-3i +4k+5k)

b)µ=-2 p=6 q=11

c) angle= 39.8 degree

bt im really stuck on d dont know wot to do at all

i thought id use
a.b=0 (coz a is perpendicular to b)
bt still cant do it!
thanks 2 any1 who can do it!

yes
take OD as a and AB as b (or vice versa)
so OD.AB = 0

Have you used the fact that the point D lies on AB?
You can express OD with only one unkown.

Reply 2

Fermat

8

tammypotato

Referred to an origin O, the points A, B and C have position vectors (9i – 2j + k), (6i + 2j + 6k) and (3i + pj + qk) respectively, where p and q are constants.

(a) Find, in vector form, an equation of the line l which passes through A and B. (2)

Given that C lies on l,
(b) find the value of p and the value of q, (2)

(c) calculate, in degrees, the acute angle between OC and AB.(3)

The point D lies on AB and is such that OD is perpendicular to AB.
(d) Find the position vector of D. (6)

i got the first 3 answers

a)l = (9i-2j+k)- µ (-3i +4k+5k)

b)µ=-2 p=6 q=11

c) angle= 39.8 degree

bt im really stuck on d dont know wot to do at all

i thought id use
a.b=0 (coz a is perpendicular to b)
bt still cant do it!
thanks 2 any1 who can do it!

OA = 9i - 2j + k
OB = 6i + 2j + 6k
OC = 3i + pj + qk

a)
l = OA + µ(OB - OA)
l = 9i - 2j + k + µ(-3i + 4j + 5k)
=====================

b)
l = (9 - 3µ)i + (-2 + 4µ)j + ( 1 + 5µ)

equating the coordinates for OC,

9 - 3µ = 3
-2 + 4µ = p
1 + 5µ = q

giving,

µ = 2
p = 6
q = 11
====

c)
OC = 3i + 6j + 11k
|OC| = √(9 + 36 + 81) = √(166)
AB = -3i + 4j +5 k
|AB| = √(9 + 16 + 25) = √(50)

cos ø = OC.AB/(|OC||OA|)
cos ø = (3i + 6j + 11k).(-3i + 4j + 5k)/( √(166)* √(50))
cos ø = (-9 + 24 + 55)/( √(166)* √(50))
cos ø = 70/( √(166)* √(50))
ø = 39.79°
=======

d)
OD = xi + yj + zk
OD.AB = 0
-3x + 4y + 5z = 0

OD lies in the line l, therefore

9 - 3µ = x
-2 + 4µ = y
1 + 5µ = z

substituting x = 9 - 3µ in -3x + 4y + 5z = 0

-2 + 4µ = y
1 + 5µ = z
4y + 5z + 9µ = 27

substituting y = -2 + 4µ

1 + 5µ = z
5z + 25µ = 35

substituting z = 1 + 5µ,

5 + 25µ + 25µ = 35
50µ = 30
µ = 0.6

z = 4
y = 0.4
x = 7.2

So, OD is 7.2i + 0.4j + 4k
================

Reply 3

Pixelfairy #1

tammypotato

Referred to an origin O, the points A, B and C have position vectors (9i – 2j + k), (6i + 2j + 6k) and (3i + pj + qk) respectively, where p and q are constants.
...

Doing P3 at 4 in the morning isn't the best idea... :rolleyes:

Are you sure you're not studying too hard? :eek:

P3 vectors

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