The Student Room Group

P3 Vector From Jan 05 Ed Paper

The question is:
Relative to a fixed origin O, the point A has position vector 5j + 5k and the point B has position vector 3i+2j-k.

a) Find the vector equation of the line L which passes through A and B.

The point C lies on the line L and OC is perpendicular to L.

b) Find the position vector of C.

For a) I got:
(3i+2j-k)-(5j+5k)=3i+2j-k-5j-5k=3i-3j-6k

I am stuck with part b).
I think it's to do with a.b=|a||b|cos@
Because it's perpendicular, it's going to be cos 90 = 0
so a.b= 0
and a is 3i-3j-6k
Putting this in I get (3i-3j-6k).b=0
(3i-3j-6k).(xi+yj+zk)=0
3x-3y-6z=0
I don't think I'm getting anywhere lol. Vectors and Integration is my weakness in P3; I don't understand. I've got a crap teacher. :rolleyes:
Reply 1
a)
B-A = (3, -3, -6)
So L: (0, 5, 5) + t(3, -3, -6)
[or (3, 2, -1) + t(3, -3, -6)]

b)
Since C lies on L, then it has position vector:
(3t, 5-3t, 5-6t)
OC . direction vector of L = 0
(3t, 5-3t, 5-6t) . (3, -3, -6) = 0
9t - 3(5-3t) - 6(5-6t) = 0
9t - 15 + 9t - 30 + 36t = 0 => t=5/6
Substituting in (3t, 5-3t, 5-6t) gives:
(5/2, 5/2, 0)
So position vector of C is:
2.5i + 2.5j

I've forgotten most of P3 vectors, so I'm not very sure my answer's right.
Reply 2
SednaNeptune
The question is:
Relative to a fixed origin O, the point A has position vector 5j + 5k and the point B has position vector 3i+2j-k.

a) Find the vector equation of the line L which passes through A and B.


0i + 5j + 5k + t[3i -3j -6k]
The part in brackets is the direction and the magnitude doesn't matter. Hence it can be cancaelled down to:
0i + 5j + 5k + t[i - j - 2k]

The point C lies on the line L and OC is perpendicular to L.

b) Find the position vector of C.

If we let C have position ai + bj + ck.

OC perpendicular to L
Hence (directions) : OC.L = 0
(ai+bj+ck).(i-j-2k) = 0 = a-b-2c
a-b=2c or c = 0.5(a-b).
Hence the position of C can be refined to as:
i[a] + b[j] + 0.5(a-b)[k]

The equation of L can be rewritten as:
r = i[t] + j[5-t] + k[5-2t]
C lies on L. Hence:
a=t,
b=5-t
0.5(a-b) = 5-2t

As a=t, b=5-a
Hence, using the third expression:
a-b = 10 - 4t
a-5+a = 10 - 4a
6a = 15
a=(15/6) = 5/2 = 2.5
b=5-a=2.5
c=0.5[a-b] = 0.5[2.5-2.5]=0

Hence C has position 2.5i + 2.5j + 0k.