Expert: Paul Klarreich - 8/30/2009

Question
Study CIE 9709 "A LEVEL" MATHEMATICS & vectors topic. The points A & B have position vectors,relative to the origin O,given by OA=i+2j+3k & OB=2i+j+3k.The line l has vector equation r=(1-2t)i+(5+t)j+(2-t)k.I) Show that I does not intersect the line passing through A & B. II) The point P lies on I & is such that angle PAB is equal to 60degrees. Given that the position vector of P is (1-2t)i+ (5+ t)j- (2-t)k, show that 3t^2+ 7t+2=0. Hence find the only possible position vector of P. For part i) I got AB=-i-2j-3k+2i+j+3k =i-j. Vector equation for line through AB=> r=i+2j+3k+s(i-j). I dont know how to continue the rest. For part ii) I got direction vector for AP is (-2t, 3+t, -1+t). The rest I dont know to do. Can you please help me? Thank you very much

Answer
Questioner: Warm
Country: China
Category: Advanced Math
Private: No
Subject: Vectors
Question: Study CIE 9709 "A LEVEL" MATHEMATICS & vectors topic.
The points A & B have position vectors,relative to the origin O,given by OA=i+2j+3k & OB=2i+j+3k.

The line L has vector equation r=(1-2t)i+(5+t)j+(2-t)k.

I) Show that L does not intersect the line passing through A & B.

II) The point P lies on L & is such that angle PAB is equal to 60degrees.

Given that the position vector of P is (1-2t)i+ (5+ t)j- (2-t)k,
show that 3t^2+ 7t+2=0.

Hence find the only possible position vector of P.

For part i) I got AB = -i-2j-3k+2i+j+3k =i-j.

>> Correct, but I like the <1,-1,0> notation.

Vector equation for line through AB=> r=i+2j+3k+s(i-j).
I dont know how to continue the rest.

For part ii) I got direction vector for AP is (-2t, 3+t, -1+t). The rest I dont know to do. Can you please help me? Thank you very much

........................................
Hi, again, Warm,

I did not do a good job on the last two questions, so I hope this one is better.  (It took a while to work it all out.)

OA=i+2j+3k & OB=2i+j+3k.

So  A = (1,2,3)  and  B = (2,1,3)

Now vector AB = < 1, -1, 0>  and the line AB = A + tAB

<x,y,z> = <1,2,3> + t <1, -1, 0>  or:

x = 1 + t
y = 2 - t
z = 3
...................

The line L has vector equation r=(1-2t)i+(5+t)j+(2-t)k.

I assume that means:  (changing t's to s's -- you can use any letter.)

x = 1 - 2s
y = 5 + s
z = 2 - s.
............................
I) Show that L does not intersect the line passing through A & B.

If there is a point on both lines, we must have x,y,z the same:

1 + t = 1 - 2s   AND
2 - t = 5 + s    AND
   3 = 2 - s.

So we have simultaneous equations to solve for s,t.

   3 = 2 - s.
gives  s = -1

2 - t = 5 + s
2 - t = 5 - 1
2 - t = 4
gives: t = - 2

Now:

1 + t = 1 - 2s
1 + -2 = 1 - 2(-1)
  - 1 = 1 + 2

Oops -- no good.  So there is no such point.
...................................................
II) The point P lies on L & is such that angle PAB is equal to 60degrees.

Given that the position vector of P is (1-2t)i+ (5+t)j + (2-t)k,
show that 3t^2 + 7t + 2 = 0.

Hence find the only possible position vector of P.
If angle PAB is 60 degrees, then the vectors  AP and AB have that angle between them.
................
Now there is a rule for the dot-product that says, for vectors X,Y:

X dot Y = | X | | Y | cos (angle between X and Y)

If P = < 1-2t, 5+t, 2-t > then

vector AP = P - A = < 1-2t, 5+t, 2-t > - <1,2,3>

AP = < - 2t, 3 + t, - t - 1 >
AB = < 1, -1, 0 >

Now apply the dot product rule:

AP dot AB = - 2t - 3 - t = - 3 - 3t

|AP| = sqrt( 4t^2 + (3+t)^2 + (-t-1)^2 )

|AP| = sqrt( 4t^2 + 9 + 6t + t^2 + t^2 + 2t + 1 )

|AP| = sqrt( 6t^2  + 8t + 10 )

|AB | = sqrt(1 + 1) = sqrt(2)

and cos 60 = 1/2

AP dot AB = | AP | | AB | cos 60

- 3 - 3t = sqrt(6t^2 + 8t + 10 )sqrt(2) (1/2)

Do some simplifying:

-6(1 + t) = sqrt(6t^2 + 8t + 10 )sqrt(2)

Square both sides:

36(1 + 2t + t^2) = (6t^2 + 8t + 10 )(2)

18(1 + 2t + t^2) = 6t^2 + 8t + 10

18 + 36t + 18t^2 = 6t^2 + 8t + 10

28t + 12t^2 + 8 = 0

7t + 3t^2 + 2 = 0

I think that is it.