
Applied vector question
Hey there  I'm stumped with this piece of revision. Can't find many if any examples in my class notes. Can anyone offer some comments on my attempt and how I can complete this  Hopefully I can learn and use this as an example as ther are others.
The position vectors of point A,B & C, relative to a fixed origin (O), are:
a = 6i + 4j  k
b = 8i + 5j  3k
c = 2i + 8j  5k
(i)Find the vector AB, the lenght AB, the cosine of angle ABC and the area of triangle ABC
(ii) Find the vector equation of the straight line passing through A and B
(iii) Find the coordss of the point D on the line for which OD is prependicular to AB. Hence, or otherwise calculate the shortest distance from O to the line AB
OK, so for (i)
AB = b  a
AB = 2i + j  2k
Modulus of AB is 3
To find the cosine of angle ABC I need the vector and modulus of CB,
which is b c = 6i  3j +2k and mod. = 7
The anlge is cos x = (AB * CB)/(mod. AB * mod. CB)
cos x = 5/21
For the area: The length/modulus of AC and CB multiplied and divided by 2
mod. AC = 4*sqrt3
mod. CB = 7
Area therefore, 14sqrt3
(ii) vector equation of line passing through A, B
let P be any point on the line with vector r.
OP = OA + AP
AP = t*AB
AP = t*(ba)
AB = (b  a) = 2i + j  2k
AP = 2ti + tj  2tk
therfore OP = OA + AP
OP = 6i + 4j  k + 2ti + tj  2tk
OP = (6 +2t)i + (4+t)j  (1 + 2t)k
(iii) no idea.
How have I gotten on with my parts (i) & (ii)?
Thanks, D

Hi
I do not find the same as you for the area
For (iii)
D is on (AB) therefore from (ii) OD = (6 +2t)i + (4+t)j  (1 + 2t)k
OD is perpendicular to AB therefore their dot product is equal to 0
This gives you an equation to find t

Yes, I've taken a look again  Not sure why I did that. area will be (mod.AB*mod.BC)/2
So I get 21/2
As for (iii) I calculate the coords as 2,2,3 and length sqrt17

Err, got it wrong again I see. I should use 1/2*AB*BC*Sin(arc cos(5/21)) which gives 10.198(3dp)

Yes but using the cross product gives a simpler solution
BTW
sin²(arccos(5/21)) + cos²(arccos(5/21) = 1
sin²(arccos(5/21)) + (5/21)² = 1
sin²(arccos(5/21)) = 1  (5/21)² = 416/21²
sin(arcos(5/21)) = sqrt(416)/21 = 4sqrt(26)/21
Area = 1/2 * 3 * 7 * 4sqrt(26)/21 = 2sqrt(26)
You can easier find this solution using cross product