# Applied vector question

• May 3rd 2010, 08:33 AM
dojo
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 co-ordss 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*(b-a)
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
• May 3rd 2010, 11:37 AM
running-gag
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
• May 3rd 2010, 11:52 AM
dojo
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
• May 3rd 2010, 12:00 PM
dojo
Err, got it wrong again I see. I should use 1/2*AB*BC*Sin(arc cos(5/21)) which gives 10.198(3dp)
• May 3rd 2010, 12:35 PM
running-gag
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