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
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
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