Dot product help

[Stick]

Can someone please explain these questions for me? Thanks.

[polar]

is the vector from point C to point D, the information can be used in the formula for the dot product

[Stick]

Ah, I should've drawn a diagram, because I didn't think of using the cosine rule! Thanks!

If someone could still help out for question 17, that would be great.

[Special At Specialist]

I didn't think of using the cosine rule either...
I calculated that CD = -1/7 * (45i + 1209j) and then calculated the magnitude. I got the same answer, though, but I had to use a calculator.

For question 17:
DB = DO + OA + AB
DB = -OD + OA + OC
DB = -2i + j + 3j - i + j + 2k
DB = -3i + 5j + 2k
Let M = Midpoint of DB
DM = (-3/2)i + (5/2)j + k
OM = OD + DM
OM = 2i - j - (3/2)i + (5/2)j + k
OM = (1/2)j + (3/2)j + k

CE = CO + OA + AC
CE = -OC + OA + OD
CE = i - 2j - 2k + 3j + 2i - j
CE = 3i + j - 2k
Let N = Midpoint of CE
CN = (3/2)i + (1/2)j - k
ON = OC + CN
ON = -i + j + 2k + (3/2)i + (1/2)j - k
ON = (1/2)i + (3/2)j + k

Since ON and OM are the same, then the midpoints of DB and CE are the same, thus proving that they bisect each other.

To find the angle between DB and CE:
|DB| = sqrt(9 + 25 + 4)
|DB| = sqrt(38)
|CE| = sqrt(9 + 1 + 4)
|CE| = sqrt(14)
DB . CE = -9 + 5 - 4
DB . CE = -8
DB . CE = |DB||CE|cos(θ)
-8 = sqrt(14)sqrt(38)cos(θ)
-8 = 2sqrt(133)cos(θ)
cos(θ) = -4 / sqrt(133)
θ = arccos(-4 / sqrt(133))

If you want an approximation (don't write this unless the answer specifically asks for it), then here is it:
θ = 1.925 radians (to 6 decimal places)
θ = 110.2944 degrees (to 6 decimal places)
θ = 110 degrees, 17 minutes and 40 seconds (to the nearest second)

[hollowkenshin]

It says acute angle, so would it be approx. 69.71deg?

[Stick]

Whoops! I accidentally wrote out one of the vectors wrong, because I got 71 degrees as my answer. Thanks anyway.

[Special At Specialist]

It says acute angle, so would it be approx. 69.71deg?

Good point. Yes, you are right.