# Finding an angle in 3D

## Homework Statement

Determine the angle θ between the 300-lb force and the line OC.

I have attached an image of the question

P°Q = PQcos(θ)

## The Attempt at a Solution

I tried finding the vectors AB and OA as well as their magnitudes first
O is the origin
A = (9,0,0)
B = (0, 16, 6)
C =(9, 16, 0)

AB = <0-9, 16-0, 6-0>
AB = <-9, 16, 6>

|AB| = sqrt((-92) + 162 + 62)
|AB| = 19.31

OA = <9-0, 0-0, 0-0>
OA = <9, 0, 0>

|OA| = 9

AB°OA = |AB||OA| cos(θ)

<-9, 16, 6>°<9, 0, 0> = (19.31)(9)cos(θ)

arccos(-81/173.79)

θ = 117.78

It says this answer is not correct and I realize that I haven't included the 300-lb force, but how would I take that force into account?

#### Attachments

• 19.6 KB Views: 436

Related Engineering and Comp Sci Homework Help News on Phys.org
Simon Bridge
Homework Helper
Looks like you've found the angle of AB with line OA ... but it asks for the angle with OC.

...Excellent observation...it's been a long day.

So I ran the calculations again, this time with OC. and I got an angle of 34.046 which is still wrong. :/

I ran the calculations, for the angle with OC using the given info and got a larger angle than yours, double check your calculations.

So I checked my calculations and still got 34.056°. My calculations are as follows:

O is the origin
A = (9,0,0)
B = (0, 16, 6)
C =(9, 16, 0)

AB = <0-9, 16-0, 6-0>
AB = <-9, 16, 6>

|AB| = sqrt((-92) + 162 + 62)
|AB| = 19.31

OC = <9-9, 16-0, 0-0>
OC = <0, 16, 0>

|OC| = 16

AB°OC = |AB||OC| cos(θ)

<-9, 16, 6>°<0, 16, 0> = (19.31)(16)cos(θ)

θ = arccos(256/308.96)

θ = 34.046

Shouldn't OC = <9,16,0> ?

Yes, you are correct. I think I've been at this too long :/.

My calculations are as follows

A = (9, 0, 00
B = (0, 16, 6)
C = (9, 16, 0)

AB = <0-9, 16-0, 6-0>
AB = <-9, 16, 6>

|AB| = sqrt((-9)^2 + 16^2 + 6^2)
|AB| = 19.31 kN

OC = <9-0, 16-0, 0-0>
OC = < 9, 16, 0>

|OC| = sqrt(9^2 + 16^2)
|OC| = 18.358 kN

AB°OC = |AB||OC|cos(θ)

<-9, 16, 6>°<9, 16, 0> = (19.31)(18.358) cos(θ)

θ = arccos(175/354.47)
θ = 60.42°