# 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

• statics hwk 7 2.113.png
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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°