What Is the Angle Between Position Vectors AB and AC?

[Robb]

Homework Statement

FInd the angle theta between AB and AC. a-1.7m, b=1.3m

masteringengineering.com says my answer is wrong. I'm not sure of any other way to do the problem. Please advise.

Homework Equations

A(3, 0, 0) B(0, -.75, 1.3) C(0, 1.7, 1.5)

The Attempt at a Solution

r(AC)= -3i +1.7j + 1.5k
r(AB)= -3i -.75j +1.3k

r(ACmag)*r(ABmag)= 13.3

r(AB)*r(AC)= 9.67

arccos(9.67/13.3)= 43.4

 

[Nidum]

Can you see a way of getting the lengths AB , BC and CA using right angle triangles and Pythagoras' Theorem ?

 

[Robb]

BC= sqrt(2.45^2+.2^2)= 2.46
AB= 3.54
AC= 3.76

I found AB & AC by finding the magnitude of the direction vectors r. If this is correct, I assume I take the tangent but I'm not sure which side would be considered adjacent.

 

[ehild]

r(ACmag)*r(ABmag)= 13.3

Wrong.

r(AB)*r(AC)= 9.67

Keep 4 digit during the calculations.

 

[ehild]

BC= sqrt(2.45^2+.2^2)= 2.46
AB= 3.54
AC= 3.76

BC is not needed. Your initial approach was correct, but the magnitude of the vectors are wrong.

 

[Robb]

13.3 is the magnitude of each vector multiplied together (3.76*3.54). 9.76 is obtained by multiplying the vectors together then adding the result (-3i-.75j+1.3k)(-3i+1.7k+1.5k)= (9-1.28+1.95) = 9.76. So, I figured arccos(9.76/13.3 )= 43.4

 

[Robb]

So, by carrying out my decimal to four places I get 12.6136 for a magnitude. so, arccos(9.76/12.6136)= 39.9

 

[ehild]

So, by carrying out my decimal to four places I get 12.6136 for a magnitude. so, arccos(9.76/12.6136)= 39.9

The product of the two vectors is not 9.76, but close. The result for the angle is all right. Write it out with two significant digits.