Why are they using Cosine instead of Sine for Cross Product?

  • Thread starter Thread starter yashboi123
  • Start date Start date
AI Thread Summary
The discussion centers on the use of cosine instead of sine for calculating the cross product, specifically addressing the negative sign in the answer and the use of a 25-degree angle. The negative sign is explained through the right-hand rule, indicating that the direction of the vectors leads to a negative result when following the anti-clockwise rotation. The angle AOC is derived as 90 degrees minus the marked 25-degree angle, which leads to the use of cosine in the calculations. Additionally, the relationship between sine and cosine is clarified, showing that sin(295) equals -cos(25), reinforcing the choice of cosine in the context. Understanding these trigonometric relationships is crucial for accurate vector calculations in physics.
yashboi123
Messages
17
Reaction score
0
Homework Statement
What is the cross product of A X C?
Relevant Equations
A X B = ABSin(x)
1693362565812.png
I don't understand why they are using cos and putting a negative in front of the answer, and secondly why they are using the 25 degree angle. The way I was thinking of solving it would be (96.0 m^2)sin(295). Can anyone explain this for me?
 
Physics news on Phys.org
The answer is -96 times sine of angle AOC.

The minus sign arises from the right-hand rule. Take your right hand, point the thumb in the direction of the positive z axis, and curl your fingers. The direction your fingers curl, which is anti-clockwise as we look at the diagram, must be the direction from the vector that's the first argument to the cross product to the vector that's the second argument, in order to get a positive sign. Since in this diagram ##\vec A## is clockwise from ##\vec C## we get a negative sign.

Now what about that cos? The diagram does not mark angle AOC, but we know it it is 90 degrees minus the marked 25 degree angle.
So they just use the formula for sine of the sum of two angles, as follows:
\begin{align*}
\sin\ AOC &= \sin(90 - 25) \\&= \sin(90 + (-25))\\& = \sin 90\ \cos (-25) + \cos 90\ \sin(-25)
\\&= \cos(-25)\times 1 + 0\times \sin(-25)
\\&= \cos\ 25+0
\\&= \cos\ 25\end{align*}
 
@andrewkirk , I would think @yashboi123 is looking for what is wrong with sin(295), following a standard method, rather than for an alternate method. Which method the book used is unknown.
yashboi123 said:
I don't understand why they are using cos and putting a negative in front of the answer, and secondly why they are using the 25 degree angle. The way I was thinking of solving it would be (96.0 m^2)sin(295). Can anyone explain this for me?
As your calculator will tell you, sin(295)=-cos(25).
There are some useful formulas:
cos(90-x)=sin(x)=sin(180-x)=-sin(-x)=sin(360+x)
So sin(295)=sin(360-65)=sin(-65)=-sin(65)=-cos(25).
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top