Find the value of ##\theta## when the Vx and Vy components are the same

In summary, the conversation discusses finding the value of theta when the Vx component and Vy component are the same. Various equations are presented, including the formula for calculating the length of a vector when the angle between two components is known, and the dot product equation to find the angle between two vectors. The final result is that the angle between a vector and itself is zero.
  • #1
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Homework Statement
Find value of ##\theta## when ##V_x=V_y##
Relevant Equations
Vector
1630334534902.png

The picture may be blurry. I couldn't take more less blurry picture hence, giving it.

The question is : Find value of ##\theta## when ##V_x## component and ##V_y## component same.

I was using a simple equation of vector.

$$C=\sqrt{A^2+B^2+2AB\cos\theta}$$
$$V=\sqrt{V_x^2+V_y^2+2V_xV_y\cos\theta}$$
$$=\sqrt{2V_x^2+2V_x^2\cos\theta}$$
$$=\sqrt{2V_x^2(1+\cos\theta)}$$
$$=V_x\sqrt{2 \cdot 2\cos^2(\frac{\theta}{2})}$$
$$=2V_x\cos\frac{\theta}{2}$$
$$2\cos^{-1}\frac{V}{2V_x}=\theta$$
Somehow, I took ##V=V_x## cause, I couldn't get any other way to remove them. When I took ##V=V_x## then, I got ##\theta=120 \deg##. But, which was completely wrong. The answer was ##45 \deg##. I was trying to use another equation. ##\vec A \cdot \vec B=AB\cos\theta##; Using the equation I got ##0## which is 100% wrong.
 
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  • #2
Your $$V=\sqrt{V_x^2+V_y^2+2V_xV_y\cos\theta}$$ is a formula to calculate the length of ##\vec V## as the sum of ##\vec V_x## and ##\vec V_y## when the angle between ##\vec V_x## and ##\vec V_y## is ##\theta##. That angle is ##\ \pi/2\ ##, so the result is $$V=\sqrt{V_x^2+V_y^2}$$from which you can derive ##V## in terms of ##V_x##.

Your other equation $$\vec A \cdot \vec B=AB\cos\theta$$can then be used to find the angle between ##\vec V## and ##\vec V_x##

##\ ##
 
  • #3
BvU said:
Your $$V=\sqrt{V_x^2+V_y^2+2V_xV_y\cos\theta}$$ is a formula to calculate the length of ##\vec V## as the sum of ##\vec V_x## and ##\vec V_y## when the angle between ##\vec V_x## and ##\vec V_y## is ##\theta##. That angle is ##\ \pi/2\ ##, so the result is $$V=\sqrt{V_x^2+V_y^2}$$from which you can derive ##V## in terms of ##V_x##.

Your other equation $$\vec A \cdot \vec B=AB\cos\theta$$can then be used to find the angle between ##\vec V## and ##\vec V_x##

##\ ##
I wrote that ##\vec A \cdot \vec B = AB\cos\theta##
$$=>\vec A^2=A^2\cos\theta$$
$$=>1=\cos\theta$$
$$\theta=0$$
I am sure that I am mistaking with ##\cdot## I was getting more better way to multiply them. That's why I just wrote ##\vec A^2##. And, somehow I canceled vector A with A which was wrong either. So, how can I use the equation?
 
  • #4
Istiakshovon said:
The question is : Find value of ##\theta## when ##V_x## component and ##V_y## component same.
##tan\theta = \frac {V_y}{V_x}##

Edit: If the x and y are the other way round (picture too blurry to be sure) then ##tan\theta = \frac {V_x}{V_y}##.
 
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  • #5
Istiakshovon said:
I wrote that $$\vec A \cdot \vec B = AB\cos\theta \Rightarrow \vec A^2=A^2\cos\theta $$ $$\Rightarrow 1=\cos⁡\theta \Rightarrow ⁡\theta =0 $$
Which is correct: the angle between a vector and itself is zero.

Istiakshovon said:
I am sure that I am mistaking with ⋅ I was getting more better way to multiply them. That's why I just wrote A→2. And, somehow I canceled vector A with A which was wrong either. So, how can I use the equation?
Did you solve
BvU said:
That angle is ##\pi/2## , so the result is ##V=\sqrt { V_x^2+V_y^2} \ ## from which you can derive ##V## in terms of ##V_x##.
and what did you get ?

And what do you then get for ##\vec V \cdot \vec V_x\ ## ?Of course, @Steve4Physics is showing you a direct short track to the answer.

##\ ##
 
  • #6
1630326941962-1299657806.jpg

BvU said:
Your $$V=\sqrt{V_x^2+V_y^2+2V_xV_y\cos\theta}$$ is a formula to calculate the length of ##\vec V## as the sum of ##\vec V_x## and ##\vec V_y## when the angle between ##\vec V_x## and ##\vec V_y## is ##\theta##. That angle is ##\ \pi/2\ ##, so the result is $$V=\sqrt{V_x^2+V_y^2}$$from which you can derive ##V## in terms of ##V_x##.

Your other equation $$\vec A \cdot \vec B=AB\cos\theta$$can then be used to find the angle between ##\vec V## and ##\vec V_x##

##\ ##
I got that $$V=\sqrt{2}V_x$$
so, when i put it in my main equation. I get $$2\cos^{-1}\sqrt{2}=\theta$$. But, cos inverse 1 of square root 2 is undefined.
 
  • #7
##V=\sqrt{2}V_x## is correct.

Istiakshovon said:
when i put it in my main equation
Your main equation being which one ? Is it applicable ?

BvU said:
Your other equation ##\vec A\cdot \vec B = A B\cos \theta \ ## can then be used to find the angle between ##\vec V## and ##\vec V_x##

BvU said:
And what do you then get for ## \vec V \cdot \vec V_x\ ## ?
You did not answer that. Why not ?

##\ ##
 
  • #8
BvU said:
Your main equation being which one ? Is it applicable ?
Forgot what you said... :)
BvU said:
You did not answer that. Why not ?
I was in mobile when writing my last reply. Umm! ##\vec V \cdot \vec V_x =V V_x \cos\theta##
$$=>\sqrt{2}\vec V_x \cdot \vec V_x=\sqrt{2} V_x^2 \cos\theta$$
$$=>\cos\theta=1$$

Blah! Still not working. I think I had missed LHS.

BvU said:
Which is correct: the angle between a vector and itself is zero.
Isn't that unit vector? ##\hat x=\frac{\vec x}{|x|}##
 
  • #9
$$=>\sqrt{2}\vec V_x \cdot \vec V_x$$ NOOOOO ! :frown:

You want to review the inner product !

Let ##\ \vec V = (V_x, V_y)\ ## and ##\ \vec V_x = (V_x, 0)\ ##
What is ##\vec V\cdot \vec V_x\ ## ?

##\ ##
 
Last edited:
  • #10
BvU said:
What is ##\vec V\cdot \vec V_x\ ## ?
##V V_x \cos\theta##
 
  • #11
Istiakshovon said:
View attachment 288322

I got that $$V=\sqrt{2}V_x$$
so, when i put it in my main equation. I get $$2\cos^{-1}\sqrt{2}=\theta$$. But, cos inverse 1 of square root 2 is undefined.
I assume you mean you plugged it into this in post #1:
##2\cos^{-1}\frac{V}{2V_x}=\theta##
But you made a mistake in the algebra. If you do it correctly you will get the answer.
 
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  • #12
haruspex said:
I assume you mean you plugged it into this in post #1:
##2\cos^{-1}\frac{V}{2V_x}=\theta##
But you made a mistake in the algebra. If you do it correctly you will get the answer.
I found where the mistake is. Although, I got ##90 \deg## instead of ##45 \deg##. If there was no constant ##2## outside of cosine then, I would get ##45 \deg##. But, it wasn't accurate way to remove ##2##. According #2 the equation won't work here.
 
  • #13
Istiakshovon said:
I found where the mistake is. Although, I got ##90 \deg## instead of ##45 \deg##. If there was no constant ##2## outside of cosine then, I would get ##45 \deg##. But, it wasn't accurate way to remove ##2##. According #2 the equation won't work here.
Yes, you are right ... I did not study post #1 carefully enough.
There you wrote
##V=\sqrt{V_x^2+V_y^2+2V_xV_y\cos\theta}##
but in that equation theta would be the angle between ##V_x## and ##V_y##, i.e. 90 degrees. So you should have deduced ##V=\sqrt{V_x^2+V_y^2}##, which leads to the same equation, ##V=\sqrt{2}V_x##, as you had in post #6, which gets you no further forward.

Instead, start again with ##\vec V=\vec V_x+\vec V_y## and take the dot product of each side with ##\vec V_x##.
 

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