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

[Istiak]

Homework Statement

Find value of $\theta$ when $V_x=V_y$

Relevant Equations

Vector

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.

 

[BvU]

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 $\vec V_x$

$\$

 

[Istiak]

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?

 

[Steve4Physics]

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 $tan\theta = \frac {V_x}{V_y}$.

 

[BvU]

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.

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

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.

$\$

 

[Istiak]

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.

 

[BvU]

$V=\sqrt{2}V_x$ is correct.

when i put it in my main equation

Your main equation being which one ? Is it applicable ?

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

And what do you then get for $\vec V \cdot \vec V_x\$ ?

You did not answer that. Why not ?

$\$

 

[Istiak]

Your main equation being which one ? Is it applicable ?

Forgot what you 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.

Which is correct: the angle between a vector and itself is zero.

Isn't that unit vector? $\hat x=\frac{\vec x}{|x|}$

 

[BvU]

$$=>\sqrt{2}\vec V_x \cdot \vec V_x$$ NOOOOO !

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\$ ?

$\$

 

[Istiak]

What is $\vec V\cdot \vec V_x\$ ?

$V V_x \cos\theta$

 

[haruspex]

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.

 

[Istiak]

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.

 

[haruspex]

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 $\vec V_x$.