Proving Parallel Vectors: AB & CD

[adjacent]

Homework Statement

This is not a homework question,but I will post here anyway.
$\vec{AB}=\left(\begin{matrix} 1 \\ 2 \end{matrix}\right)$
$\vec{CD}=\left(\begin{matrix} 2 \\ 4 \end{matrix}\right)$
(a)Prove that $|\vec{CD}|$ is double of $|\vec{AB}|$
(b)Prove that $\vec{CD}$ is parallel to $\vec{AB}$

The Attempt at a Solution

(a) The sides of right angle triangle are doubled here.
$c^2=b^2+a^2$
$2(a^2+b^2)=2c^2$
If a and b double,c will also double.

(b) $\theta=\tan^{-1}\left(b \over a \right)$
$\theta=\tan^{-1}\left(\frac{b}{a} \right)$
Which means even if the two sides double,the angle will not double,therefore the slope of vector will not change.

I don't think this is a proof.Can anyone help me?
Also what are the uses of Vectors like this?( I just learned them today)

 

[Mentallic]

Denote the length of AB as d₁ and the length of CD as d₂. Can you show what d₁ is equal to? Now show what d₂ is equal to.

For these 2 vectors to be parallel, their gradients must be equal which you have correctly used the right expression to calculate it, but let's do it slowly to avoid any confusion. What is $\theta _1$ equal to? And now $\theta _2$?

Essentially I want you to first show that you can calculate the lengths and angles of each vector separately, and once you can handle that, then we'll move on to comparing them.

 

[adjacent]

Denote the length of AB as d₁ and the length of CD as d₂. Can you show what d₁ is equal to? Now show what d₂ is equal to.

Do I have to use variables like a and b or the numbers(1,2 , 2,4)?

 

[Mentallic]

Given a vector

$$v=\left(\begin{matrix} a \\ b \end{matrix}\right)$$

the length is given by

$$\sqrt{a^2+b^2}$$

and angle by

$$\tan^{-1}\left(\frac{b}{a}\right)$$

So regardless of what values your vectors have, these two properties never change. So I want you to do the numerical calculations with the vectors you were given since you've already shown that you know the formulae.

 

[HallsofIvy]

Homework Statement

This is not a homework question,but I will post here anyway.
$\vec{AB}=\left(\begin{matrix} 1 \\ 2 \end{matrix}\right)$
$\vec{CD}=\left(\begin{matrix} 2 \\ 4 \end{matrix}\right)$
(a)Prove that $|\vec{CD}|$ is double of $|\vec{AB}|$
(b)Prove that $\vec{CD}$ is parallel to $\vec{AB}$

The Attempt at a Solution

(a) The sides of right angle triangle are doubled here.
$c^2=b^2+a^2$
$2(a^2+b^2)=2c^2$
If a and b double,c will also double.

What right angle triangle are you talking about? What are "a", "b", "c", and "d"? You need to say that. You haven't shown that "If a and b double, c will also double". Multiplying $a^2+ b^2$ by 2 has nothing to do with doubling a and b. What is $(2a)^2+ (2b)^2$?

(b) $\theta=\tan^{-1}\left(b \over a \right)$
$\theta=\tan^{-1}\left(2\frac{b}{a} \right)$
Which means even if the two sides double,the angle will not double,therefore the slope of vector will not change.

Again, you have not said what "a" and "b" are here. I Nor have you said what "$\theta$" is. The two "$\theta$s" here are NOT the same. $tan^{-1}\left(b \over a\right)$ is NOT equal to $tan^{-1}\left(2 b\over a\right)$. You have made an obvious error- perhaps just a typo. But there is no reason to introduce trig functions here. Just the fact that the angles are the same in similar is sufficient.

I don't think this is a proof.Can anyone help me?

With proofs like these, think in terms of basic definitions. What does it mean for a vector to be "doubled"? (I can think of two equivalent definitions and what proof I give depends on which definition I use.) What does it mean for two vectors to be "parallel"?

Also what are the uses of Vectors like this?( I just learned them today)[/QUOTE]
In physics such things as "velocity", "force", and "momentum" are vectors.

 

[adjacent]

So regardless of what values your vectors have, these two properties never change. So I want you to do the numerical calculations with the vectors you were given since you've already shown that you know the formulae.

$d_1=\sqrt{1^2+2^2} \approx 2.236$
$d_2=\sqrt{2^2+4^2} \approx 4.472$

$\theta_1$=Angle of vector AB(Assuming right is 0°,I mean horizontal right)
$\theta_2$=Angle of vector CD(Assuming right is 0°,I mean horizontal right)

$\theta_1=tan^{-1}\left( 2 \over 1 \right) \approx 63.435°$
$\theta_2=tan^{-1}\left( 4 \over 2 \right) \approx 63.435°$

 

[adjacent]

Again, you have not said what "a" and "b" are here. I Nor have you said what "$\theta$" is. The two "$\theta$s" here are NOT the same. $tan^{-1}\left(b \over a\right)$ is NOT equal to $tan^{-1}l\left(2 b\over a\right)$. You have made an obvious error- perhaps just a typo. But there is no reason to introduce trig functions here. Just the fact that the angles are the same in similar is sufficient.

Oh.It should be $tan^{-1}\left(\frac{2b}{2a}\right)$.What's the $l$ you included?What you and Mentallic are saying is contradictory.

 

[HallsofIvy]

No, what Mentallic and I are saying are not contradictory. And I did not say that your calculations were wrong. Only that you cannot change from "u" and "v" and start talking about "a", "b", and "c", and $\theta$ without defining them. The "l" was a typo. I have removed it.

 

[adjacent]

I have done the numerical calculations.Now what?

 

[Mentallic]

Sorry about taking so long to respond, life gets in the way sometimes.

$d_1=\sqrt{1^2+2^2} \approx 2.236$
$d_2=\sqrt{2^2+4^2} \approx 4.472$

It's best if you leave the answers in surd form which is not approximate. Can you now show that d₂ is twice that of d₁ while not approximating the lengths of your vectors?

$\theta_1$=Angle of vector AB(Assuming right is 0°,I mean horizontal right)
$\theta_2$=Angle of vector CD(Assuming right is 0°,I mean horizontal right)

$\theta_1=tan^{-1}\left( 2 \over 1 \right) \approx 63.435°$
$\theta_2=tan^{-1}\left( 4 \over 2 \right) \approx 63.435°$

Not to be pedantic, but something more like the following is what you should be aiming for:

$\theta_1=\tan^{-1}\left(2 \over 1\right) = \tan^{-1}(2)$
$\theta_2=\tan^{-1}\left(4\over 2\right) = \tan^{-1}(2) = \theta_1$
Therefore since both vectors have equal arguments (angle) then they are parallel.

The use of a calculator here is unnecessary. You don't need to know exactly what the angle is in terms of degrees, all you need is to show that the two angles (whatever they may be) are equal.

Now, since we've got the archaic approach out of the way, how about we try a more simple and elegant solution?

$\vec{CD}=2\cdot \vec{AB}$

since

$\left(\begin{matrix} 2 \\ 4 \end{matrix}\right)=2\left(\begin{matrix} 1 \\ 2 \end{matrix}\right)$

and there must be some simple properties of vectors that you've been taught such that you can easily prove what you need without getting into the calculations.

 

[adjacent]

Thank you.
Is that a proof?
What about all vectors(Scalar multiples) in general?

 

[Mentallic]

Thank you.
Is that a proof?

Yes.

What about all vectors(Scalar multiples) in general?

In general, if $\vec{v_1}=\alpha\vec{v_2}$ for some non-zero constant $\alpha$ then these vectors are parallel and $\vec{v_1}$ is a scalar multiple of $\vec{v_2}$ by a value of $\alpha$. Notice that if $\alpha<0$ then the vectors point in opposite directions but they are still parallel. The angle calculation also agrees with this.

Another useful question that you might want to try answer is what would the unit vector parallel to these two vectors be? Remember that a unit vector has length 1.

 

[Mark44]

In general, if $\vec{v_1}=\alpha\vec{v_2}$ for some non-zero constant $\alpha$ then these vectors are parallel and $\vec{v_1}$ is a scalar multiple of $\vec{v_2}$ by a value of $\alpha$.

This is how I would have done the problem.
$$\vec{CD} = \begin{pmatrix} 2 \\ 4\end{pmatrix} = 2\begin{pmatrix} 1 \\ 2\end{pmatrix} = 2\vec{AB}$$

Clearly the magnitude of CD is twice that of AB, and since CD is a scalar multiple of AB, AB is parallel to CD.

 

[Mentallic]

This is how I would have done the problem.
$$\vec{CD} = \begin{pmatrix} 2 \\ 4\end{pmatrix} = 2\begin{pmatrix} 1 \\ 2\end{pmatrix} = 2\vec{AB}$$

I would have too, but I prefer to first help the student with the method that they've brought to the thread - of course only if the method is valid.