Finding the value of lambda so that two lines are parallel

[ChiralSuperfields]

Homework Statement

I am try to understand how they the solution got their answer as my method is not giving the correct answer.

Relevant Equations

$\textbf {PR} = -3\hat i + 6\hat j - 2\hat k$

$\textbf {RS} = 2\hat i - 4\hat j + \lambda \hat k$

For this (ii),

The solution is $\lambda = \frac{4}{3}$, however when I tried solving the problem I did not get their answer. Dose somebody please guide me to their solution and tell me what I did wrong with my method below:

$\textbf {PR} = -3\hat i + 6\hat j - 2\hat k$
$\textbf {RS} = 2\hat i - 4\hat j + \lambda \hat k$
$\hat {PR} = \frac{\textbf {PR}}{|PR|} = \hat {RS} = \frac{\textbf {RS}}{|RS|}$
$\hat {PR} = \frac{-3\hat i + 6\hat j - 2\hat k}{\sqrt{49}} = \frac{2\hat i - 4\hat j + \lambda \hat k}{\sqrt{20 + \lambda^2}}$

Then square both sides (which I think we then use the scalar product) giving:
$\frac{9 + 36 + 4}{49} = \frac{4 + 16 + \lambda^2}{20 + \lambda^2}$
$\frac{49}{49} = \frac{20 + \lambda^2}{20 + \lambda^2}$
$1 = 1$

Many thanks!

[Moderator's note: moved from a technical forum.]

 

[Frabjous]

You found that the length of a unit vector is 1 (twice).
How do you show that two vectors are parallel?

 

[ChiralSuperfields]

You found that the length of a unit vector is 1 (twice).
How do you show that two vectors are parallel?

Thank you for your reply @Frabjous!

The two vectors should be parallel if they are multiples of each other by some constant.

Many thanks!

 

[Frabjous]

The two vectors should be parallel if they are multiples of each other by some constant.

Don’t normalize the vectors. Find the constant for each component. They should all be equal.

 

[ChiralSuperfields]

Don’t normalize the vectors. Find the constant for each component. They should all be equal.

Thank you for your reply @Frabjous!

I have gotten the correct solution :) However, I am wondering why my first method did not work. Is it because there is no other way of solving for lambda without squaring both sides which makes the numerator equal to the denominator for each unit vector.

Many thanks!

 

[Mark44]

I have gotten the correct solution :) However, I am wondering why my first method did not work. Is it because there is no other way of solving for lambda without squaring both sides which makes the numerator equal to the denominator for each unit vector.

As already noted, you absolutely don't need to generate unit vectors. You did a lot of work for no noticeable gain.

$\textbf {PR} = -3\hat i + 6\hat j - 2\hat k$
$\textbf {RS} = 2\hat i - 4\hat j + \lambda \hat k$

Another point that was already noted is that for two vectors to be parallel (or antiparallel -- pointing in opposite directions), each one must be a nonzero scalar multiple of the other.
For the vectors above one can determine by nothing more than inspection that the scalar multiple must be -3/2. So $-2 = -3/2 \times \lambda$, or $\lambda = (-2)(-2/3) = 4/3$.

The solution is
$\lambda = \frac{4}{3}$, however when I tried solving the problem I did not get their answer. Dose somebody please guide me to their solution and tell me what I did wrong with my method below:

$\textbf {PR} = -3\hat i + 6\hat j - 2\hat k$
$\textbf {RS} = 2\hat i - 4\hat j + \lambda \hat k$

You don't need to do any of the stuff below.

$\hat {PR} = \frac{\textbf {PR}}{|PR|} = \hat {RS} = \frac{\textbf {RS}}{|RS|}$
$\hat {PR} = \frac{-3\hat i + 6\hat j - 2\hat k}{\sqrt{49}} = \frac{2\hat i - 4\hat j + \lambda \hat k}{\sqrt{20 + \lambda^2}}$

Then square both sides (which I think we then use the scalar product) giving:
$\frac{9 + 36 + 4}{49} = \frac{4 + 16 + \lambda^2}{20 + \lambda^2}$
$\frac{49}{49} = \frac{20 + \lambda^2}{20 + \lambda^2}$
$1 = 1$

Aside from the fact that the above work is not useful, it's even less useful to end up with an equation that is trivially true; i.e., that 1 is equal to itself.

 

[Ibix]

Is it because there is no other way of solving for lambda without squaring both sides which makes the numerator equal to the denominator for each unit vector.

It was (almost) fine up to here:

$\hat {PR} = \frac{-3\hat i + 6\hat j - 2\hat k}{\sqrt{49}} = \frac{2\hat i - 4\hat j + \lambda \hat k}{\sqrt{20 + \lambda^2}}$

What went wrong after that is that squaring both sides is equivalent to insisting that both vectors have the same length - but you'd already made them unit vectors, so that was true independent of $\lambda$.

You could have compared the coefficients of $\hat i$, which would hopefully have flagged to you why I wrote "almost", which is that $\hat{PR}=-\hat{RS}$ because they point in opposite directions. Or you could have required that $\hat{PR}\cdot\hat{RS}=\pm 1$. Either is more work than the method already mentioned of noting that for parallel vectors the coefficients of the basis vectors need to be in the same ratio, so $\frac{-3}2=\frac 6{-4}=\frac{-2}\lambda$.