Vector problem

[jonnburton]

I've got a question in my book which I've attempted several times and not been able to resolve. I was wondering whether anyone could tell me whether I'm going about this in the right way, and if so, what I'm doing wrong?

The line L1 passes through the point A (2,0,2) and has equation \(\displaystyle r = \begin{bmatrix}2\\0\\2\end{bmatrix} + t\begin{bmatrix}2\\6\\-3\end{bmatrix}\)


Show that the line L2, which passes through the points B(4,4,-5) and C(0,-8,1) is parallel to the line L1.



It took me a while to figure out how to approach this as I knew how to show two lines are perpendicular to one another, but now how to show they are parallel. Then it struck me that if a line bisects both L2 and L2 at 90 degrees, they must be parallel.


So, in order to make a line bisect both L1 and L2, I thought find a point P on L1 closest to point C (0, -8,1) on L2.


To find coordinates of point P, use \(\displaystyle \vec{CP}\):

\(\displaystyle \begin{bmatrix}0\\-8\\1\end{bmatrix} - \begin{bmatrix}2+2t\\6t\\2-3t\end{bmatrix}\)

\(\displaystyle \begin{bmatrix}-2-2t\\-8-6t\\-1+3t\end{bmatrix}\)



\(\displaystyle \vec{CP} * \begin{bmatrix}2\\6\\-3\end{bmatrix} = 0\) so the line is perpendicular to L1.

\(\displaystyle \begin{bmatrix}-2-2t\\-8-6t\\-1+3t\end{bmatrix} * \begin{bmatrix}2\\6\\-3\end{bmatrix}\)

This leads to t = 1

So, substituting this value t=1 back into the original equation for L1, the coordinates of P are (4,6, -1).




Getting a vector equation for \(\displaystyle \vec{CP}\)

Direction vector will be \(\displaystyle \begin{bmatrix}4\\6\\-1\end{bmatrix} - \begin{bmatrix}0\\-8\\1\end{bmatrix} = \begin{bmatrix}4\\14\\-2\end{bmatrix}\)

\(\displaystyle \begin{bmatrix}0\\-8\\1\end{bmatrix} + \lambda\begin{bmatrix}4\\14\\-2\end{bmatrix}\)



But when multiplying the direction vectors of \(\displaystyle \vec{CP}\) and L1 does not give zero:

\(\displaystyle \begin{bmatrix}4\\-14\\-2\end{bmatrix} * \begin{bmatrix}2\\0\\2\end{bmatrix} = 4\)

 

[serds]

Just have a look at the normel vectors of the two lines, which are perpendicular to them.
If they have the same direction, the lines must be parallel.

 

[pka]

The line L1 passes through the point A (2,0,2) and has equation \(\displaystyle r = \begin{bmatrix}2\\0\\2\end{bmatrix} + t\begin{bmatrix}2\\6\\-3\end{bmatrix}\)

Show that the line L2, which passes through the points B(4,4,-5) and C(0,-8,1) is parallel to the line L1.

Given \(\displaystyle {\ell _1}= \left\langle {2,0,2} \right\rangle + t\left\langle {2,6, - 3} \right\rangle \) and \(\displaystyle \;B:\,\,(4,4, - 5)\;\& \;C:\,(0. - 8.1)\)

Note that \(\displaystyle <2,6,-3>\) is the direction vector of \(\displaystyle \ell_1\) and \(\displaystyle \overrightarrow {BC} =<-4,-12,6>\) is the direction vector of \(\displaystyle \ell_2\)

Are they parallel?

 

[jonnburton]

Given \(\displaystyle {\ell _1}= \left\langle {2,0,2} \right\rangle + t\left\langle {2,6, - 3} \right\rangle \) and \(\displaystyle \;B:\,\,(4,4, - 5)\;\& \;C:\,(0. - 8.1)\)

Note that \(\displaystyle <2,6,-3>\) is the direction vector of \(\displaystyle \ell_1\) and \(\displaystyle \overrightarrow {BC} =<-4,-12,6>\) is the direction vector of \(\displaystyle \ell_2\)

Are they parallel?

Thanks for the replies.

I'm not sure how to tell if the normal vectors have the same direction.

\(\displaystyle \begin{bmatrix}2\\6\\-3\end{bmatrix} \cdot \begin{bmatrix}-4\\-12\\6\end{bmatrix} = (-8)+(-72)+(-18) = -98\), so according to this, they are not parallel...!

 

[jonnburton]

Thanks for the replies.

I'm not sure how to tell if the normal vectors have the same direction.

\(\displaystyle \begin{bmatrix}2\\6\\-3\end{bmatrix} \cdot \begin{bmatrix}-4\\-12\\6\end{bmatrix} = (-8)+(-72)+(-18) = -98\), so according to this, they are not parallel...!

Actually, I was getting confused there... The fact that the product doesn't come to zero means they're not perpendicular.

What is clear is that the direction vector of L1 is multiplied by -2 but I wasn't aware that this results in a parallel line...

 

[DrPhil]

Thanks for the replies.

I'm not sure how to tell if the normal vectors have the same direction.

\(\displaystyle \begin{bmatrix}2\\6\\-3\end{bmatrix} \cdot \begin{bmatrix}-4\\-12\\6\end{bmatrix} = (-8)+(-72)+(-18) = -98\), so according to this, they are not parallel...!

They are parallel if one is a real number times the other.
Is there an \(\displaystyle \alpha\) such that

\(\displaystyle \begin{bmatrix}2\\6\\-3\end{bmatrix} = \alpha \begin{bmatrix}-4\\-12\\6\end{bmatrix} = \begin{bmatrix}-4\ \alpha\\-12\ \alpha\\6\ \alpha\end{bmatrix} \)

 

[pka]

Thanks for the replies.
I'm not sure how to tell if the normal vectors have the same direction.
\(\displaystyle \begin{bmatrix}2\\6\\-3\end{bmatrix} \cdot \begin{bmatrix}-4\\-12\\6\end{bmatrix} = (-8)+(-72)+(-18) = -98\), so according to this, they are not parallel...!

These are not normal vectors. Lines have direction vectors.
Two line are parallel if and only if their direction vectors are parallel.

Two vectors are parallel it one is a scalar multiple of the other.

Can you multiply \(\displaystyle <-4,-12,6>\) some number to get \(\displaystyle <2,6,-3>~?\) If you can the they are parallel.

So \(\displaystyle \ell_1\|\ell_{2}\).

 

[jonnburton]

Thanks a lot for the replies; I think I've got it now...