linear algebra

[renegade05]

First I'm not to sure where to post linear algebra stuff. I don't see a section on the main page. Should there be one maybe?

Anyway, I am having troubles with this question: Find all vector(s) of magnitude 13 that are parallel to the line of intersection between the planes 2x-3y+z=7 and x+4y-7z=2.

I think if someone gave me the first step or a hint in the right direction I may be able to figure it out. I'm really not too sure how to find the line of intersection though.

Please help.

 

[tkhunny]

You'll need some direction cosines. Can you find the line of intersection?

 

[renegade05]

No, thats really my problem i think. I totally am drawing a black when it comes to that. I mean in calc i would just make the equations equal each other to find the intersections. However, this vector stuff and working in R^3 is new to me, so im not sure how to go about it. Or is that how you go about it?

 

[tkhunny]

You should note that the line of intersection is orthogonal to BOTH plane's Normals. This should suggest a Cross Product.

 

[renegade05]

Ah i think i see now.

\(\displaystyle \vec{n1}=\begin{bmatrix} 2 \\-3 \\1 \end{bmatrix}\)

\(\displaystyle \vec{n2}=\begin{bmatrix} 1 \\4 \\-7 \end{bmatrix}\)

\(\displaystyle \vec{n3} = \vec{n1} \times \vec{n2}=\begin{vmatrix} \vec{e1} & \vec{e2} & \vec{e3} \\2 & -3 & 1 \\1 & 4 & -7 \end{vmatrix}=\begin{bmatrix} 17 \\15 \\11 \end{bmatrix}\)

So now I just find \(\displaystyle \hat{n3}\) and multiply by (+/-)13 ?

 

[tkhunny]

Better try that again. I get ( -5, 17, 9)

 

[soroban]

Hello, renegade05!

Find all vector(s) of magnitude 13
. . that are parallel to the line of intersection of the planes .\(\displaystyle \begin{Bmatrix}2x-3y+z&=&7 \\ x+4y-7z&=&2\end{Bmatrix}\)

Your cross product is correct.

\(\displaystyle \begin{vmatrix}i & j & k \\ 2 & \text{-}3 & 1 \\ 1 & 4 & \text{-}7 \end{vmatrix} \;=\; 17i + 15j + 11k \;=\;\langle 17,15,11\rangle\)

The vector \(\displaystyle \vec v \,=\,\langle17,15,11\rangle\) is parallel to the line of intersection.

Vector \(\displaystyle \vec v\) has magnitude: .\(\displaystyle |\vec v| \:=\:\sqrt{17^2+15^2+11^2} \:=\:\sqrt{635}\)

Hence, \(\displaystyle \dfrac{\vec v}{|\vec v|} \:=\:\frac{1}{\sqrt{635}}\langle17,15,11\rangle\) is a unit vector parallel to the line of intersection.


Therefore: .\(\displaystyle 13 \cdot \frac{1}{\sqrt{635}}\langle 17,15,11\rangle\) has magnitude 13.
The vector parallel to the intersection with magnitude 13 is: .\(\displaystyle \left\langle \frac{221}{\sqrt{635}},\,\frac{195}{\sqrt{635}},\,\frac{143}{\sqrt{635}}\right\rangle \)

 

[renegade05]

Therefore: .

has magnitude 13.
The vector parallel to the intersection with magnitude 13 is: .

Wouldn't it be \(\displaystyle \left\langle \frac{221}{\sqrt{635}},\,\frac{195}{\sqrt{635}},\,\frac{143}{\sqrt{635}}\right\rangle\) and \(\displaystyle \left\langle -\frac{221}{\sqrt{635}},\,-\frac{195}{\sqrt{635}},\,-\frac{143}{\sqrt{635}}\right\rangle\) ?

 

[tkhunny]

Making up my own problems again? Wonder how I did that... Okay, now I get (17,15,11).

"All Vectors of Magnitude 13". Did you find them all? Check your defintion of "parallel" and see if it includes those in the opposite direction. Be sure. Don't guess.

 

[renegade05]

I'm positive it does!