equation of a line in three dimensions

[mathstresser]

I'm supposed to find the equation for:
The line through (2,1,0) and perpendicular to both i+j and j+k


I don't even know where to start. I know that it needs to be perpendicular to <1,1,0> and <0,1,1>, but that's as far as I get (yes, I realize that's not far.)

What are the parametric and symmetric equations?

 

[pka]

Here are the calculations: \(\displaystyle \L
\left( \begin{array}{l}
1 \\
1 \\
0 \\
\end{array} \right) \times \left( \begin{array}{l}
0 \\
1 \\
1 \\
\end{array} \right) = \left( \begin{array}{r}
1 \\
- 1 \\
1 \\
\end{array} \right)\)

Use that vector as your direction vector of the line.

 

[mathstresser]

So, I have a direction vector of <1,-1,1> took me a minute to figure that out, we do our matrices rotated from yours!

Anyway, so do I use the original point for my equations? like:
x= 2+t
y=1-t
z=0+t or just z=t

So, then does that make the symmetric equation:

(x-2)/1=(y-1)/1=(z-0)/1
or just
(x-2)=(y-1)=z

Did I do those right?

 

[soroban]

Hello, mathstresser!

Just one teensy slip . . .

So, I have a direction vector of \(\displaystyle \langle1,-1,1\rangle\)

Anyway, so do I use the original point for my equations? . . . yes!

\(\displaystyle \;\;\begin{array}{ccc}x\:=\:2\,+\,t \\ y\:=\:1\,-\,t \\ z\,=\,t\end{array}\;\) . . . right!

So, then does that make the symmetric equation:

\(\displaystyle \frac{x\,-\,2}{1}\:=\:\frac{y\,-\,1}{1}\:=\:\frac{z\,-\,0}{1}\)
. . . . . . . . . ↑
. . . . . . .should be -1

Otherwise, great work!

 

[mathstresser]

Thank you!