1. ## vectors

Show that the line $\displaystyle x=t\begin{pmatrix} 2\\ 1\\ 3 \end{pmatrix}$ is parallel to $\displaystyle 3x-3y-z=2$

2. Originally Posted by vuze88
Show that the line $\displaystyle x=t\begin{pmatrix} 2\\ 1\\ 3 \end{pmatrix}$ is parallel to $\displaystyle 3x-3y-z=2$
What you have given isn't the equation of a line...

You need to have a point that the line passes through...

3. Hello, vuze88!

Show that the line $\displaystyle L:\,x\,=\,t \begin{pmatrix}2\\ 1\\ 3 \end{pmatrix}$ .is parallel to the plane .$\displaystyle P:\,3x-3y-z\:=\:2$

The line has direction vector: .$\displaystyle \vec v \:=\:\langle 2t,t,3t\rangle$

The plane has normal vector: .$\displaystyle \vec n \:=\:\langle 3,\text{-}3,\text{-}1\rangle$

The line is parallel to the plane if $\displaystyle \vec v\perp \vec n$

And we find that: .$\displaystyle \vec v \cdot\vec n \;=\;\langle2t,t,3t\rangle \cdot \langle 3,\text{-}3,\text{-}1\rangle \;=\;6t - 3t - 3t \;=\;0$

. . Q. E. D.

4. Originally Posted by Prove It
What you have given isn't the equation of a line...

You need to have a point that the line passes through...
No, that's perfectly valid- it is a line that passes through (0, 0, 0) when t= 0.

Although I think it is bad practice to use "x" to represent the (x,y,z) coordinates of a point on the line and also use "x" to represent the x coordinate of a point on the plane.

Better would have been $\displaystyle r= \begin{pmatrix}x \\ y \\ z\end{pmatrix}= t\begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix}$.

From that x= 2t, y= t, and z= 3t. Put those into the equation of the plane. If you can solve for t, the line intersects the plane at that point. If you can't solve for t, then the line is parallel to the plane.

Of course, since the equation is linear, the only reason you couldn't solve for t would be that the coefficient of t is 0 which is exactly the same as saying that the dot product of the two vectors Soroban gives is 0- that those two vectors are perpendicular.

5. Originally Posted by HallsofIvy
No, that's perfectly valid- it is a line that passes through (0, 0, 0) when t= 0.

Although I think it is bad practice to use "x" to represent the (x,y,z) coordinates of a point on the line and also use "x" to represent the x coordinate of a point on the plane.

Better would have been $\displaystyle r= \begin{pmatrix}x \\ y \\ z\end{pmatrix}= t\begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix}$.

From that x= 2t, y= t, and z= 3t. Put those into the equation of the plane. If you can solve for t, the line intersects the plane at that point. If you can't solve for t, then the line is parallel to the plane.

Of course, since the equation is linear, the only reason you couldn't solve for t would be that the coefficient of t is 0 which is exactly the same as saying that the dot product of the two vectors Soroban gives is 0- that those two vectors are perpendicular.
OK good point - I'm just used to seeing a point that the line goes through...