1. ## parallel

which of the following lines is parallel to the plane $4x+y-z-10=0$

a) $\vec{r}=(3,0,2)+t(1,-2,2)$

b) $x=-3t,y=-5+2t,z=-10t$

c) $\frac{x-1}{4}=\frac{y+6}{-1}=\frac{z}{1}$

2. A line is parallel to a plane if and only if the direction of the line is perpendicular to the normal of the plane.

3. Originally Posted by Plato
A line is parallel to a plane if and only if the direction of the line is perpendicular to the normal of the plane.
i know the normal is (4,1,-1) although i'm not sure if i know what you mean by direction of the line

4. Originally Posted by william
which of the following lines is parallel to the plane $4x+y-z-10=0$

a) $\vec{r}=(3,0,2)+t(1,-2,2)$ the direction vector is $\color{red}<1,-2,2>$

b) $x=-3t,y=-5+2t,z=-10t$ the direction vector is $\color{red}<-3,2,-10>$

c) $\frac{x-1}{4}=\frac{y+6}{-1}=\frac{z}{1}$ the direction vector is $\color{red}<4,-1,1>$
.

5. Originally Posted by Plato
.
i'm sorry im still really stumped are you saying the the third one would be parallel to the plane? wouldn't it have to be <-4,-1,1> Note* the back of my book says the second and only the second like is parallel to the plane. Also, how would i determine if any of the lines lies in the plane.

6. No, neither <4, -1, 1>.<4, 1, -1> nor <-4, -1, 1>.<4, 1, -1> is equal to 0.

What is <1, -2, 2>.<4, 1, -1>?

What is <-3, 2, -10>.<4, 1, -1>?

A line lies in the plane if and only if (x, y, z) satisifes the equation of the plane for all t. That is, put the expressions for x, y, and z in the line, in terms of t, into the equation of the plane. If the entire line is in the plane, the "t" terms will cancel out, giving a equation that is always true, like "0 = 0". If a line is parallel to the plane but not in it, the "t" terms will also cancel but give an equation that is always false, like "1= 0".

7. Originally Posted by HallsofIvy
No, neither <4, -1, 1>.<4, 1, -1> nor <-4, -1, 1>.<4, 1, -1> is equal to 0.

What is <1, -2, 2>.<4, 1, -1>?

What is <-3, 2, -10>.<4, 1, -1>?

A line lies in the plane if and only if (x, y, z) satisifes the equation of the plane for all t. That is, put the expressions for x, y, and z in the line, in terms of t, into the equation of the plane. If the entire line is in the plane, the "t" terms will cancel out, giving a equation that is always true, like "0 = 0". If a line is parallel to the plane but not in it, the "t" terms will also cancel but give an equation that is always false, like "1= 0".
THANK YOU!!!! this makes it much, much more clear!

must be line 2