# Thread: Intersection of Lines and Planes

1. ## (Solved) Intersection of Lines and Planes

Hi, I also could not figure out how to solve this problem.

For what value of K will the line x-k/3 = y+4/2 = z+6 intersect the plane x-4y+5z+5=0

a) in a single point
b) in an infinite number of points
c) in no points

I tried plugging in the parametric equations of the line into the scalar equation, but I'm not sure how to isolate and solve for k.

If anyone has any ideas, that would be greatly appreciated.

Thanks,
Sam

2. Hello, Sam!

This is a very sneaky problem!

For what value of $k$ will the line: . $\frac{x-k}{3} \:=\:\frac{y+4}{2} \:=\: \frac{z+6}{1}$

intersect the plane $x-4y+5z+5\:=\:0$

. . (a) in a single point . . (b) in an infinite number of points . . (c) in no points
The parametric equations of the line are: . $\begin{Bmatrix}x &=& k+3t \\ y &=& 4+2t \\ z &=& 6 + t \end{Bmatrix}$

The line has direction vector: . $\vec v \:=\:\langle 3,2,1\rangle$

The plane has normal vector: . $\vec n \:=\:\langle 1,\text{-}4,5\rangle$

We find that: . $\vec v\cdot\vec n \:=\:\langle3,2,1\rangle\cdot\langle1,\text{-}4,5\rangle \:=\:0$

. . That is: . $\vec v \perp \vec n$

Hence. the line and the plane are parallel.

In most cases, there are no points of intersection.

The sole exception is when the line lies in the plane.
. . When does this happen?

Substitute the parametric equations into the equation of the plane:

. . $(k+3t) - 4(4+2t) + 5(6+t) + 5 \:=\:0 \quad\Rightarrow\quad k+3t-16-8t + 30 + 5t + 5 \:=\:0$

The $t$'s cancel out and we have: . $k = -19$

Therefore: . $\begin{Bmatrix}(a) & \text{a single point} & \text{Never} \\
(b) & \text{infinite points} & k\:=\:\text{-}19 \\
(c) & \text{no points} & k \:\neq\:\text{-}19 \end{Bmatrix}$

3. Thanks so much for your help