# Vector Plane Problem

• May 23rd 2010, 04:34 AM
Potato
Vector Plane Problem
This question has stumped me, it is the only question like this in my textbook and i can't figure out how to solve this, i'm not the best at maths so any help people could give me would be great!

The vector $A = 3i + j - k$ is normal to the plane $M1$and the vector $B = 2i - j + k$ is normal to a second plane $M2$

Do the two planes necessarily intersect if they are both extended indefinetly? If the two planes do intersect, find a vector which is parallel to their line of intersection.

Thanks guys!
• May 23rd 2010, 05:19 AM
Soroban
Hello, Potato!

Quote:

The vector $\vec A \:=\: 3i + j - k$ is normal to the plane $M_1$
and the vector $\vec B \:=\: 2i - j + k$ is normal to a second plane $M_2$.

Do the two planes necessarily intersect if they are both extended indefinetly?
If the two planes do intersect, find a vector which is parallel to their line of intersection.

Two planes are parallel if their normal vectors are parallel (equal).

. . Since $\vec A \:\neq\:\vec B$, the planes will intersect.

Their line of intersection has the vector: . $\vec A \times \vec B$
. .
Do you see why?

$\vec v \;=\;\vec A \times \vec B \;=\;\left|\begin{array}{ccc} i & j & k \\ 3 & 1 & \text{-}1 \\ 2 & \text{-}1 & 1 \end{array}\right| \;=\;i(1-1) - j(3+2) + k(\text{-}3-2) \;=\;\langle 0,\text{-}5,\text{-5}\rangle$

. . Therefore: . $\vec v \:=\:j + k$

• May 23rd 2010, 06:17 AM
HallsofIvy
Quote:

Originally Posted by Soroban
Hello, Potato!

Two planes are parallel if their normal vectors are parallel (equal).

. . Since $\vec A \:\neq\:\vec B$, the planes will intersect.

"parallel" for vectors does not necessarily mean "equal"- one might be a scalar multiple of the other. Of course, here, since the i components are 3 and 2, such a multiple would have to be 2/3 or 3/2. And then (2/3)(1) is not equal to -1.

Quote:

Their line of intersection has the vector: . $\vec A \times \vec B$
. .
Quote:

Do you see why?

$\vec v \;=\;\vec A \times \vec B \;=\;\left|\begin{array}{ccc} i & j & k \\ 3 & 1 & \text{-}1 \\ 2 & \text{-}1 & 1 \end{array}\right| \;=\;i(1-1) - j(3+2) + k(\text{-}3-2) \;=\;\langle 0,\text{-}5,\text{-5}\rangle$

. . Therefore: . $\vec v \:=\:j + k$