Vector Plane Problem

May 2010
7
0
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 \(\displaystyle A = 3i + j - k\) is normal to the plane \(\displaystyle M1 \)and the vector \(\displaystyle B = 2i - j + k\) is normal to a second plane \(\displaystyle M2\)

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!
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, Potato!

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

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 \(\displaystyle \vec A \:\neq\:\vec B\), the planes will intersect.


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

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


. . Therefore: .\(\displaystyle \vec v \:=\:j + k\)

 
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HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Hello, Potato!


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

. . Since \(\displaystyle \vec A \:\neq\:\vec B\), 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.


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

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


. . Therefore: .\(\displaystyle \vec v \:=\:j + k\)