# Vector Plane Problem

#### Potato

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

• HallsofIvy

#### HallsofIvy

MHF Helper
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: .$$\displaystyle \vec v \:=\:j + k$$