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Math Help - Vector Plane Problem

  1. #1
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    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!
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  2. #2
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    Hello, Potato!

    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

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  3. #3
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    Quote Originally Posted by Soroban View Post
    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.


    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

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