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Math Help - a true and false question

  1. #1
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    Question a true and false question

    The graph of the equation z=r*cos(theta) is a plane passing through the origin with normal vector <-1,0,1>.

    The answer is true.

    Why? Please teach me . Thank you very much.
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  2. #2
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    Hello, Jenny!

    There are a number of approaches . . . here's one.


    The graph of the equation z = r\cos\theta is a plane
    passing through the origin with normal vector \langle-1,0,1\rangle

    The answer is True . . . why?

    Converting from Cylindrical to Rectangular, we have: . z \:=\:x

    If you sketch (or visualize) this plane, its trace is the 45 line on the "left wall".
    . . The plane extends infinitely to the left and right.
    So it does pass through the origin.


    Now we must find its normal vector.

    Pick three points on the plane: A(0,0,0),\;B(1,0,1),\;C(0,1,0)

    We have two vectors in the plane: . \begin{array}{cc}\vec{u}\:=\:AB\:=\:\langle 1,0,1\rangle \\ \vec{v}\:=\:AC \:=\:\langle 0,1,0\rangle\end{array}

    The normal vector of the plane is perpendicular to both \vec{u} and \vec{v}.

    Hence: . \vec{n}\:=\:\vec{u} \times \vec{v} \:=\:\begin{vmatrix}i & j & k \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{vmatrix} \;= \;-i + 0j + k

    Therefore: . \vec{n} \:=\:\langle -1,0,1\rangle . . . The statement is true.

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