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  1. MHB Master
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    #1
    $\tiny{s4.854.13.5.47}$
    $\textsf{a. Find symmeteric equations for the line of intersection of planes}\\$
    $\textsf{b. Find the angle between the planes}\\$

    \begin{align}\displaystyle
    j+y-z&=2 \\
    3x-4y+5z &=6
    \end{align}
    \begin{align}\displaystyle
    n_1&=\langle 1,1,-1\rangle\\
    n_2&=\langle 3,-4,+5\rangle
    \end{align}

    \begin{align}
    \displaystyle
    \frac{n_1\cdot n_2}{|n_1||n_2|}
    &=\frac{3(1)+(-4)(1)+5(-1)}{\sqrt{3}\sqrt{50}}\\
    &=\frac{\sqrt{6}}{5}\\
    \cos^{-1}\left({\frac{\sqrt{6}}{5}}\right)
    &=119^o \textit{or} \, 61^o
    \end{align}

    \begin{align}
    \begin{bmatrix}
    i & j & k\\
    1 &1 &-1\\
    3 &-4 &5
    \end{bmatrix} &=\textbf{i-8j-7k}
    \end{align}

    $\textsf{the symmetric equations are:}$
    \begin{align}
    x-2&=\frac{y}{(-8)}=\frac{z}{(-7)}
    \end{align}

    suggestions?
    Last edited by karush; January 7th, 2017 at 03:01.

  2. MHB Master
    MHB Math Helper

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    #2
    Quote Originally Posted by karush View Post
    $\tiny{s4.854.13.5.47}$
    $\textsf{a. Find symmeteric equations for the line of intersection of planes}\\$
    $\textsf{b. Find the angle between the planes}\\$

    \begin{align}\displaystyle
    j+y-z&=2 \\
    3x-4y+5z &=6
    \end{align}
    \begin{align}\displaystyle
    n_1&=\langle 1,1,-1\rangle\\
    n_2&=\langle 3,-4,+5\rangle
    \end{align}

    \begin{align}
    \displaystyle
    \frac{n_1\cdot n_2}{|n_1||n_2|}
    &=\frac{3(1)+(-4)(1)+5(-1)}{\sqrt{3}\sqrt{50}}\\
    &=\frac{\sqrt{6}}{5}\\
    \cos^{-1}\left({\frac{\sqrt{6}}{5}}\right)
    &=119^o \textit{or} \, 61^o
    \end{align}

    \begin{align}
    \begin{bmatrix}
    i & j & k\\
    1 &1 &-1\\
    3 &-4 &5
    \end{bmatrix} &=\textbf{i-8j-7k}
    \end{align}

    $\textsf{the symmetric equations are:}$
    \begin{align}
    x-2&=\frac{y}{(-8)}=\frac{z}{(-7)}
    \end{align}

    suggestions?
    Notice that

    $\displaystyle \begin{align*} -\frac{6}{\sqrt{3}\,\sqrt{50}} &= -\frac{6}{5\,\sqrt{3}\,\sqrt{2}} \\ &= -\frac{6}{5\,\sqrt{6}} \\ &= -\frac{6\,\sqrt{6}}{5 \cdot 6} \\ &= -\frac{\sqrt{6}}{5} \end{align*}$

    and when you end up with $\displaystyle \begin{align*} \cos{ \left( \theta \right) } < 0 \end{align*}$ we can assume that the angle will be obtuse.

  3. MHB Master
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    #3 Thread Author
    now I'm beginning to believe that sign errors are the most common mistake....😰

  4. MHB Master
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    #4 Thread Author
    Quote Originally Posted by Prove It View Post
    Notice that

    $\displaystyle \begin{align*} -\frac{6}{\sqrt{3}\,\sqrt{50}} &= -\frac{6}{5\,\sqrt{3}\,\sqrt{2}} \\ &= -\frac{6}{5\,\sqrt{6}} \\ &= -\frac{6\,\sqrt{6}}{5 \cdot 6} \\ &= -\frac{\sqrt{6}}{5} \end{align*}$

    and when you end up with $\displaystyle \begin{align*} \cos{ \left( \theta \right) } < 0 \end{align*}$ we can assume that the angle will be obtuse.
    there are 2 angles of an intersecting plane....they are supplementary

  5. MHB Master
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    #5
    Quote Originally Posted by karush View Post
    there are 2 angles of an intersecting plane....they are supplementary
    I'm not saying they're not, I'm saying what you should be EXPECTING for your answer...

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