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  1. MHB Master
    karush's Avatar
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    #1
    $\tiny{{s4}.{13}.{5}.{41}}$

    $\textsf{find if planes are $\parallel, \perp$ or $\angle$ of intersection }\\$
    \begin{align}
    \displaystyle
    {P_1}&={x+z=1}\\
    \therefore n_1&=\langle 1,0,1 \rangle\\
    \\
    {P_2}&={y+z=1}\\
    \therefore n_2&=\langle 0,1,1 \rangle\\
    \\
    \cos(\theta)&=
    \frac{n_1\cdot n_2}{|n_1||n_2|}\\
    &=\frac{1(0)+0(1)+1(1)}
    {\sqrt{1+1}\cdot\sqrt{1+1}}
    =\frac{1}{2}\\
    \cos^{-1}\left({\frac{1}{2}}\right)&=
    \color{red}{60^o}
    \end{align}

    $\textit{there are 2 more problems like this so presume this is best method.. }\\$
    $\textit{didn't know if it is
    common notation to call a plane $P_1$}$

  2. MHB Master
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    #2
    Quote Originally Posted by karush View Post
    $\tiny{{s4}.{13}.{5}.{41}}$

    $\textsf{find if planes are $\parallel, \perp$ or $\angle$ of intersection }\\$
    \begin{align}
    \displaystyle
    {P_1}&={x+z=1}\\
    \therefore n_1&=\langle 1,0,1 \rangle\\
    \\
    {P_2}&={y+z=1}\\
    \therefore n_2&=\langle 0,1,1 \rangle\\
    \\
    \cos(\theta)&=
    \frac{n_1\cdot n_2}{|n_1||n_2|}\\
    &=\frac{1(0)+0(1)+1(1)}
    {\sqrt{1+1}\cdot\sqrt{1+1}}
    =\frac{1}{2}\\
    \cos^{-1}\left({\frac{1}{2}}\right)&=
    \color{red}{60^o}
    \end{align}

    $\textit{there are 2 more problems like this so presume this is best method.. }\\$
    $\textit{didn't know if it is
    common notation to call a plane $P_1$}$
    Everything you've posted is fine. You can give a plane any name you like, but I would write something like this:

    $\displaystyle \begin{align*} P_1 : x + z = 1 \end{align*}$

    That way we can see that $\displaystyle \begin{align*} P_1 \end{align*}$ is DEFINED as the relationship "the sum of the x and z values needs to be 1", not that it is some variable that has something to do with the equation.

  3. MHB Journeyman
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    #3
    The problem asks you to do three things:
    1) determine if the planes are parallel.
    2) determine if the planes are perpendicular.
    3) if neither of those, determine the angle of intersection of the two planes.

    Yes, by using that formula to determine the angle, you can then answer all three questions but it should be simpler to determine the first two without using that formula:

    The two planes are parallel if and only if the two normal vectors are parallel- if one is a multiple of the other.

    The two planes are perpendicular if and only if the two normal vectors are perpendicular: if their dot product is 0.

    If neither of those is true, then you can use the dot product you found as the numerator in the formula to determine the angle.

  4. MHB Master
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    #4 Thread Author
    that's very helpful
    hard to see that in their examples

    the next 2 problems are probably
    $ \displaystyle \parallel , \perp$

  5. MHB Master
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    #5 Thread Author
    $\tiny{s4.854.13.5.43}$
    $\textsf{Determine if the 2 given planes are perpendicular, parallel or at an angle to each other}$
    \begin{align}
    \displaystyle
    {p_1}&:{x+4y-3z=1}
    \therefore n_1=\langle 1,4,-3 \rangle\\
    \nonumber\\
    {p_2}&:{-3x+6y+7z=0} \therefore n_2=\langle -3,6,7 \rangle
    \end{align}
    \begin{align}
    \displaystyle
    \cos(\theta)&= \frac{n_1\cdot n_2}{|n_1||n_2|}=0\\
    \therefore p_1 &\perp p_2
    \end{align}

    $\tiny{s4.854.13.5.45}$
    \begin{align}
    \displaystyle
    {p_1}&:{2x+4y-2z=1}
    \therefore n_1=\langle 2,4,-2 \rangle\\
    \nonumber\\
    {p_2}&:{-3x+6y+3z=0} \therefore n_2=\langle -3,6,7 \rangle \\
    n_1&=-\frac{2}{3} n_2 \\
    &\therefore n_1\parallel n_2
    \end{align}

    $\textit{btw what is the input string to see the 2 plane on a W|A graph?}$
    Last edited by karush; January 4th, 2017 at 18:55.

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