# Thread: s4.854.13.5.47 Find symmetric equations, angle between the planes

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?

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

\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.

now I'm beginning to believe that sign errors are the most common mistake....ðŸ˜°

Originally Posted by Prove It
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. Originally Posted by karush
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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