MHB 243.12.5.21 acute angle btw vectors

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\tiny{243.12.5.21}
$\textsf{Use a calculator to find the acute angle between the planes to the nearest thousandth of a radian.}$
$$\textit{$3x+10y+7z=9$ and $7x+ 2y + 9z = 3$}$$
\begin{align*}\displaystyle
u&=3x+10y+7z=9\\
v&=7x+2y+9z=3\\
u \cdot v&=3\cdot7+10\cdot2+7\cdot9\\
&=21+20+63=104\\
|u|&=\sqrt{3^2+10^2+7^2}=\sqrt{158}\\
|v|&=\sqrt{7^2+2^2+9^2}=\sqrt{134}\\
\end{align*}
$\textit{therefore}$
\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{u\cdot v}{|u||v|} \right]\\
&=\cos^{-1}\left[\frac{104}
{|\sqrt{158}||\sqrt{134}|} \right]\\
W|A &=\color{red}{\textbf{$0.775$ rad}}
\end{align*}

think this is ok so any suggestions

also why can we drop the 9 and 3 for dot product?
 
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karush said:
\tiny{243.12.5.21}
$\textsf{Use a calculator to find the acute angle between the planes to the nearest thousandth of a radian.}$
$$\textit{$3x+10y+7z=9$ and $7x+ 2y + 9z = 3$}$$
\begin{align*}\displaystyle
u&=3x+10y+7z=9\\
v&=7x+2y+9z=3\\
u \cdot v&=3\cdot7+10\cdot2+7\cdot9\\
&=21+20+63=104\\
|u|&=\sqrt{3^2+10^2+7^2}=\sqrt{158}\\
|v|&=\sqrt{7^2+2^2+9^2}=\sqrt{134}\\
\end{align*}
$\textit{therefore}$
\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{u\cdot v}{|u||v|} \right]\\
&=\cos^{-1}\left[\frac{104}
{|\sqrt{158}||\sqrt{134}|} \right]\\
W|A &=\color{red}{\textbf{$0.775$ rad}}
\end{align*}

think this is ok so any suggestions

Looks good to me, other than the absolute value bars in the last arccos(...) being unnecessary.

karush said:
also why can we drop the 9 and 3 for dot product?

$ax+by+cz=\alpha$ and $ax+by+cz=\beta$ are parallel planes (where $\alpha,\beta$ are real numbers that may be distinct),
 
karush said:
also why can we drop the 9 and 3 for dot product?

The Normal Vector to the plane is described by the three coefficients. You are finding the angle between the two Normal Vectors.

If you move one plane or the other up or down, such movement will not change the Angle of Intersection. The Line of Intersection will move.

Personally, I've never been a fan of the relatively unexplained magnitudes sneaking their way into such formulae. I always thought it wise to suggest that we should be using Unit Vectors. Anyway, unique answers don't care how you find them. :-)
 

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