243.12.5.21 acute angle btw vectors

In summary, you are finding the angle between two planes that are parallel. The angle is found by calculating the cosine of the angle between the two Normal Vectors.
  • #1
karush
Gold Member
MHB
3,269
5
\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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  • #2
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),
 
  • #3
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. :-)
 

Related to 243.12.5.21 acute angle btw vectors

1. What is the definition of an acute angle between vectors?

An acute angle between two vectors is an angle that measures less than 90 degrees. This means that the two vectors are pointing towards each other, rather than away from each other.

2. How is the acute angle between vectors calculated?

The acute angle between vectors can be calculated using the dot product formula: θ = cos⁻¹((a · b) / (|a| * |b|)), where a and b are the two vectors.

3. Why is it important to know the acute angle between vectors?

Knowing the acute angle between vectors can provide insight into the relationship between the two vectors. It can also be useful in solving geometric and trigonometric problems involving vectors.

4. Can the acute angle between vectors be negative?

No, the acute angle between vectors cannot be negative. The range of values for the acute angle is 0 to 90 degrees, as it is always measured as the smallest angle between the two vectors.

5. What is the difference between an acute angle and a obtuse angle between vectors?

An acute angle is less than 90 degrees, while an obtuse angle is greater than 90 degrees. This means that the two vectors are pointing in opposite directions, rather than towards each other.

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