I'm struggling with the following exercise:
Given the 2-D vectors $<A>$ and $<B>$, where their inner product $<A> \cdot <B> = -6$ and the magnitude of the vectorial product is $ \parallel <A> \times <B> \parallel= 9$ ; find the angle between them.
- I know that the inner product is equal to $\parallel A\parallel\parallel B\parallel \cos(\Theta)$, $\Theta$ being the angle between the vectors.
- I also know that the magnitude of the vectorial product is $\parallel A\parallel\parallel B\parallel \sin(\Theta)$.
So I started reasoning like this:
$$ -6 = \parallel A\parallel\parallel B\parallel \cos(\Theta) $$ $$ \frac{(-6)}{\parallel A\parallel\parallel B\parallel} = \cos(\Theta) $$ $$ (\frac{(-6)}{\parallel A\parallel\parallel B\parallel})^2 = (\cos(\Theta))^2 $$
And then did the same with the length of the vectorial product:
$$ 9 = \parallel A\parallel\parallel B\parallel \sin(\Theta) $$ $$ \frac{9}{\parallel A\parallel\parallel B\parallel} = \sin(\Theta) $$ $$ (\frac{9}{\parallel A\parallel\parallel B\parallel})^2 = (\sin(\Theta))^2 $$
Finally I tried to use both formulas and the trigonometric identity $\left(\sin(\Theta)\right)^2 + \left(\cos(\Theta)\right)^2 = 1$ to get the value of the vectors' lengths multiplied together.
$$ (\frac{(-6)}{\parallel A\parallel\parallel B\parallel})^2 + (\frac{9}{\parallel A\parallel\parallel B\parallel})^2 = (\sin(\Theta))^2 + (\cos(\Theta))^2 $$ $$ (\frac{(-6)}{\parallel A\parallel\parallel B\parallel})^2 + (\frac{9}{\parallel A\parallel\parallel B\parallel})^2 = 1 $$ $$ \frac{36}{(\parallel A\parallel\parallel B\parallel)^2} + \frac{81}{(\parallel A\parallel\parallel B\parallel)^2} = 1 $$ $$ \frac{36 + 81}{(\parallel A\parallel\parallel B\parallel)^2} = 1 $$ $$ \frac{117}{(\parallel A\parallel\parallel B\parallel)^2} = 1 $$ $$ 117 = (\parallel A\parallel\parallel B\parallel)^2 $$ $$ \parallel A\parallel\parallel B\parallel = \sqrt{117} $$
Knowing that $ \parallel A\parallel\parallel B\parallel = \sqrt{117} $ I just plugged that in in the inner and vectorial product formulas.
$$ \sqrt{117} * \cos((\Theta)) = -6 $$ $$ \cos((\Theta)) = \frac{-6}{\sqrt{117}} $$ $$ (\Theta) = \arccos (\frac{-6}{\sqrt{117}}) $$
Same with the length of the vectorial product:
$$ \sqrt{117} * \sin((\Theta)) = 9 $$ $$ \sin((\Theta)) = \frac{9}{\sqrt{117}} $$ $$ (\Theta) = \arcsin (\frac{9}{\sqrt{117}}) $$
And here is what I don't understand:
$$ \arccos (\frac{-6}{\sqrt{117}}) \neq \arcsin (\frac{9}{\sqrt{117}}) $$
Since $ \arcsin (\frac{9}{\sqrt{117}}) \approx 56 $(degrees) and $ \arccos (\frac{-6}{\sqrt{117}}) \approx 124 $ (degrees).
The textbook gives 124 degrees as the correct answer but I don't understand. Why am I getting 2 results for the same angle? Why is 124 the correct one? Did I make a mistake?
$$\frac{\lvert A\times B\lvert}{A\cdot B}=\frac{\lvert A \lvert\cdot\lvert B\lvert\sin\theta}{\lvert A \lvert\cdot\lvert B\lvert\cos\theta}=\tan\theta=\frac{9}{-6}$$ then $$\theta=\arctan(-3/2)$$ then how the dot product is negative,the angle is between 90 and 180 degrees, we have:
$$\theta=180+\arctan(-3/2)=180-56=124$$