The exact value of cos(θ) in the form of p/q

[karush]

View attachment 1401
calculations and boxed answers are mine

my question is with (iii) (provided the (i) and (ii) are correct.

from the diagram $cos \theta$ would be $\frac{OC}{OA}$ or $\frac{11.2}{13}$

but would need the the length of $OC$ to do so, what would be the best approach to get point $C$

we could convert $a$ and $b$ and to lines and find the intersection at $C$ hence $OC$.

but not sure how this could be done with just vectors

 

[topsquark]

Re: the exact value of cos theta in the form of p/q

View attachment 1401
calculations and boxed answers are mine

my question is with (iii) (provided the (i) and (ii) are correct.

from the diagram $cos \theta$ would be $\frac{OC}{OA}$ or $\frac{11.2}{13}$

but would need the the length of $OC$ to do so, what would be the best approach to get point $C$

we could convert $a$ and $b$ and to lines and find the intersection at $C$ hence $OC$.

but not sure how this could be done with just vectors

Consider a triangle OAB. You can find the length of AB as you know the position vectors for A and B. Then use the Law of Cosines to find your angle.

-Dan

 

[Prove It]

Re: the exact value of cos theta in the form of p/q

View attachment 1401
calculations and boxed answers are mine

my question is with (iii) (provided the (i) and (ii) are correct.

from the diagram $cos \theta$ would be $\frac{OC}{OA}$ or $\frac{11.2}{13}$

but would need the the length of $OC$ to do so, what would be the best approach to get point $C$

we could convert $a$ and $b$ and to lines and find the intersection at $C$ hence $OC$.

but not sure how this could be done with just vectors

Use the fact that \displaystyle \begin{align*} \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{(\theta)} \end{align*}

Also note that you have evaluate \displaystyle \begin{align*} \mathbf{a}\cdot \mathbf{b} \end{align*} incorrectly.

\displaystyle \begin{align*} \mathbf{a}\cdot \mathbf{b} &= 12 \cdot 6 + 5 \cdot 8 \\ &= 72 + 40 \\ &= 112 \end{align*}

 

[karush]

Re: the exact value of cos theta in the form of p/q

$o^2=a^2+b^2-2ab\ cos\ \theta$

or $\displaystyle\frac{o^2 -a^2-b^2}{-2ab}=cos\ \theta$

plugging in $\displaystyle o=3\sqrt{3}\ a=10\ b=13$ we get $cos\ \theta = \frac{56}{65}$

well easier than finding $OB...$

 

[karush]

Re: the exact value of cos theta in the form of p/q

Use the fact that \displaystyle \begin{align*} \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{(\theta)} \end{align*}

Also note that you have evaluate \displaystyle \begin{align*} \mathbf{a}\cdot \mathbf{b} \end{align*} incorrectly.

\displaystyle \begin{align*} \mathbf{a}\cdot \mathbf{b} &= 12 \cdot 6 + 5 \cdot 8 \\ &= 72 + 40 \\ &= 112 \end{align*}

$\displaystyle\frac{a\cdot b}{10\times 13} = \frac{112}{130}=\frac{56}{65}$
i reduced the fraction..

 

[Deveno]

Re: the exact value of cos theta in the form of p/q

This may be something along the lines of what you were looking for:

Suppose we write the point $C$ as: $(6t,8t)$.

Clearly the length of $OC$ is $10t$.

The length of $AC$ is: $\sqrt{(8t - 5)^2 + (6t - 12)^2} = \sqrt{100t^2 - 224t + 169}$.

Because we have a right triangle, we know that:

$100t^2 + 100t^2 - 224t + 169 = 169$
$200t^2 - 224t = 0$
$t(25t - 28) = 0$

We can discount the solution $t = 0$, leaving us with $t = \frac{28}{25} = \frac{112}{100} = 1.12$, which establishes that the length of $OC$ is 11.2, as we surmised by the other method.