1. ## [SOLVED] Vectors Question

Question:
The position vectors of points $A$ and $B$ are $\left(\begin{array}{c}-3\\6\\3\end{array}\right)$ and $\left(\begin{array}{c}-1\\2\\4\end{array}\right)$ respectively, relative to an origin $O$.
(i) Calculate angel AOB.
(ii) The point C is such that $\vec{AC} = 3\vec{AB}$. Find the unit vector in the direction of $\vec{OC}$.

Attempt:

(i) $\theta = 36.7^o$

(ii) Need help!

2. Originally Posted by looi76
Question:
The position vectors of points $A$ and $B$ are $\left(\begin{array}{c}-3\\6\\3\end{array}\right)$ and $\left(\begin{array}{c}-1\\2\\4\end{array}\right)$ respectively, relative to an origin $O$.
(i) Calculate angel AOB.
(ii) The point C is such that $\vec{AC} = 3\vec{AB}$. Find the unit vector in the direction of $\vec{OC}$.

Attempt:

(i) $\theta = 36.7^o$

Correct
Originally Posted by looi76
(ii) Need help!
$\vec{AB} = \vec{OB} - \vec{OA}$

Thus:
$\vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} - \vec{OA} = 3(\vec{OB} - \vec{OA}) \Rightarrow \vec{OC}= 3\vec{OB} - 2\vec{OA}$
So complete it now

3. Originally Posted by Isomorphism

Correct

$\vec{AB} = \vec{OB} - \vec{OA}$

Thus:
$\vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} - \vec{OA} = 3(\vec{OB} - \vec{OA}) \Rightarrow \vec{OC}= 3\vec{OB} - 2\vec{OA}$
So complete it now
Thanks Isomorphism
$
\vec{OC} = 3\vec{OB} - 2\vec{OA}$

$\vec{OC} = 3\left(\begin{array}{c}-1\\2\\4\end{array}\right) - 2\left(\begin{array}{c}-3\\6\\3\end{array}\right)$

$\vec{OC} = \left(\begin{array}{c}-3\\2\\12\end{array}\right) - \left(\begin{array}{c}-6\\12\\6\end{array}\right)$

$\vec{OC} = \left(\begin{array}{c}3\\-6\\6\end{array}\right)$

4. Almost.... You still did not find the unit vector

But did you understand how I got the expression:
$\vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} = 3\vec{OB} - 2\vec{OA}$

That is more important than substituting and computing the answer

5. Originally Posted by Isomorphism
Almost.... You still did not find the unit vector

But did you understand how I got the expression:
$\vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} = 3\vec{OB} - 2\vec{OA}$

That is more important than substituting and computing the answer
Yes, I have understood, thanks to you . By unit vector you mean i,j,k?
I mean $3i - 6j + 6k$, is this right?

6. Originally Posted by looi76
Yes, I have understood, thanks to you . By unit vector you mean i,j,k?
I mean $3i - 6j + 6k$, is this right?
No..No..A unit vector is a vector whose magnitude is 1.

(ii) The point C is such that . Find the unit vector in the direction of .
You have found OC but not the unit vector in the direction of
To get that vector divide $\vec{OC}$ by magnitude of $\vec{OC}$

7. Originally Posted by Isomorphism
No..No..A unit vector is a vector whose magnitude is 1.

You have found OC but not the unit vector in the direction of
To get that vector divide $\vec{OC}$ by magnitude of $\vec{OC}$
I got the formula from the text book: $\hat{v} = \frac{v}{|v|}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 - 6^2 + 6^2}}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{81}}$

$\hat{OC} = \frac{3i - 6j + 6k}{9}$

$\hat{OC} = \frac{3}{9}i - \frac{6}{9}j + \frac{6}{9}k$

$\hat{OC} = \frac{1}{2}i - \frac{2}{3}j + \frac{2}{3}k$

Is this correct?

8. Originally Posted by looi76
I got the formula from the text book: $\hat{v} = \frac{v}{|v|}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 - 6^2 + 6^2}}$

Is this correct?
Ya right

Except for a typo...
$|v| = \sqrt{3^2 + (-6)^2 + 6^2}$

9. Originally Posted by Isomorphism
Wrong

$|v| = \sqrt{3^2 + (-6)^2 + 6^2}$
$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 + (-6)^2 + 6^2}}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{81}}$

$\hat{OC} = \frac{3i - 6j + 6k}{9}$

$\hat{OC} = \frac{3}{9}i - \frac{6}{9}j + \frac{6}{9}k$

$\hat{OC} = \frac{1}{2}i - \frac{2}{3}j + \frac{2}{3}k$

Is this correct?

10. Originally Posted by looi76
$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 + (-6)^2 + 6^2}}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{81}}$

$\hat{OC} = \frac{3i - 6j + 6k}{9}$

$\hat{OC} = \frac{3}{9}i - \frac{6}{9}j + \frac{6}{9}k$

$\hat{OC} = \frac{1}{2}i - \frac{2}{3}j + \frac{2}{3}k$

Is this correct?
You were right all along... It was just a typing mistake..

Well done