1. ## scalar products

The position vectors of three points A, B, and C with respect to a fixed origin O are 2i - 2j + k , 4i + 2j + k and i + j + 3k respectively.
Find the unit vectors in the directions of CA and CB. <
Calculate the ange ACB in degrees.

2. Originally Posted by Oasis1993
The position vectors of three points A, B, and C with respect to a fixed origin O are 2i - 2j + k , 4i + 2j + k and i + j + 3k respectively.
Find the unit vectors in the directions of CA and CB. <
Calculate the ange ACB in degrees.

$CA = CO + OA$

$= -OC + OA$

$= OA - OC$

$= 2\mathbf{i} - 2\mathbf{j} + \mathbf{k} - (4\mathbf{i} + 2\mathbf{j} + \mathbf{k})$

$= -2\mathbf{i} - 4\mathbf{j} + 0\mathbf{k}$.

To find the unit vector in this direction, divide it by its length.

$|CA| = \sqrt{(-2)^2 + (-4)^2 + 0^2}$

$= \sqrt{4 + 16 + 0}$

$= \sqrt{20}$

$= 2\sqrt{5}$.

So $\frac{OC}{|OC|} = \frac{-2\mathbf{i} - 4\mathbf{j} + 0\mathbf{k}}{2\sqrt{5}}$

$= \frac{-\mathbf{i} - 2\mathbf{j} + 0\mathbf{k}}{\sqrt{5}}$

$= \frac{-\sqrt{5}\mathbf{i} - 2\sqrt{5}\mathbf{j} + 0\mathbf{k}}{5}$

$= -\frac{\sqrt{5}}{5}\mathbf{i} - \frac{2\sqrt{5}}{5}\mathbf{j} + 0\mathbf{k}$.

Can you do the same to find the unit vector in the direction of CB?

3. Btw once you have both unit vectors, use the fact that

$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}$ (where $\theta$ is the angle in between $\mathbf{a}$ and $\mathbf{b}$) to find $\theta$.

4. hello. Thank you for your help but my book has a different answer than yours. For CA it says (i - 3j - 2k)/ sqrt 14

5. Oops. It looks like I actually worked out $BA$.

The process is basically the same though...