1. Vector's

I've been given this question and am struggling to complete it.

Relative to a fixed origin, O, the points A and B have position vectors

(i + 5j -1k) and (6i + 3j - 6k) respectively

Find in exact, simplified form,
(i) the cosine of angle AOB - i have done this and it comes out as root 3 over 3
(ii) the area of the triangle
(iii) the shortest distance from A to the line OB

Any help on this would be very much appreciated. Thanks

2. Originally Posted by lemmingsrevolt
I've been given this question and am struggling to complete it.

Relative to a fixed origin, O, the points A and B have position vectors

(i + 5j -1k) and (6i + 3j - 6k) respectively

Find in exact, simplified form,
(i) the cosine of angle AOB - i have done this and it comes out as root 3 over 3
(ii) the area of the triangle
(iii) the shortest distance from A to the line OB

Any help on this would be very much appreciated. Thanks
(ii) If you know about the cross product you can calculate $\displaystyle \frac{1}{2} | \vec{OA} \times \vec{OB} |$.

Otherwise use the sine rule: $\displaystyle \frac{1}{2} | OA | \cdot | OB | \sin (\angle AOB)$.

(iii) Calculate the scalar resolute of the vector $\displaystyle \vec{OA}$ perpendicular to the vector $\displaystyle \vec{OB}$.

3. Thank you very much that's really helpful.
My main problem is how to find the position vector of O so i can do these things? would you be able to tell me how to do this?
thanks a lot

4. Hi

O is the origin therefore its position vector is 0i + 0j + 0k

The cross product of $\displaystyle \overrightarrow{OA}$ and $\displaystyle \overrightarrow{OB}$ is -27i - 27k

The area of the triangle is $\displaystyle \frac12\:\sqrt{27^2+27^2} = \frac{27\:\sqrt{2}}{2}$