# Thread: vector..again

1. ## vector..again

the 4 points A,B,C and D have the following co-ordinates in a Cartesian coordinate system A = (2,2,2), B = (5,2,4), C = (-2,4,-1), D = (0,0,6).
i) find the volume of the parallelepiped whose edges are the lines AB,AC and AD.
ii)calculate the angle BAC

2. Do you know

$V=|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|$ ?

where

$\mathbf{a}=3\mathbf{i}+2\mathbf{k}$

$\mathbf{b}=-4\mathbf{i}+2\mathbf{j}-3\mathbf{k}$

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

3. Originally Posted by ashes
the 4 points A,B,C and D have the following co-ordinates in a Cartesian coordinate system A = (2,2,2), B = (5,2,4), C = (-2,4,-1), D = (0,0,6)
....
ii)calculate the angle BAC
Use the transscription of the dot-product:

$\cos(\theta) = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{ | \overrightarrow{AB}| \cdot |\overrightarrow{AC}|}$

For confrimation only: I've got $\cos(\theta) = -\frac{18}{\sqrt{13} \cdot \sqrt{29}} ~\implies~ \theta \approx 157.979^\circ$

But check my results!! When it comes to calculation I'm a master of desaster.

4. ## reply

k got the first part..but which formula to use to get angle BAC? i could not map it out on my drawing..

5. Originally Posted by ashes
k got the first part..but which formula to use to get angle BAC? i could not map it out on my drawing..
Consider the stationary vectors pointing at the points A, B, C and D.

Then the vector

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

According to the coordinates of the points you get:

$\overrightarrow{AB} = (5, 2, 4) - (2, 2, 2,) = (3, 0, 2)$

etc. ...

I've attached a very rough sketch of the 4 points and I've marked the angle $\angle(BAC)$

6. ## reply

Is your rough sketch a parallelpiped, cos i thought a parallelepiped is a polyhedron with all its faces being a parallelogram.. care to enlighten me cos im very bad at 3d visualisation..

7. ## reply

by formula i got the same answer as yours..its only the visualizing thingy which i cant seem to get everytime i do vectors..thanks btw..

8. Originally Posted by ashes
Is your rough sketch a parallelpiped, cos i thought a parallelepiped is a polyhedron with all its faces being a parallelogram.. care to enlighten me cos im very bad at 3d visualisation..
No of course not. I only placed the 4 points in a 3-D-coordinate system to give an impression of the distances.

I've attached a sketch of the complete parallelepiped.