1. ## scalar products

The vectors AB and AC are (-1, 0, 3) and ( 2, 4, 3) respectively.
The vector AD is the sum of vector AB and AC. Determine the acute angle between the diagonals of the parallelogram defined by the points A,B,C and D.

THANKS!

2. Hello, Oasis1993!

Given vectors: .$\displaystyle \overrightarrow{AB} = \langle -1,0,3\rangle,\;\;\overrightarrow{AC} = \langle 2, 4, 3\rangle$

. . . . . . . . . . . $\displaystyle \overrightarrow{AD} \:=\:\overrightarrow{AB} + \overrightarrow{AC}$

Determine the acute angle between the diagonals
of the parallelogram defined by the points $\displaystyle A,B,C, D.$
Code:
          (-1,0,3)           (1,4,6)
B o - - - - - - - - o D
/  *           *  /
/     *     *     /
/        * θ      /
/     *     *     /
/  *           *  /
A o - - - - - - - - o C
(0,0,0)           (2,4,3)

We have: .$\displaystyle \begin{array}{c}\vec u \:=\: \overrightarrow{AD} \:=\:\langle 1,4,6\rangle \\ \\[-3mm] \vec v \:=\:\overrightarrow{BC} \:=\:\langle 3,4,0\rangle \end{array}$

$\displaystyle \cos\theta \:=\:\frac{u \cdot v}{|u||v|} \;=\;\frac{\langle 1,4,6\rangle\cdot\langle3,4,0\rangle}{\sqrt{1+16+3 6}\,\sqrt{9+16+0}} \:=\:\frac{19}{5\sqrt{53}} \:=\:0.521970143$

Therefore: .$\displaystyle \theta \;\approx\;58.5^o$

3. Hello! Thank you very much for your answre but im confused on how you got BC?

4. AB+BC=AC, therefore AB-AC=-BC.