Hello Oasis1993 Originally Posted by

**Oasis1993** The points A, B and C have position vectors a = ( 2, 1, 2) b= (-3, 2, 5)

c= ( 4, 5, -2) respectively with respect to a fixed origin. The point D is such that ABCD, in that order is a parallelogram.

a) Find the position vector of D

b) Fİnd the position vector of the point at which the diagonals of the parallelogram interesct

c) Calculate the angle BAC, giving your answer to the nearst tenth of a degree.

Thank you in advance! I would really appreciate it if you showed the steps.

I agree with alexmahone's answer to (b), but there's a simple typo in (a): Originally Posted by

**alexmahone** ...

$\displaystyle (2+4, 1+5, 2-2)=(-3+x, 2+y, \color{red}3\color{black}+z)$

$\displaystyle (x, y, z)=(9, 4, \color{red}-3\color{black})$

...

It should be $\displaystyle (9,4,-5)$.

Here's the outline for part (c). Use the dot (scalar) product formula:$\displaystyle \vec{AB}.\vec{AC} = |AB||AC|\cos(\angle BAC)$

$\displaystyle \Rightarrow\cos(\angle BAC)=\frac{\vec{AB}.\vec{AC}}{|AB||AC|}$

Now$\displaystyle \vec{AB} = \vec{b}-\vec{a}=\begin{pmatrix}-3\\2\\5\end{pmatrix}-\begin{pmatrix}2\\1\\2\end{pmatrix}=\begin{pmatrix }-5\\1\\3\end{pmatrix}$

and$\displaystyle \vec{AC} = \vec{c}-\vec{a}=...$

Also$\displaystyle |AB|=\sqrt{(-5)^2+1^2+3^2}= \sqrt{35}$

and$\displaystyle |AC| = ...$

The dot product is:$\displaystyle \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} .\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}=x_1x_2+ y_1y_2+z_1z_2$

So

$\displaystyle \vec{AB}.\vec{AC}$ = ...

Can you complete it now?

Grandad