1. ## vectors

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.

2. 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.
a) Let the position vector of D be $(x, y, z)$.

Midpoint of AC=Midpoint of BD

$(2+4, 1+5, 2-2)=(-3+x, 2+y, 5+z)$

$(x, y, z)=(9, 4, -5)$

b) The point at which the diagonals of the parallelogram intersect=Midpoint of AC= $(\frac{2+4}{2}, \frac{1+5}{2}, \frac{2-2}{2})=(3, 3, 0)$

3. 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
...

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

$(x, y, z)=(9, 4, \color{red}-3\color{black})$
...
It should be $(9,4,-5)$.

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

$\Rightarrow\cos(\angle BAC)=\frac{\vec{AB}.\vec{AC}}{|AB||AC|}$
Now
$\vec{AB} = \vec{b}-\vec{a}=\begin{pmatrix}-3\\2\\5\end{pmatrix}-\begin{pmatrix}2\\1\\2\end{pmatrix}=\begin{pmatrix }-5\\1\\3\end{pmatrix}$
and
$\vec{AC} = \vec{c}-\vec{a}=...$
Also
$|AB|=\sqrt{(-5)^2+1^2+3^2}= \sqrt{35}$
and
$|AC| = ...$
The dot product is:
$\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} .\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}=x_1x_2+ y_1y_2+z_1z_2$
So
$\vec{AB}.\vec{AC}$ = ...
Can you complete it now?

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# the position vector of point D if ABCD is a parallelogram

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