Question

The point D is such that ABCD is a parallelogram.
Find the position vector of D.

( The position vectors is at the image below)

Ans : (8i+j+8k)

Answer 1

my idea for the angle ABC is to use dot product of two vectors...

Answer 2

given position vectors...can be in the form\[\vec{OA}=\vec{A}=2\hat{i}+3\hat{j}+5\hat{k}\]\[\vec{OB}=\vec{B}=4\hat{i}+2\hat{j}+3\hat{k}\]\[\vec{OC}=\vec{C}=10\hat{i}+6\hat{k}\]

Answer 3

i thought someone have finish this... btw just to continue... we can determine the vectors from each position vectors that create a triangle such as...
From \(\vec{A}\) to \(\vec{B}\) it will be...\[\vec{r_{AB}}=\vec{B}-\vec{A}\]\[\vec{r_{AB}}=4\hat{i}+2\hat{j}+3\hat{k}-(2\hat{i}+3\hat{j}+5\hat{k})\]\[\vec{r_{AB}}=4\hat{i}+2\hat{j}+3\hat{k}-2\hat{i}-3\hat{j}-5\hat{k}\]\[\vec{r_{AB}}=(4-2)\hat{i}+(2-3)\hat{j}+(3-5)\hat{k}\]\[\vec{r_{AB}}=2\hat{i}-\hat{j}-2\hat{k}\]

Answer 4

From \(\vec{A}\) to \(\vec{C}\):\[\vec{r_{AC}}=10\hat{i}+6\hat{k}-(2\hat{i}+3\hat{j}+5\hat{k})\]\[\vec{r_{AC}}=10\hat{i}+6\hat{k}-2\hat{i}-3\hat{j}-5\hat{k}\]\[\vec{r_{AC}}=(10-2)\hat{i}+(0-3)\hat{j}+(6-5)\hat{k}\]\[\vec{r_{AC}}=8\hat{i}-3\hat{j}+\hat{k}\]

Answer 5

From \(\vec{B}\) to \(\vec{C}\):\[\vec{r_{BC}}=10\hat{i}+6\hat{k}-(4\hat{i}+2\hat{j}+3\hat{k})\]\[\vec{r_{BC}}=10\hat{i}+6\hat{k}-4\hat{i}-2\hat{j}-3\hat{k}\]\[\vec{r_{BC}}=(10-4)\hat{i}+(0-2)\hat{j}+(6-3)\hat{k}\]\[\vec{r_{BC}}=6\hat{i}-2\hat{j}+3\hat{k}\]

Answer 6

To find the measure of angle ABC... i assume that this is the angle formed at position vector \(\vec{B}\) which is the dot product of vectors \(\vec{r_{BC}}\) and \(\vec{r_{BA}}\) such that \[\vec{r_{BC}}\bullet\vec{r_{BA}}=\left|\vec{r_{BC}}\right|\left|\vec{r_{BA}}\right|\cos\theta_B\]where \(\vec{r_{BA}}=-~\vec{r_{AB}}\) therefore... \[\vec{r_{BA}}=-(2\hat{i}-\hat{j}-2\hat{k})\]\[\vec{r_{BA}}=-2\hat{i}+\hat{j}+2\hat{k}\]

Answer 7

The Dot Product of the said vectors are...\[\vec{r_{BC}}\bullet\vec{r_{BA}}=\left|\vec{r_{BC}}\right|\left|\vec{r_{BA}}\right|\cos\theta_B\]or\[\vec{r_{BC}}\bullet\vec{r_{BA}}=(6\hat{i}-2\hat{j}+3\hat{k})\bullet(-2\hat{i}+\hat{j}+2\hat{k})\]\[\vec{r_{BC}}\bullet\vec{r_{BA}}=(6)(-2)+(-2)(1)+(3)(2)\]\[\vec{r_{BC}}\bullet\vec{r_{BA}}=-12-2+6=-8\]Then find magnitudes of individual vectors:\[\left|\vec{r_{BC}}\right|=\sqrt{6^2+(-2)^2+3^2}=7\]\[\left|\vec{r_{BA}}\right|=\sqrt{(-2)^2+1^2+2^2}=3\]Therefore...\[-8=(7)(3)\cos\theta_B\]\[\cos\theta_B=\frac{-8}{21}=-0.38095238095238...\]\[\theta_B=\cos^{-1}(-0.38095238095238...)\]\[\theta_B\approx112.393^\circ\]

Answer 8

Let us check other magnitudes and angles... to come up with the location of position vector \(\vec{OD}\), we refer here also as vector \(\vec{D}\).
Magnitude of vector \(\vec{r_{AC}}\):\[\left|\vec{r_{AC}}\right|=\sqrt{8^2+(-3)^2+1^2}=\sqrt{74}\approx8.60...\]obviously this is the longest diagonal in our parallelogram and opposite to angle ABC (or \(\angle\theta_B\))... let us draw now our parallelogram ABCD (imagine it in space)...

Answer 9

For a parallelogram, we all know that we have two opposite side are parallel and equal in dimension. From above illustration, we can easily locate the position vector \(\vec{D}\). We can conclude the following:
1. Vector from A to D, \(\vec{r_{AD}}=\vec{r_{BC}}=6\hat{i}-2\hat{j}+3\hat{k}\)
2. Vector from C to D, \(\vec{r_{CD}}=\vec{r_{BA}}=-2\hat{i}+\hat{j}+2\hat{k}\)
From these conclusions, there are two ways we can determine the unknown position vector.

Answer 10

@dan815 check it out, this chick is schooling u in math. Best explanation ever!

Answer 11

Using conclusion 1...this is just finding the resultant of position vector \(\vec{A}\) and \(\vec{r_{AD}}\) which is the position vector we're looking.\[\vec{D}=\vec{A}+\vec{r_{AD}}\]\[\vec{D}=2\hat{i}+3\hat{j}+5\hat{k}+6\hat{i}-2\hat{j}+3\hat{k}\]\[\vec{D}=(2+6)\hat{i}+(3-2)\hat{j}+(5+3)\hat{k}\]\[\vec{D}=8\hat{i}+\hat{j}+8\hat{k}\]Therefore our position vector \(\vec{OD}\) to complete the parallelogram is \[\large{\vec{OD}=\left(\begin{matrix}8 \\1\\ 8\end{matrix}\right)}\]

Answer 12

hope you get it @eunnice ...