# Vector Geometry

In the given figure, AD = DC and the point G is the centroid of the triangle ABC. If the positive vectors of the points B and D are 3$\overrightarrow{i}$ + 7$\overrightarrow{j}$ and 3$\overrightarrow{i}$ - 2$\overrightarrow{j}$ respectively, find the position vector of G.

Let O be the origin and $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$, $\overrightarrow{OD}$ be the position vector of the vertices A, B, C and D.
$\overrightarrow{OB}$ = 3$\overrightarrow{i}$ + 7$\overrightarrow{j}$
$\overrightarrow{OD}$ = 3$\overrightarrow{i}$ - 2$\overrightarrow{j}$

Since AD = DC, from mid-point theorem, $\overrightarrow{OD}$ = $\frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC})$ or, $\overrightarrow{OA} + \overrightarrow{OC}$ = 2$(\overrightarrow{OD})$
so, $\overrightarrow{OA} + \overrightarrow{OC}$ = 2(3$\overrightarrow{i}$ - 2$\overrightarrow{j}$)

Now, $\overrightarrow{OG}$ is the centroid, so,
$\overrightarrow{OG}$ = $\frac{\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} }{3}$ = $\frac{2(3\overrightarrow{i} - 2\overrightarrow{j}) \, + \, 3\overrightarrow{i} + 7\overrightarrow{j}}{3}$
= $\frac{ 9\overrightarrow{i} + 3\overrightarrow{j} }{3}$

$\overrightarrow{OG}$ = 3$\overrightarrow{i}$ + $\overrightarrow{j}$

If the position vectors of the vertices A, B, and C of △ABC are (3$\overrightarrow{i}$ + 5$\overrightarrow{j}$), (5$\overrightarrow{i}$ - $\overrightarrow{j}$) and ($\overrightarrow{i}$ - 8$\overrightarrow{j}$) respectively, find the position vectors of the enroid G of the triangle.

Let O be the origin and $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$ be the position vector of the vertices A, B and C.
$\overrightarrow{OA}$ = 3$\overrightarrow{i}$ + 5$\overrightarrow{j}$
$\overrightarrow{OB}$ = 5$\overrightarrow{i}$ - $\overrightarrow{j}$
$\overrightarrow{OC}$ = $\overrightarrow{i}$ - 8$\overrightarrow{j}$

Now, for centroid,
$\overrightarrow{OG}$ = $\frac{\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} }{3}$ = $\frac{3\overrightarrow{i} + 5\overrightarrow{j} + 5\overrightarrow{i} - \overrightarrow{j} + \overrightarrow{i} - 8\overrightarrow{j}}{3}$
= $\frac{9\overrightarrow{i} - 4\overrightarrow{j}}{3}$

$\overrightarrow{OG}$ = $3\overrightarrow{i}$ - $\frac 43 \overrightarrow{j}$

If the position vectors of two points A and B are $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} -5 \\ 4 \end{pmatrix}$ respectively then find position vector of the mid-point C of AB.

Let O be the origin and $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be the position vector of the vertices A and B.
$\overrightarrow{OA}$ = $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ = 3$\overrightarrow{i}$ + 2$\overrightarrow{j}$
$\overrightarrow{OA}$ = $\begin{pmatrix} -5 \\ 4 \end{pmatrix}$ = -5$\overrightarrow{i}$ + 4$\overrightarrow{j}$

Now, for mid-point,
$\overrightarrow{OC}$ = $\frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}$ = $\frac{3\overrightarrow{i} + 2\overrightarrow{j} - 5\overrightarrow{i} + 4\overrightarrow{j}}{2}$
= $\frac{-2\overrightarrow{i} + 6\overrightarrow{j}}{2}$
= -$\overrightarrow{i}$ + 3$\overrightarrow{j}$

$\overrightarrow{OC}$ = $\begin{pmatrix} -1 \ 3 \end{pmatrix}$

In the given parallelogram $\overrightarrow{SR}$ = 4$\overrightarrow{i}$ - 2$\overrightarrow{j}$ and $\overrightarrow{PR}$ = 6$\overrightarrow{i}$ + 5$\overrightarrow{j}$, find $\overrightarrow{QR}$.

Here,
$\overrightarrow{SR}$ = $\overrightarrow{i}$ - 2$\overrightarrow{j}$
$\overrightarrow{PR}$ = 6$\overrightarrow{i}$ + 5$\overrightarrow{j}$
$\overrightarrow{QR}$ = ?

Now, From triangle law of vector addition,
$\overrightarrow{QR}$ = $\overrightarrow{QP}$ + $\overrightarrow{PR}$ ($\overrightarrow{QP}$ = $\overrightarrow{RS}$) = $\overrightarrow{RS}$ + $\overrightarrow{PR}$
= -$\overrightarrow{SR}$ + $\overrightarrow{PR}$
= $-4\overrightarrow{i} + 2\overrightarrow{j} + 6\overrightarrow{i} + 5\overrightarrow{j}$

$\overrightarrow{QR}$ = 2$\overrightarrow{i}$ + 7$\overrightarrow{j}$

The position vectors of A and B are 9$\overrightarrow{i}$ + 7$\overrightarrow{j}$ and $\overrightarrow{i}$ - 3$\overrightarrow{j}$ respectively. If M is the mid-point of AB, find the position vector of M.

Let O be the origin and $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be the position vector of A and B.
$\overrightarrow{OA}$ = 9$\overrightarrow{i}$ + 7$\overrightarrow{j}$
$\overrightarrow{OB}$ = $\overrightarrow{i}$ - 3$\overrightarrow{j}$

Now, position vector of mid-point M,
$\overrightarrow{OM}$ = $\frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}$ = $\frac{9\overrightarrow{i} + 7\overrightarrow{j} + \overrightarrow{i} - 3\overrightarrow{j}}{2}$

$\overrightarrow{OM}$ = 5$\overrightarrow{i}$ + 2$\overrightarrow{j}$

If the position vectors of two points A and B are $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ 5 \end{pmatrix}$ respectively, then find the position vector of K which is mid-point of AB.

Let O be the origin and $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be the position vector of the vertices A and B.
$\overrightarrow{OA}$ = $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ = 3$\overrightarrow{i}$ + 4$\overrightarrow{j}$
$\overrightarrow{OA}$ = $\begin{pmatrix} 4 \\ 5 \end{pmatrix}$ = 4$\overrightarrow{i}$ + 5$\overrightarrow{j}$

Now, for mid-point,
$\overrightarrow{OK}$ = $\frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}$ = $\frac{3\overrightarrow{i} + 4\overrightarrow{j} + 4\overrightarrow{i} + 5\overrightarrow{j}}{2}$
= $\frac{7}{2} \overrightarrow{i}$ + $\frac{9}{2}\overrightarrow{j}$

$\overrightarrow{OC}$ = $\begin{pmatrix} 7/2 \ 9/2 \end{pmatrix}$ = $\begin{pmatrix} 3.5 \ 4.5 \end{pmatrix}$

IF the position vector of the mid-point of the line segment EF is 4$\overrightarrow{i}$ + 7$\overrightarrow{j}$ and the position vector of the point E is -3$\overrightarrow{i}$ - 4$\overrightarrow{j}$, find the position vector of the point F.

Let O be the origin. Position vector of E, F and mid-point M is,
$\overrightarrow{OE}$ = -3$\overrightarrow{i}$ - 4$\overrightarrow{j}$
$\overrightarrow{OF}$ = ?
$\overrightarrow{OM}$ = 4$\overrightarrow{i}$ + 7$\overrightarrow{j}$

Now, For $\overrightarrow{OF}$, $\overrightarrow{OM}$ = $\frac{\overrightarrow{OE} + \overrightarrow{OF}}{2}$ or, 4$\overrightarrow{i}$ + 7$\overrightarrow{j}$ = $\frac{3\overrightarrow{i} - 4\overrightarrow{j} + \overrightarrow{OF}}{2}$
or, 8$\overrightarrow{i}$ + 14$\overrightarrow{j}$ + 3$\overrightarrow{i}$ + 4$\overrightarrow{j}$ = $\overrightarrow{OF}$
$\overrightarrow{OF}$ = 11$\overrightarrow{i}$ + 18$\overrightarrow{j}$

If the position vectors of M and N are 9$\overrightarrow{i}$ + 3$\overrightarrow{j}$ and 3$\overrightarrow{i}$ + 5$\overrightarrow{j}$ respectively. Find the position vector of a point P such that $\overrightarrow{MP}$ = $\overrightarrow{PN}$.

Let O be the origin and position vector of M, N, and P are,
$\overrightarrow{OM}$ = 9$\overrightarrow{i}$ + 3$\overrightarrow{j}$
$\overrightarrow{ON}$ = 3$\overrightarrow{i}$ + 5$\overrightarrow{j}$
$\overrightarrow{OP}$ = ?

Now, We have given, $\overrightarrow{MP}$ = $\overrightarrow{PN}$ or, $\overrightarrow{MO}$ + $\overrightarrow{OP}$ = $\overrightarrow{PO}$ + $\overrightarrow{ON}$
or, -$\overrightarrow{OM}$ + $\overrightarrow{OP}$ = -$\overrightarrow{OP}$ + $\overrightarrow{ON}$
or, 2$\overrightarrow{OP}$ = $\overrightarrow{ON}$ + $\overrightarrow{OM}$
or, 2$\overrightarrow{OP}$ = 3$\overrightarrow{i}$ + 5$\overrightarrow{j}$ + 9$\overrightarrow{i}$ + 3$\overrightarrow{j}$
$\overrightarrow{OP}$ = 6$\overrightarrow{i}$ + 4$\overrightarrow{j}$

The position vector of A and B are 2$\overrightarrow{i}$ + 7$\overrightarrow{j}$ and 7$\overrightarrow{i}$ - 3$\overrightarrow{j}$. Find the position vector of a point which divides AB internally in the ration of 2:3.

Let O be the origin and position vector of A and B are,
$\overrightarrow{OA}$ = 2$\overrightarrow{i}$ + 7$\overrightarrow{j}$
$\overrightarrow{OB}$ = 7$\overrightarrow{i}$ - 3$\overrightarrow{j}$

Let $\overrightarrow{OP}$ be the position vector which divides AB internally in the ratio of 2:3 (m:n). Then,
$\overrightarrow{OP}$ = $\frac{m\overrightarrow{OA} + n\overrightarrow{OB}}{m + n}$ = $\frac{2(7\overrightarrow{i} - 3\overrightarrow{j}) + 3(2\overrightarrow{i} + 7\overrightarrow{j})}{5}$
= $\frac{20\overrightarrow{i} + 15\overrightarrow{j}}{5}$

$\overrightarrow{OP}$ = 4$\overrightarrow{i}$ + 3$\overrightarrow{j}$

Another method Let O be the origin and position vector of A and B are,
$\overrightarrow{OA}$ = 2$\overrightarrow{i}$ + 7$\overrightarrow{j}$ = (2, 7) = (x1, y1)
$\overrightarrow{OB}$ = 7$\overrightarrow{i}$ - 3$\overrightarrow{j}$ = (7, -3) = (x2, y2)

Let $\overrightarrow{OP}$ = (x, y) be the position vector which divides AB internally in the ratio of 2:3 (m1:m2). Then,
(x, y) = $(\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}$, $\frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}})$ = $(\frac{2(7) + 3(2)}{5}$, $\frac{2(-3) + 3(7)}{5})$
= $(\frac{14 + 6}{5}$, $\frac{-6 + 21}{5})$
= (4, 3)

$\overrightarrow{OP}$ = 4$\overrightarrow{i}$ + 3$\overrightarrow{j}$

D is the mid-point of the side BC of △ABC. If the position vectors of the points A and D are 3$\overrightarrow{i}$ + 5$\overrightarrow{j}$ and 3$\overrightarrow{i}$ - 4$\overrightarrow{j}$ respectively, find the position vector of the centroid(G) of △ABC.

Let O be the origin and position vectors of the points A and D are,
$\overrightarrow{OA}$ = 3$\overrightarrow{i}$ + 5$\overrightarrow{j}$
$\overrightarrow{OD}$ = 3$\overrightarrow{i}$ - 4$\overrightarrow{j}$

D is the mid-point of side BC. So, $\overrightarrow{OD}$ = $\frac{\overrightarrow{OB} + \overrightarrow{OC}}{2}$ or, $\overrightarrow{OB} + \overrightarrow{OC}$ = 6$\overrightarrow{i}$ - 8$\overrightarrow{j}$ --- (i)

Finally, position vector of centroid(G),
$\overrightarrow{OG}$ = $\frac{\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}}{3}$ = $\frac{3\overrightarrow{i} + 5\overrightarrow{j} + 6\overrightarrow{i} - 8\overrightarrow{j}}{3}$ (from 'i')
= $\frac{9\overrightarrow{i} - 3\overrightarrow{j}}{3}$

$\overrightarrow{OG}$ = 3$\overrightarrow{i}$ - $\overrightarrow{j}$

If 3$\overrightarrow{i}$ + 6$\overrightarrow{j}$ and 5$\overrightarrow{i}$ + 2$\overrightarrow{j}$ are the position vectors of the points P and Q respectively, find the position vectors of the point M which divides the line PQ internally in the ratio of 3:2.

Let O be the origin and position vector of P and Q are,
$\overrightarrow{OP}$ = 3$\overrightarrow{i}$ + 6$\overrightarrow{j}$
$\overrightarrow{OQ}$ = 5$\overrightarrow{i}$ + 2$\overrightarrow{j}$

Let $\overrightarrow{OP}$ be the position vector which divides AB internally in the ratio of 3:2 (m:n). Then,
$\overrightarrow{OP}$ = $\frac{m\overrightarrow{OA} + n\overrightarrow{OB}}{m + n}$ = $\frac{3(5\overrightarrow{i} + 2\overrightarrow{j}) + 2(3\overrightarrow{i} + 6\overrightarrow{j})}{5}$
= $\frac{15\overrightarrow{i} + 6\overrightarrow{j} + 6\overrightarrow{i} + 12\overrightarrow{j}}{5}$

$\overrightarrow{OP}$ = $\frac{21}{5}\overrightarrow{i}$ + $\frac{18}{5}\overrightarrow{j}$
Another method Let O be the origin and position vector of A and B are,
$\overrightarrow{OP}$ = 3$\overrightarrow{i}$ + 6$\overrightarrow{j}$ = (3, 6) = (x1, y1)
$\overrightarrow{OQ}$ = 5$\overrightarrow{i}$ + 2$\overrightarrow{j}$ = (5, 2) = (x2, y2)

Let $\overrightarrow{OP}$ = (x, y) be the position vector which divides AB internally in the ratio of 3:2 (m1:m2). Then,
(x, y) = $(\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}$, $\frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}})$ = $(\frac{3(5) + 2(3)}{5}$, $\frac{3(2) + 2(6)}{5})$
= $(\frac{15 + 6}{5}$, $\frac{6 + 12}{5})$
= $(\frac{21}{5}$, $\frac{18}{5})$

$\overrightarrow{OP}$ = $\frac{21}{5}\overrightarrow{i}$ + $\frac{18}{5}\overrightarrow{j}$

The given figure is semi-circle with the center O, prove that: $(\overrightarrow{QO} + \overrightarrow{OP}).(\overrightarrow{QO} + \overrightarrow{OR})$ = 0.

We know,
$\overrightarrow{PO}$ = $\overrightarrow{OR}$ = $\overrightarrow{QO}$ = radius --- (i)

To prove, $(\overrightarrow{QO} + \overrightarrow{OP}).(\overrightarrow{QO} + \overrightarrow{OR})$ = 0 or, $(\overrightarrow{QO} - \overrightarrow{PO}).(\overrightarrow{QO} + \overrightarrow{PO})$ = 0 (from 'i')
or, $\overrightarrow{QO}$2 - $\overrightarrow{PO}$2 = 0
or, $\overrightarrow{QO}$2 - $\overrightarrow{QO}$2 = 0 (from 'i')
∴ 0 = 0

In the given figure, $\overrightarrow{OA}$ = $\overrightarrow{a}$ and $\overrightarrow{OB}$ = $\overrightarrow{b}$. If $\overrightarrow{AC}$ = 3$\overrightarrow{AB}$, find $\overrightarrow{OC}$.

Here,
$\overrightarrow{OA}$ = $\overrightarrow{a}$
$\overrightarrow{OB}$ = $\overrightarrow{b}$

We have given, $\overrightarrow{AC}$ = 3$\overrightarrow{AB}$
or, $\overrightarrow{AO}$ + $\overrightarrow{OC}$ = 3$(\overrightarrow{AO}$ + $\overrightarrow{OB})$ (From triangle law of vector addition)
or, -$\overrightarrow{OA}$ + $\overrightarrow{OC}$ = 3$(-\overrightarrow{OA}$ + $\overrightarrow{OB})$
or, -$\overrightarrow{a}$ + $\overrightarrow{OC}$ = 3$(-\overrightarrow{a} + \overrightarrow{b})$
or, $\overrightarrow{OC}$ = -3$\overrightarrow{a}$ + 3$\overrightarrow{b}$ + $\overrightarrow{a}$
$\overrightarrow{OC}$ = 3$\overrightarrow{b}$ - 2$\overrightarrow{a}$