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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 3i→\overrightarrow{i} + 7j→\overrightarrow{j} and 3i→\overrightarrow{i} - 2j→\overrightarrow{j} respectively, find the position vector of G.

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

Since AD = DC, from mid-point theorem, OD→\overrightarrow{OD} = 12(OA→+OC→)\frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC}) or, OA→+OC→\overrightarrow{OA} + \overrightarrow{OC} = 2(OD→)(\overrightarrow{OD})
so, OA→+OC→\overrightarrow{OA} + \overrightarrow{OC} = 2(3i→\overrightarrow{i} - 2j→\overrightarrow{j})

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

∴ OG→\overrightarrow{OG} = 3i→\overrightarrow{i} + j→\overrightarrow{j}

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

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

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

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

If the position vectors of two points A and B are (32)\begin{pmatrix} 3 \\ 2 \end{pmatrix} and (−54)\begin{pmatrix} -5 \\ 4 \end{pmatrix} respectively then find position vector of the mid-point C of AB.

Let O be the origin and OA→\overrightarrow{OA} and OB→\overrightarrow{OB} be the position vector of the vertices A and B.
OA→\overrightarrow{OA} = (32)\begin{pmatrix} 3 \\ 2 \end{pmatrix} = 3i→\overrightarrow{i} + 2j→\overrightarrow{j}
OA→\overrightarrow{OA} = (−54)\begin{pmatrix} -5 \\ 4 \end{pmatrix} = -5i→\overrightarrow{i} + 4j→\overrightarrow{j}

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

∴ OC→\overrightarrow{OC} = (−13)\begin{pmatrix} -1 \ 3 \end{pmatrix}

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

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

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

∴ QR→\overrightarrow{QR} = 2i→\overrightarrow{i} + 7j→\overrightarrow{j}

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

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

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

∴ OM→\overrightarrow{OM} = 5i→\overrightarrow{i} + 2j→\overrightarrow{j}

If the position vectors of two points A and B are (34)\begin{pmatrix} 3 \\ 4 \end{pmatrix} and (45)\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 OA→\overrightarrow{OA} and OB→\overrightarrow{OB} be the position vector of the vertices A and B.
OA→\overrightarrow{OA} = (34)\begin{pmatrix} 3 \\ 4 \end{pmatrix} = 3i→\overrightarrow{i} + 4j→\overrightarrow{j}
OA→\overrightarrow{OA} = (45)\begin{pmatrix} 4 \\ 5 \end{pmatrix} = 4i→\overrightarrow{i} + 5j→\overrightarrow{j}

Now, for mid-point,
OK→\overrightarrow{OK} = OA→+OB→2\frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} = 3i→+4j→+4i→+5j→2\frac{3\overrightarrow{i} + 4\overrightarrow{j} + 4\overrightarrow{i} + 5\overrightarrow{j}}{2}
= 72i→\frac{7}{2} \overrightarrow{i} + 92j→\frac{9}{2}\overrightarrow{j}

∴ OC→\overrightarrow{OC} = (7/29/2)\begin{pmatrix} 7/2 \ 9/2 \end{pmatrix} = (3.54.5)\begin{pmatrix} 3.5 \ 4.5 \end{pmatrix}

IF the position vector of the mid-point of the line segment EF is 4i→\overrightarrow{i} + 7j→\overrightarrow{j} and the position vector of the point E is -3i→\overrightarrow{i} - 4j→\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,
OE→\overrightarrow{OE} = -3i→\overrightarrow{i} - 4j→\overrightarrow{j}
OF→\overrightarrow{OF} = ?
OM→\overrightarrow{OM} = 4i→\overrightarrow{i} + 7j→\overrightarrow{j}

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

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

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

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

The position vector of A and B are 2i→\overrightarrow{i} + 7j→\overrightarrow{j} and 7i→\overrightarrow{i} - 3j→\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,
OA→\overrightarrow{OA} = 2i→\overrightarrow{i} + 7j→\overrightarrow{j}
OB→\overrightarrow{OB} = 7i→\overrightarrow{i} - 3j→\overrightarrow{j}

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

∴ OP→\overrightarrow{OP} = 4i→\overrightarrow{i} + 3j→\overrightarrow{j}

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

Let OP→\overrightarrow{OP} = (x, y) be the position vector which divides AB internally in the ratio of 2:3 (m1:m2). Then,
(x, y) = (m1x2+m2x1m1+m2(\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}, m1y2+m2y1m1+m2)\frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}) = (2(7)+3(2)5(\frac{2(7) + 3(2)}{5}, 2(−3)+3(7)5)\frac{2(-3) + 3(7)}{5})
= (14+65(\frac{14 + 6}{5}, −6+215)\frac{-6 + 21}{5})
= (4, 3)

∴ OP→\overrightarrow{OP} = 4i→\overrightarrow{i} + 3j→\overrightarrow{j}

D is the mid-point of the side BC of △ABC. If the position vectors of the points A and D are 3i→\overrightarrow{i} + 5j→\overrightarrow{j} and 3i→\overrightarrow{i} - 4j→\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,
OA→\overrightarrow{OA} = 3i→\overrightarrow{i} + 5j→\overrightarrow{j}
OD→\overrightarrow{OD} = 3i→\overrightarrow{i} - 4j→\overrightarrow{j}

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

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

∴ OG→\overrightarrow{OG} = 3i→\overrightarrow{i} - j→\overrightarrow{j}

If 3i→\overrightarrow{i} + 6j→\overrightarrow{j} and 5i→\overrightarrow{i} + 2j→\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,
OP→\overrightarrow{OP} = 3i→\overrightarrow{i} + 6j→\overrightarrow{j}
OQ→\overrightarrow{OQ} = 5i→\overrightarrow{i} + 2j→\overrightarrow{j}

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

∴ OP→\overrightarrow{OP} = 215i→\frac{21}{5}\overrightarrow{i} + 185j→\frac{18}{5}\overrightarrow{j}
Another method Let O be the origin and position vector of A and B are,
OP→\overrightarrow{OP} = 3i→\overrightarrow{i} + 6j→\overrightarrow{j} = (3, 6) = (x1, y1)
OQ→\overrightarrow{OQ} = 5i→\overrightarrow{i} + 2j→\overrightarrow{j} = (5, 2) = (x2, y2)

Let OP→\overrightarrow{OP} = (x, y) be the position vector which divides AB internally in the ratio of 3:2 (m1:m2). Then,
(x, y) = (m1x2+m2x1m1+m2(\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}, m1y2+m2y1m1+m2)\frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}) = (3(5)+2(3)5(\frac{3(5) + 2(3)}{5}, 3(2)+2(6)5)\frac{3(2) + 2(6)}{5})
= (15+65(\frac{15 + 6}{5}, 6+125)\frac{6 + 12}{5})
= (215(\frac{21}{5}, 185)\frac{18}{5})

∴ OP→\overrightarrow{OP} = 215i→\frac{21}{5}\overrightarrow{i} + 185j→\frac{18}{5}\overrightarrow{j}

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

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

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

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

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

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

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