Porisma

Worked solution

Apply the section formula and preserve vertex order

The position vectors of vertices AA, BB and CC, relative to the origin OO, are −3i+4k-3\mathbf i+4\mathbf k, −2i+4j−3k-2\mathbf i+4\mathbf j-3\mathbf k and 11i−4j−9k11\mathbf i-4\mathbf j-9\mathbf k respectively.

(i) Find a unit vector parallel to OA→\overrightarrow{OA}.

(ii) A point PP divides ACAC in the ratio 1:31:3. Find the position vector of PP.

(iii) Show that the points OO, BB and PP are collinear.

(iv) OBDCOBDC forms a parallelogram. Find the position vector of DD.

[2010/RVHS/Promo/6]

CJC · Discussion · Question 3

Answer

(i) 15(−3,0,4)\frac 15(-3,0,4); (ii) OP→=(12,−1,34)\overrightarrow{OP}=(\frac 12,-1,\frac 34); (iii) OP→=−14OB→\overrightarrow{OP}=-\frac 14\overrightarrow{OB}; (iv) OD→=(9,0,−12)\overrightarrow{OD}=(9,0,-12).

(i) Normalise OA

∣OA→∣=(−3)2+42=5|\overrightarrow{OA}|=\sqrt{(-3)^2+4^2}=5. One unit vector parallel to it is 15(−304)\frac 15\begin{pmatrix}-3\\0\\4\end{pmatrix}. Its negative is also parallel, in the opposite orientation; the source gives the positive-direction choice.

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(ii) Divide AC internally

With AP:PC=1:3AP:PC=1:3,

OP→=3OA→+OC→4=(12−134).\overrightarrow{OP}=\frac{3\overrightarrow{OA}+\overrightarrow{OC}}4=\begin{pmatrix}\frac 12\\-1\\\frac 34\end{pmatrix}.

The endpoint nearer P receives the larger opposite weight.

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(iii) Compare OB and OP

OP→=−14OB→\overrightarrow{OP}=-\frac 14\overrightarrow{OB}, since OB→=(−2,4,−3)\overrightarrow{OB}=(-2,4,-3). The vectors are parallel and both start at O, so O, B and P are collinear. The negative multiple places P on the opposite side of O from B.

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(iv) Complete the specified parallelogram

In the order O–B–D–C, BD→=OC→\overrightarrow{BD}=\overrightarrow{OC}. Thus

OD→=OB→+OC→=(90−12).\overrightarrow{OD}=\overrightarrow{OB}+\overrightarrow{OC}=\begin{pmatrix}9\\0\\-12\end{pmatrix}.

The independent schematic retains this order and the equal directed sides.

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Compare your sketch

Expand diagram

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Schematic parallelogram OBDC: O bottom left, B bottom right, D top right, C top left. The equal vectors O to C and B to D are marked. This is a labelled schematic, not a projection of the three-dimensional coordinates.
Schematic parallelogram OBDC: O bottom left, B bottom right, D top right, C top left. The equal vectors O to C and B to D are marked. This is a labelled schematic, not a projection of the three-dimensional coordinates.
Schematic parallelogram OBDC: O bottom left, B bottom right, D top right, C top left. The equal vectors O to C and B to D are marked. This is a labelled schematic, not a projection of the three-dimensional coordinates.

Common mistakes

OBDC has D opposite O. It is not the same ordering as OBCD.

Solution notes

All four printed asks retained. The source diagram is schematic; the redrawing is explicitly not a coordinate projection. A parallel unit-vector answer is not unique unless orientation is specified.

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