A perpendicular foot organizes four shapes

The position vectors of A, B and C relative to origin O are −i+5j+k-\mathbf i+5\mathbf j+\mathbf k, i−j+5k\mathbf i-\mathbf j+5\mathbf k and pi−4j+7kp\mathbf i-4\mathbf j+7\mathbf k respectively, where pp is constant. Given A, B and C are collinear, show p=2p=2. marks

(i) N divides AC in ratio AN:NC=λ:1−λAN:NC=\lambda:1-\lambda. Find λ\lambda given ON is perpendicular to AC. marks

(ii) Hence find the exact area of triangle OAC. marks

(iii) Find the position vector of D such that OADC is a parallelogram. marks

(iv) Find the coordinates of E such that OAEC is a kite. marks

[modified 2017/DHS/Promo/6]

Total marks: 12

Solution

Collinearity

Collinear points

A,B,CA,B,C have position vectors −i+5j+k-\mathbf i+5\mathbf j+\mathbf k, i−j+5k\mathbf i-\mathbf j+5\mathbf k and pi−4j+7kp\mathbf i-4\mathbf j+7\mathbf k. Given they are collinear, show p=2p=2.

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Hint

  1. Collinear means AB→\overrightarrow{AB} is a multiple of AC→\overrightarrow{AC}.

  2. Compare components.

Answer · Collinearity

p=2p=2.

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The three points lie on one line

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(i)

Perpendicular foot by dot product

NN divides ACAC with AN:NC=λ:1−λAN:NC=\lambda:1-\lambda and ON⊥ACON\perp AC. Find λ\lambda.

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Hint

  1. ON→=(1−λ)OA→+λOC→\overrightarrow{ON}=(1-\lambda)\overrightarrow{OA}+\lambda\overrightarrow{OC}.

  2. Set ON→⋅AC→=0\overrightarrow{ON}\cdot\overrightarrow{AC}=0.

Answer · (i)

λ=1/3\lambda=1/3; N=(0,2,3)N=(0,2,3).

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Slide N along AC until ON is perpendicular

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(ii)

Area of a triangle

Find the exact area of triangle OACOAC.

Hint

  1. Area =12×AC×ON=\frac12\times AC\times ON since ON⊥ACON\perp AC.

Answer · (ii)

3182/23\sqrt{182}/2.

(iii)

Parallelogram

Find DD such that OADCOADC is a parallelogram.

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Hint

  1. In parallelogram OADCOADC, AD→=OC→\overrightarrow{AD}=\overrightarrow{OC}.

Answer · (iii)

D=(1,1,8)D=(1,1,8).

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The fourth vertex of OADC

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(iv)

Kite

Find EE such that OAECOAEC is a kite.

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Hint

  1. The kite is symmetric about ACAC: EE is the reflection of OO in ACAC.

Answer · (iv)

E=(0,4,6)E=(0,4,6), the reflection of OO in ACAC.

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The kite OAEC

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