Porisma

Worked solution

Find a perpendicular foot and use it as a triangle height

Referred to the origin OO, points AA and BB have position vectors a=−i−6j−5k\mathbf a=-\mathbf i-6\mathbf j-5\mathbf k and b=4i+4j+10k\mathbf b=4\mathbf i+4\mathbf j+10\mathbf k respectively. Point PP lies on ABAB, between AA and BB, such that AP:PB=2:3AP:PB=2:3.

(i) Find the position vector OP→\overrightarrow{OP}.

(ii) Show that AB→\overrightarrow{AB} and OP→\overrightarrow{OP} are perpendicular.

The vector c\mathbf c is a unit vector in the direction of OP→\overrightarrow{OP}.

(iii) Find the value of ∣a⋅c∣|\mathbf a\cdot\mathbf c| and give the geometrical interpretation of ∣a⋅c∣|\mathbf a\cdot\mathbf c|. Hence, or otherwise, find the area of triangle OABOAB.

[2016/TPJC/Promo/4]

CJC · Discussion · Question 4

Answer

OP→=(1,−2,1)\overrightarrow{OP}=(1,-2,1); AB⋅OP=0AB\cdot OP=0; ∣a⋅c∣=6|\mathbf a\cdot\mathbf c|=\sqrt 6; triangle area 5215\sqrt{21}.

(i) Find P by the section formula

OP→=3a+2b5=(1−21).\overrightarrow{OP}=\frac{3\mathbf a+2\mathbf b}{5}=\begin{pmatrix}1\\-2\\1\end{pmatrix}.

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(ii) Test perpendicularity

AB→=b−a=(5,10,15)\overrightarrow{AB}=\mathbf b-\mathbf a=(5,10,15). Its dot product with OP→\overrightarrow{OP} is 5−20+15=05-20+15=0. Both are nonzero, so they are perpendicular.

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(iii) Projection length

∣OP→∣=6|\overrightarrow{OP}|=\sqrt 6 and c=(1,−2,1)/6\mathbf c=(1,-2,1)/\sqrt 6. Thus

∣a⋅c∣=∣−1+12−5∣6=6.|\mathbf a\cdot\mathbf c|=\frac{|-1+12-5|}{\sqrt 6}=\sqrt 6.

This is the length of the projection of OA→\overrightarrow{OA} onto the direction of OP→\overrightarrow{OP}.

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Area, method 1: base and height

Because P lies on AB and OP is perpendicular to AB, OP is the altitude from O. Therefore

area(OAB)=12∣AB∣ ∣OP∣=123506=521.\begin{aligned}&\text{area}(OAB)\\ & =\frac 12|AB|\,|OP|\\ & =\frac 12\sqrt{350}\sqrt 6\\ & =5\sqrt{21}.\end{aligned}

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Area, method 2: cross product

a×b=(−40−1020).\mathbf a\times\mathbf b=\begin{pmatrix}-40\\-10\\20\end{pmatrix}.

Hence the area is 12(−40)2+(−10)2+202=521\frac 12\sqrt{(-40)^2+(-10)^2+20^2}=5\sqrt{21}. Both printed area methods give the same exact result.

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Common mistakes

The prompt has absolute-value bars around the scalar product. Projection length is nonnegative; a signed scalar component is a different quantity.

Solution notes

Magnitude bars and radicals restored from page images. Both printed area methods on PDF 4 retained; no source diagram appears for this question.

Read the hint