Author Topic: Vector proof question (Read 719 times) Tweet Share

TrueTears · « **on:** December 23, 2008, 02:39:08 am »

In the figure, O is the centre of the circle, radius r. The circle is inscribed in the square ABCD. P is any point on the circumference of the circle.

a) show that $\overrightarrow{AP}.\overrightarrow{AP}$ = $3$ $r^2$ - $2$ $\overrightarrow{OP}.\overrightarrow{OA}$

b) Hence find $AP^2 +BP^2 + CP^2 +DP^2$ in terms of r.

(the sides of the square are tangent to the circle)

Mao · « **Reply #1 on:** December 23, 2008, 02:45:41 am »

a)

$\vec{AP}=\vec{AO}+\vec{OP}$

$\vec{AP}\cdot \vec{AP} = (\vec{AO}+\vec{OP})\cdot (\vec{AO}+\vec{OP}) = |\vec{AO}|^2 + |\vec{OP}|^2 + 2\cdot \vec{AO}\cdot \vec{OP} = (\sqrt{2}\cdot r)^2 + r^2 + (-2\cdot \vec{OA}\cdot \vec{OP}) = 3r^2 - 2\cdot \vec{OA}\cdot \vec{OP}$

b)

$AP^2 = \vec{AP}\cdot \vec{AP} = 3r^2 - 2\cdot \vec{OA} \cdot \vec{OP}$
$BP^2 = \vec{BP}\cdot \vec{BP} = 3r^2 - 2\cdot \vec{OB} \cdot \vec{OP}$
$CP^2 = \vec{CP}\cdot \vec{CP} = 3r^2 - 2\cdot \vec{OC} \cdot \vec{OP}$
$DP^2 = \vec{DP}\cdot \vec{DP} = 3r^2 - 2\cdot \vec{OD} \cdot \vec{OP}$

now, since ABCD is a square and O is the centre, we realise that $\vec{OC}=-\vec{OA},\; \vec{OD}=-\vec{OD}$

hence, when the above are added, all the dot products at the end cancel out, and we end up with $12r^2$

TrueTears · « **Reply #2 on:** December 23, 2008, 02:12:05 pm »

ahhh thanks mao

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