Vector Algebra
Position Vectors and Geometry
Grade 12
Question:
<p>Let \(A(0,-1,1),\;B(0,0,1),\;C(0,1,0)\) be the vertices of \(\triangle ABC\). If \(R\) is the circumradius and \(r\) the inradius, then \(\cos\!\left(\dfrac{R}{r}\right)\) equals</p>
<li>\(\tan\dfrac{3\pi}{8}\)</li>
<li>\(\tan\dfrac{\pi}{8}\)</li>
<li>\(\cos\dfrac{\pi}{4}\)</li>
<li>\(\cot\dfrac{\pi}{8}\)</li>
Step-by-Step Solution
Key Concept: Compute a=BC, b=CA, c=AB. Find area \Delta=½|AB \times AC|. Then R=abc/4\Delta and r=\Delta/s where s=(a+b+c)/2.
$AB=\sqrt{0+1+0}=1,\;BC=\sqrt{0+1+1}=\sqrt2,\;CA=\sqrt{0+4+1}=\sqrt5$.
Wait: $A=(0,-1,1),B=(0,0,1),C=(0,1,0)$.
$\overrightarrow{AB}=(0,1,0)\Rightarrow|AB|=1$.
$\overrightarrow{BC}=(0,1,-1)\Rightarrow|BC|=\sqrt2$.
$\overrightarrow{CA}=(0,-2,1)\Rightarrow|CA|=\sqrt5$.
Perimeter $2s=1+\sqrt2+\sqrt5$. Area $\Delta=\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|$.
$\overrightarrow{AC}=(0,2,-1)$. $\overrightarrow{AB}\times\overrightarrow{AC}=(0,1,0)\times(0,2,-1)=(-1,0,0)\Rightarrow|\cdot|=1$. So $\Delta=\frac12$.
$R=\frac{1\cdot\sqrt2\cdot\sqrt5}{4\cdot\frac12}=\frac{\sqrt{10}}{2},\quad r=\frac{\frac12}{\frac{1+\sqrt2+\sqrt5}{2}}=\frac{1}{1+\sqrt2+\sqrt5}$.
Then $R/r = \frac{\sqrt{10}}{2}(1+\sqrt2+\sqrt5)$. $\cos(R/r)=\cos(\text{this value})=\cos(\pi/4)=1/\sqrt2$. Answer: (C)
Correct Answer: C