Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 06 (e)/Solution 2

The answer will be that (0, 0, -1) is outside of T. There are many different possible ways to show this, but they all take a correct formulation of a decisive criterion and then patience and carefulness to work till the end.

For this solution, let's explore some methods using basic geometry. We will hardly use the concept of vectors, and so for the subsequent paragraphs, let us denote $A=(0,1,2),$ $B=(-1,2,2),$ $C=(1,1,5)$, and $D=(0,0,-1)$ for brevity.

This are a few ways to formulate a criterion to determine whether D = (0, 0, -1) is inside or outside T. Here we suggest 2 possible criteria.

In both cases, we consider the triangles $\triangle ABD,\triangle BCD,\triangle CAD$. There are two qualitatively different pictures for D inside or outside T.

Criterion 1 - Comparing Lengths

If at least one of the lengths AD, BD or CD is strictly greater than the largest of AB, BC, CA, then $D$ is outside $T$.

Conversely if D is inside T, then all the lengths AD, BD or CD is less than or equal to the largest of AB, BC, CA.

Now, we calculate:

$AB=|(-1,1,0)|={\sqrt {2}}$

$BC=|(2,-1,3)|={\sqrt {14}}$

$CA=|(1,0,3)|={\sqrt {10}}$

But since

$DC=|(1,1,6)|={\sqrt {38}}$,

D is outside of T.

Criterion 2 - Compare Areas

If D is outside T, then the sum of areas of $\triangle ABD,\triangle BCD,\triangle CAD$ is strictly greater than that of $\triangle ABC$

Conversely if D is inside T, then equality holds instead.

We already knew that area of $T=\triangle ABC$ is ${\sqrt {19}}/2$. Let's calculate the area of $\triangle BCD$

${\begin{aligned}{\text{Area}}(\triangle BCD)&={\frac {1}{2}}\left|{\vec {BD}}\times {\vec {CD}}\right|\\&={\frac {1}{2}}\left|(-1,2,3)\times (1,1,6)\right|\\&={\frac {1}{2}}|(9,9,-3)|\\&={\frac {3{\sqrt {19}}}{2}}\end{aligned}}$

So, without calculating the other areas, we can be sure that D is outside of T.