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

MATH152 April 2010

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Question B 06 (e)
Consider the triangle T in three dimensions with vertices (0, 1, 2), (1, 1, 5) and (-1,2, 2). Consider also the plane P that contains T. Is (0, 0,-1) in the triangle T? Justify.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Part (d) already shows that (0, 0, -1) is in P which contains T. So (0, 0, -1) is in the same plane as the vertices of the triangle T. From here, two of the possible routes would be to: i) express every point in T by a parametrization to check whether (0, 0, -1) is in T ii) sketch both pictures when (0, 0, -1) is inside or outside T, then deduce a geometric criterion from the picture For i), one needs to think of one of the vertices as a base point and walk to a point inside T by following the 2 directions toward the other two vertices. For ii), one can think of the paper being written on to be the plane P, then sketch the vertices of T as you would for points in 2 dimensional space. Do not assume any particular order of the vertices or relative positions. After the sketch, one needs to construct geometrical figures relating the point (0, 0, -1) to the vertices of the triangle T. For instance, one can consider triangles with this new point as a vertex.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The answer will be that (0, 0, 1) is outside of T. There can be many different ways to show this, and it takes a correct formulation of a decisive criterion and then patience and carefulness to work till the end. For the subsequent paragraphs, let us denote ${\vec {OA}}=(0,1,2),{\vec {OB}}=(-1,2,2),{\vec {OC}}=(1,1,5)$ for brevity. This solution explores the parametrization method. This method is somewhat a more sophisticated but it works beautifully. Starting from point A, we can travel to B and C by the vectors ${\vec {AB}}={\vec {OB}}-{\vec {OA}}=(-1,2,2)-(0,1,2)=(-1,1,0)$ and ${\vec {AC}}={\vec {OC}}-{\vec {OA}}=(1,1,5)-(0,1,2)=(1,0,3)$ respectively. Therefore, the expressions ${\vec {OA}}+t{\vec {AB}}$ and ${\vec {OA}}+s{\vec {AC}}$ with $0\leq s,t\leq 1$, represents all points from A to B and A to C respectively. It turns out we can combine them and write: ${\vec {OA}}+t{\vec {AB}}+s{\vec {AC}}=(0,1,2)+t(-1,1,0)+s(1,0,3)=(s-t,1+t,2+3s)$ with the extra requirement that $0\leq s+t\leq 1$ to describe all points in the triangle T. Now $(s-t,1+t,2+3s)=(0,0,-1)$ implies $s=t=-1$ Thus, it's not in T. Remark 1: (0, 0, -1) is not even a point in the parallelogram: $(s-t,1+t,2+3s)$ with $0\leq s,t\leq 1$) Remark 2: The fact that it was solvable for s and t shows that the point in the plane P. If it weren't in the plane, e.g. the point (0, 0, 1), it wouldn't be solvable and some inconsistencies in the linear equations will result.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

Solution 3

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. In this particular example there is an even faster way to see that $D$ must be outside of the triangle $T$, because $D$ is even outside the cuboid spanned by the three corners of $T$. Indeed, for all points $(x,y,z)$ in $T$ it must be true that ${\begin{aligned}\min(0,1,-1)&\leq x\leq \max(0,1,-1)\\\min(1,1,2)&\leq y\leq \max(1,1,2)\\\min(2,5,2)&\leq z\leq \max(2,5,2)\end{aligned}}$ However, since the $y$ and $z$ coordinates of $D$ do not satisfy the inequalities above, $D$ is outside the cuboid, and hence outside of $T$.