- Thread starter
- #1

- Feb 5, 2012

- 1,621

**Title: How do you do this problem?**

Hi mortifiedpenguin1,Please offer step by step solutions

There are three vectors. They are v= (3,-1,2), b= (4,2,-5) and n= (1,3,-7). Please prove that they form a closed triangle. What type of triangle is it?

Thanks!!!

I presume what you meant by a "closed triangle" is in fact to show that the given three points are non-collinear; so that a triangle is uniquely determined.

You can show that the points are not collinear by showing that the two vectors \(\overrightarrow{vb}\mbox{ and }\overrightarrow{vn}\) are not parallel. That is their cross product, \(\overrightarrow{vb}\times\overrightarrow {vn}\neq\underline{0}\)

\[\overrightarrow{vb}=\mathbf{b}-\mathbf{v}=(4,2,-5)-(3,-1,2)=(1,3,-7)\]

\[\overrightarrow {vn}=\mathbf{n}-\mathbf{v}=(1,3,-7)-(3,-1,2)=(-2,4,-9)\]

\[\Rightarrow\overrightarrow {vb}\times\overrightarrow {vn}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 3 & -7\\-2 & 4 & -9 \end{vmatrix}=(1,23,10)\neq\underline{0}\]

Therefore the three points are not collinear, and hence they determine a unique triangle.

We shall also find the vector, \(\overrightarrow {bn}\)

\[\overrightarrow {bn}=\mathbf{n}-\mathbf{b}=(1,3,-7)-(4,2,-5)=(-3,1,-2)\]

Consider the length of the vectors, \(\overrightarrow{vb},\, \overrightarrow {vn}\mbox{ and }\overrightarrow {bn}\).

\[|\overrightarrow{vb}|=\sqrt{1^2+3^2+7^2}=\sqrt{59}\]

\[|\overrightarrow {vn}|=\sqrt{2^2+4^2+9^2}=\sqrt{101}\]

\[|\overrightarrow {bn}|=\sqrt{3^2+1^2+2^2}=\sqrt{14}\]

Let \(\theta\) be the angle between the sides, \(\mbox{vb}\) and \(\mbox{bn}\). By the Cosine rule,

\[|\overrightarrow {vn}|^2=|\overrightarrow{vb}|^2+|\overrightarrow {bn}|^2-2|\overrightarrow{vb}||\overrightarrow {bn}|\cos\theta\]

\[\Rightarrow\cos\theta=-\frac{(101-59-14)}{2\sqrt{14}\sqrt{59}}\approx -0.4871\]

\[\Rightarrow\theta\approx 119.1^{0}\]

Therefore this is an Obtuse triangle.

Kind Regards,

Sudharaka.