Sudharaka
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Title: How do you do this problem?
Hi mortifiedpenguin1, :)
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.
Hi mortifiedpenguin1, :)
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.