Question #342837

Three points with position vectors, b and c are said to be colinear. If the parallelogram with adjacent sides a - b and a - c has zero geometry area. Use this fact to check whether or not the following triples of points are collinear


(a) (2,2,3), (6,1,5) (2,4,3)


(b) (2,3,3), (3,7,5), (0,-5,-1)


(c) (1,3,2), (4,2,1), (1,0,2)


1
Expert's answer
2022-05-20T08:14:39-0400

(a)


ab=(4,1,2)\vec{a}-\vec{b}=(-4, 1, -2)

ac=(0,2,0)\vec{a}-\vec{c}=(0, -2, 0)

(ab)×(ac)=ijk412020(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -4 & 1 & -2 \\ 0 & -2 & 0 \end{vmatrix}

=2ik42=4i+8k=2\begin{vmatrix} \vec{i} & \vec{k} \\ -4 & -2 \end{vmatrix}=-4\vec{i}+8\vec{k}

(ab)×(ac)=(4)2+(8)2=450\big|(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})\big|=\sqrt{(-4)^2+(8)^2}=4\sqrt{5}\not=0

Three points are not collinear.


(b)


ab=(1,4,2)\vec{a}-\vec{b}=(-1, -4, -2)

ac=(2,8,4)\vec{a}-\vec{c}=(2, 8, 4)

(ab)×(ac)=ijk142284(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & -4 & -2 \\ 2 & 8 & 4 \end{vmatrix}

=2jk428ik12+4ij14=2\begin{vmatrix} \vec{j} & \vec{k} \\ -4 & -2 \end{vmatrix}-8\begin{vmatrix} \vec{i} & \vec{k} \\ -1 & -2 \end{vmatrix}+4\begin{vmatrix} \vec{i} & \vec{j} \\ -1 & -4 \end{vmatrix}


=4j+8k+16i8k16i+4j=0=-4\vec{j}+8\vec{k}+16\vec{i}-8\vec{k}-16\vec{i}+4\vec{j}=\vec{0}


(ab)×(ac)=(4)2+(8)2=0\big|(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})\big|=\sqrt{(-4)^2+(8)^2}=0

Three points are collinear.


(c)


ab=(3,1,1)\vec{a}-\vec{b}=(-3, 1, 1)

ac=(0,3,0)\vec{a}-\vec{c}=(0, -3, 0)

(ab)×(ac)=ijk311030(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -3 & 1 &1 \\ 0 & -3 & 0 \end{vmatrix}

=3ik31=3i9k=-3\begin{vmatrix} \vec{i} & \vec{k} \\ -3 & 1 \end{vmatrix}=-3\vec{i}-9\vec{k}




(ab)×(ac)=(3)2+(9)2\big|(\vec{a}-\vec{b})\times (\vec{a}-\vec{c})\big|=\sqrt{(-3)^2+(-9)^2}

=3100=3\sqrt{10}\not=0

Three points are not collinear.



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