Answer to Question #342837 in Analytic Geometry for guy

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)


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

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

"(\\vec{a}-\\vec{b})\\times (\\vec{a}-\\vec{c})=\\begin{vmatrix}\n \\vec{i} & \\vec{j} & \\vec{k} \\\\\n -4 & 1 & -2 \\\\\n0 & -2 & 0\n\\end{vmatrix}"

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

"\\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)


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

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

"(\\vec{a}-\\vec{b})\\times (\\vec{a}-\\vec{c})=\\begin{vmatrix}\n \\vec{i} & \\vec{j} & \\vec{k} \\\\\n -1 & -4 & -2 \\\\\n2 & 8 & 4\n\\end{vmatrix}"

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


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


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

Three points are collinear.


(c)


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

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

"(\\vec{a}-\\vec{b})\\times (\\vec{a}-\\vec{c})=\\begin{vmatrix}\n \\vec{i} & \\vec{j} & \\vec{k} \\\\\n -3 & 1 &1 \\\\\n0 & -3 & 0\n\\end{vmatrix}"

"=-3\\begin{vmatrix}\n \\vec{i} & \\vec{k} \\\\\n -3 & 1\n\\end{vmatrix}=-3\\vec{i}-9\\vec{k}"




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

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

Three points are not collinear.



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