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Worked solution

Prove collinearity from two proportional displacements

O,A,B,CO,A,B,C are four distinct points with OA→=10a\overrightarrow{OA}=10\mathbf a, OB→=5b\overrightarrow{OB}=5\mathbf b and OC→=4a+3b\overrightarrow{OC}=4\mathbf a+3\mathbf b. Show that A,B,CA,B,C are collinear.

CJC · Vectors Basic · Example 3

Answer

AC→=35AB→\overrightarrow{AC}=\tfrac35\overrightarrow{AB}; the common point is A, so A,B,C are collinear.

Compare your reasoning

Find AB from the position vectors

AB→=OB→−OA→=5b−10a=5(b−2a)\begin{aligned}&\overrightarrow{AB}\\ & =\overrightarrow{OB}-\overrightarrow{OA}\\ & =5\mathbf b-10\mathbf a\\ & =5(\mathbf b-2\mathbf a)\end{aligned}

Subtract the initial position from the terminal position.

Find and factor AC

AC→=OC→−OA→=4a+3b−10a=3(b−2a)=35[5(b−2a)]=35AB→.\begin{aligned}\overrightarrow{AC}&=\overrightarrow{OC}-\overrightarrow{OA}\\&=4\mathbf a+3\mathbf b-10\mathbf a\\&=3(\mathbf b-2\mathbf a)\\&=\frac35[5(\mathbf b-2\mathbf a)]\\&=\frac35\overrightarrow{AB}.\end{aligned}

Keep the full source substitution chain to expose the same direction.

Use the common endpoint

AC∥AB⟹A,B,C are collinear.AC\parallel AB\quad\Longrightarrow\quad A,B,C\text{ are collinear.}

A,B,C are distinct, so AB is nonzero. The two parallel segments have A in common, which gives one straight line.

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