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Worked solution

Distinguish collinear points from a failed scalar relation

Given A(1,2,2)A(1,2,2), B(4,5,7)B(4,5,7), C(7,8,12)C(7,8,12) and D(1,2,4)D(1,2,4): (i) show that A,B,C are collinear; (ii) show that A,B,D are not collinear.

CJC · Vectors Basic · Example 4

(i) Collinear A,B,C

Answer

A,B,C are collinear.

Subtract positions and compare

AB→=(457)−(122)=(335),AC→=(7812)−(122)=(6610)=2(335)=2AB→.\begin{aligned}\overrightarrow{AB}&=\begin{pmatrix}4\\5\\7\end{pmatrix}-\begin{pmatrix}1\\2\\2\end{pmatrix}=\begin{pmatrix}3\\3\\5\end{pmatrix},\\\overrightarrow{AC}&=\begin{pmatrix}7\\8\\12\end{pmatrix}-\begin{pmatrix}1\\2\\2\end{pmatrix}=\begin{pmatrix}6\\6\\10\end{pmatrix}=2\begin{pmatrix}3\\3\\5\end{pmatrix}=2\overrightarrow{AB}.\end{aligned}

Both displacements start at A; AC is twice the nonzero AB.

Identify the common point

AC∥ABAC\parallel AB

Since A is common and the displacements are parallel, A,B,C are collinear.

How did this part go?

(ii) Non-collinear A,B,D

Answer

A,B,D are not collinear.

Find the second direction

AD→=(124)−(122)=(002)\overrightarrow{AD}=\begin{pmatrix}1\\2\\4\end{pmatrix}-\begin{pmatrix}1\\2\\2\end{pmatrix}=\begin{pmatrix}0\\0\\2\end{pmatrix}

Retain the source subtraction.

Rule out every possible scalar

(002)=k(335) ⟹ k=0 and 2=5k=0, a contradiction.\begin{pmatrix}0\\0\\2\end{pmatrix}=k\begin{pmatrix}3\\3\\5\end{pmatrix}\ \Longrightarrow\ k=0\text{ and }2=5k=0,\text{ a contradiction.}

If AD=kAB, the first coordinate forces k=0, while the third would give 2=0. Thus no real scalar works. The source says “not equal … for some k”; the required non-collinearity claim is that no such k exists.

How did this part go?

Solution notes

The original Teacher quantifier wording is retained in the native text. A single unsuccessful scalar is not a proof of non-collinearity; the component contradiction rules out all scalars.