2022TrialShort answer

Q13 (d)

3 marks

Question

The points $X$ , $Y$ and $Z$ bisect $AB$ , $BC$ and $CA$ respectively. Show that
$\overrightarrow{AY}+\overrightarrow{BZ}+\overrightarrow{CX}=\vec0.$

Try the question before revealing the answer.

Solution

\overrightarrow{CA}+\overrightarrow{AB}+\overrightarrow{BC}=0

2\times\overrightarrow{CZ}+2\times\overrightarrow{AX}+2\times\overrightarrow{BY}=0

2\times\left(\overrightarrow{CZ}+\overrightarrow{AX}+\overrightarrow{BY}\right)=0

\overrightarrow{CZ}+\overrightarrow{AX}+\overrightarrow{BY}=0

\overrightarrow{AY}=\overrightarrow{CY}-\overrightarrow{CA}

\overrightarrow{BZ}=\overrightarrow{AZ}-\overrightarrow{AB}

\overrightarrow{CX}=\overrightarrow{BX}-\overrightarrow{BC}

By adding:

\overrightarrow{AY}+\overrightarrow{BZ}+\overrightarrow{CX} =(\overrightarrow{CY}-\overrightarrow{CA})+(\overrightarrow{AZ}-\overrightarrow{AB})+(\overrightarrow{BX}-\overrightarrow{BC})

=\overrightarrow{CY}+\overrightarrow{AZ}+\overrightarrow{BX}-(\overrightarrow{CA}+\overrightarrow{AB}+\overrightarrow{BC})

=0+0

=0

Alternative Solution:

\overrightarrow{AY}=\overrightarrow{AC}+\frac12\overrightarrow{CB}

\overrightarrow{BZ}=\overrightarrow{BA}+\frac12\overrightarrow{AC}

\overrightarrow{CX}=\overrightarrow{CB}+\frac12\overrightarrow{BA}

\overrightarrow{AY}+\overrightarrow{BZ}+\overrightarrow{CX} =\overrightarrow{AC}+\overrightarrow{CB}+\overrightarrow{BA}+\frac12\overrightarrow{AC}+\frac12\overrightarrow{CB}+\frac12\overrightarrow{BA}

=\frac32(\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{AC})

=\frac32(0)

=0

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Syllabus alignment

Vectors

Introduction to three-dimensional vectorsRepresent and operate with vectors in three dimensions using unit, component, ordered-triple and column-vector forms.

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