2021HSCShort answer

Q16 (a) (i)

2 marks

Question

The point $P(x,y,z)$ lies on the sphere of radius $1$ centred at the origin $O$ . Using $\overrightarrow{OP}=x\vec i+y\vec j+z\vec k$ and the triangle inequality, or otherwise, show that
$|x|+|y|+|z|\ge1.$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	2
States that $\overrightarrow{OP}$ is a unit vector, or equivalent merit	1

Sample answer:

Point $P$ is on the unit sphere, so

$1=|\overrightarrow{OP}|=|x\mathbf i+y\mathbf j+z\mathbf k|\le|x\mathbf i|+|y\mathbf j+z\mathbf k|\le|x\mathbf i|+|y\mathbf j|+|z\mathbf k|=|x|+|y|+|z|$

$\therefore |x|+|y|+|z|\ge1$

Answers could include:

$x^2+y^2+z^2=1$
$|x|^2+|y|^2+|z|^2=1$
$|x|^2\le|x|,\quad |y|^2\le|y|,\quad |z|^2\le|z|$
$1\le|x|+|y|+|z|$

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Syllabus alignment

ProofVectors

The Nature of ProofUse formal proof language, contradiction, counter-examples and rigorous arguments involving numbers and inequalities.
Further operations with three-dimensional vectorsUse vector magnitude, unit vectors and scalar products to solve problems and prove geometric results.

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