2021HSCShort answer

Q16 (a) (ii)

3 marks

Question

Given vectors $\vec a=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ and $\vec b=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$ , show that
$|a_1b_1+a_2b_2+a_3b_3|\le\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}.$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Obtains $a_1b_1+a_2b_2+a_3b_3$	2
Attempts to apply the dot product, or equivalent merit	1

Sample answer:

$\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta$
$a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta$
$-1\le\cos\theta\le1$
$-|\mathbf a||\mathbf b|\le a_1b_1+a_2b_2+a_3b_3\le|\mathbf a||\mathbf b|$
$|a_1b_1+a_2b_2+a_3b_3|\le|\mathbf a||\mathbf b|$
$=\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}$

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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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