21M.1.AHL.TZ2.5

pestleMathematicsAAHLPaper 121M· ahl-3-13-scalar-(dot)-productsource ↗

Given any two non-zero vectors, $a$ and $b$ , show that ${|a \times b|}^{2} = {|a|}^{2} {|b|}^{2} - {(a \cdot b)}^{2}$ .

Markscheme / solution

METHOD 1

use of $|a \times b| = |a| |b| \sin θ$ on the LHS (M1)

${|a \times b|}^{2} = {|a|}^{2} {|b|}^{2} \sin^{2} θ$ A1

$= {|a|}^{2} {|b|}^{2} (1 - \cos^{2} θ)$ M1

$= {|a|}^{2} {|b|}^{2} - {|a|}^{2} {|b|}^{2} \cos^{2} θ$ OR $= {|a|}^{2} {|b|}^{2} - {(|a| |b| \cos θ)}^{2}$ A1

$= {|a|}^{2} {|b|}^{2} - {(a \cdot b)}^{2}$ AG

METHOD 2

use of $a \cdot b = |a| |b| \cos θ$ on the RHS (M1)

$= {|a|}^{2} {|b|}^{2} - {|a|}^{2} {|b|}^{2} \cos^{2} θ$ A1

$= {|a|}^{2} {|b|}^{2} (1 - \cos^{2} θ)$ M1

$= {|a|}^{2} {|b|}^{2} \sin^{2} θ$ OR $= {(|a| |b| \sin θ)}^{2}$ A1

$= {|a \times b|}^{2}$ AG

Note: If candidates attempt this question using cartesian vectors, e.g

$a = (\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix})$ and $b = (\begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix})$ ,

award full marks if fully developed solutions are seen.
Otherwise award no marks.

[4 marks]

Examiners’ report

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