2010 A-level H2 Mathematics (9740) Paper 1 Question 1 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Since |a| = |b|

\Rightarrow \sqrt{(2p)^2+(3p)^2+(6p)^2} = \sqrt{1+2^2+2^2}

49p^2 = 9

p = \frac{3}{7} since p > 0

(ii)
(a+b) \bullet (a-b)

= a \bullet a + b \bullet a - a \bullet b - b \bullet b

= |a|^2 - |b|^2

= 0 since |a| = |b|

KS Comments:

A very straight forward question coming out of Vectors I. Students should be well versed with the expansions.

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