# 2014 STEP 3 Q7

for 7ii - I used the method that since Q lies on both lines $P_1P_3$ and $P_2P_4$

$q= p_1 + t(p_1 - p_3)$
$q=p_2 + s(p_4 - p_2)$

hence equating the above two we find that
$p_1 (1-t) + p_2 ( s-1) + tp_3 - sp_4 = 0$

since there must exist real numbers t and s such that the above holds and as the sum of coefficients on the vectors is 0 it implies that
$a_1 = 1-t$
$a_2 = s-1$
$a_3 = t$
$a_4 = -s$

for 7iii for the first part can't you just say that, using the above values, $a_1 + a_3 = 1-t + t = 1$ hence it is not equal to 0
also I do not understand the next expression given for $q$ - using the values for the constants above and subbing them into the first equation for q I find that $q=a_1 p_1 + a_3 p_3$ - why do I have to divide by $a_1 + a_3$ ?

### In part (iii) you need to

In part (iii) you need to consider any numbers $a_i$ which satisfy (*). You showed existence by explicitly constructing $a_i$ which satisfy (*), but unless you've shown uniqueness also, you can't use these values to demonstrate general results about $a_i$.

### A quick example

Just to emphasise what's been said about uniqueness, it's obvious there can't be a unique solution: if $a_1,...a_4$ satisfy the equations, it's obvious that so do $ka_1,...ka_4$ for any constant $k$. This also hints at where the $a_1+a_3$ factor comes from: we need to divide by something to make the factors of $k$ disappear.

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