O is the origin, OP = p and OQ = q. Z is a point on PQ such that PZ : ZQ = 5 : 2. How to work out, in terms of p and q, the position vector of Z?

Question

I don't understand why the fraction part of vector PZ= 5/7 (-p+q) is 5/7, shouldn't it be 2/7 as the route taken is PO + OQ + QZ?

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Answer 1 · 2018-02-16T09:55:18

see below

$P Z : Z Q : : 5 : 7$ $PZ:ZQ::5:7$

so $P Z$ $PZ" "$ is $5$ $5" "$ parts and $Z Q$ $" "ZQ" "$ is $2$ $2" "$ parts.

so $P Q$ $PQ" "$ is $5 + 2 = 7 parts altogether$ $" "5+2=7" parts altogether"$

$\Rightarrow P Z = \frac{5}{7} P Q$ $=>PZ=5/7PQ$

we want the position vector of $Z$ $Z" "$ which is $- - \to O Z$ $vec(OZ)$

$- - \to O Z = - - \to O P + - - \to P Z$ $vec(OZ)=vec(OP)+vec(PZ)$

$- - \to O Z = - - \to O P + \frac{5}{7} - - \to P Q$ $vec(OZ)=vec(OP)+5/7vec(PQ)$

$- - \to O Z = - - \to O P + \frac{5}{7} (- - \to P O + - - \to O Q)$ $vec(OZ)=vec(OP)+5/7(vec(PO)+vec(OQ))$

$- - \to O Z = \to p + \frac{5}{7} (- \to p + \to q)$ $vec(OZ)=vecp+5/7(-vecp+vecq)$

$- - \to O Z + \frac{2}{7} \to p + \frac{5}{7} \to q = \frac{1}{7} (2 \to p + 5 \to q)$ $vec(OZ)+2/7vecp+5/7vecq=1/7(2vecp+5vecq)$

on simplification

the route you have put down

$- - \to P O + - - \to O Q + - - \to Q Z$ $vec(PO)+vec(OQ)+vec(QZ)$

gives $- - \to P Z$ $vec(PZ)$

and not the position vector of $Z$ $Z " "$ which is $- - \to O Z$ $vec(OZ)$