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8.6.2 Vector Short Questions (Question 4 & 5)


Question 4:
Diagram below shows a parallelogram ABCD with BED as a straight line.


Given that AB=7p˜, AD=5q˜ and DE=3EB, express, in terms of p˜ and q˜.(a) BD(b) EC

Solution:

(a)
Note: for parallelogram,AB=DC=7p˜,AD=BC=5q˜.BD=BA+ADBD=7p˜+5q˜  


(b)

DE=3EB EB DE=13EB:DE=1:3EB=14DB=14(BD)=14[(7p˜+5q˜ )]From (a)=74p˜54q˜

EC=EB+BCEC=74p˜54q˜+5q˜EC=74p˜+154q˜


Question 5:

Use the above information to find the values of h and k when r = 2p – 3q.

Solution:
r=2p3q(h1)a˜+(h+k)b˜=2(5a˜7b˜)3(2a˜+3b˜)(h1)a˜+(h+k)b˜=10a˜14b˜+6a˜9b˜(h1)a˜+(h+k)b˜=16a˜23b˜Comparing vector:h1=16h=17h+k=2317+k=23k=40

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