iGCSE Mathematics (0580) :7.4 Vector geometry.iGCSE Style Questions Paper 2

Question

S is a point on PQ such that PS : SQ = 4 : 5.

Find \(\overrightarrow{OS}\), in terms of a and b, in its simplest form.

▶️ Answer/Explanation

Solution

(5/9)a + (4/9)b

PS:SQ = 4:5 means S divides PQ in ratio 4:5

OS = OP + (4/9)PQ = a + (4/9)(b-a)

Simplify: a + (4/9)b – (4/9)a = (5/9)a + (4/9)b

Question

The diagram shows a trapezium OPQR.

O is the origin, \(\overrightarrow{OR}=a\) and \(\overrightarrow{OP}=b\).
\(\left | \overrightarrow{RQ} \right |=\frac{3}{5}|\overrightarrow{OP}|\)

(a) Find \(\overrightarrow{PQ}\) in terms of a and b in its simplest form.

(b) When PQ and OR are extended, they intersect at W. Find the position vector of W.

▶️ Answer/Explanation

Answer:

(a) \(a – \frac{2}{5}b\) (or equivalent simplified form)

(b) \(\frac{5}{2}a\) (or equivalent)

Explanation:

Part (a):

Given \(|\overrightarrow{RQ}| = \frac{3}{5}|\overrightarrow{OP}| = \frac{3}{5}|b|\)
Since OPQR is a trapezium, RQ is parallel to OP, so \(\overrightarrow{RQ} = \frac{3}{5}b\)
Using vector addition: \(\overrightarrow{PQ} = \overrightarrow{PR} + \overrightarrow{RQ} = (a – b) + \frac{3}{5}b = a – \frac{2}{5}b\)

Part (b):

When extended, PW and OW are scalar multiples: \(\overrightarrow{OW} = k\overrightarrow{OR} = ka\)
Also \(\overrightarrow{PW} = t\overrightarrow{PQ} = t(a – \frac{2}{5}b)\)
Equating expressions: \(ka = b + t(a – \frac{2}{5}b)\)
Solving gives \(t = \frac{5}{2}\), so position vector of W is \(\frac{5}{2}a\)

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