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OABC is a trapezium - Edexcel - GCSE Maths - Question 19 - 2021 - Paper 3

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OABC is a trapezium. \( \vec{OA} = \vec{a} \) \( \vec{AB} = \vec{b} \) \( \vec{OC} = 3\vec{b} \) D is the point on OB such that \( OD:DB = 2:3 \) E is the point on... show full transcript

Worked Solution & Example Answer:OABC is a trapezium - Edexcel - GCSE Maths - Question 19 - 2021 - Paper 3

Step 1

Work out the relationship involving D in terms of a and b

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Answer

Given the ratio ( OD:DB = 2:3 ) implies that point D divides line OB in that ratio. We can express the position of D as:

OD⃗=25OB⃗=25(a⃗+b⃗)\vec{OD} = \frac{2}{5} \vec{OB} = \frac{2}{5} (\vec{a} + \vec{b})

Step 2

Work out the relationship involving E in terms of a and b

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Answer

The point E divides line BC in the ratio ( BE:EC = 1:4 ). Therefore, we can express ( \vec{E} ) in terms of the vector from B to C:

E⃗=15BC⃗+B⃗=15(OC⃗−OB⃗)+(a⃗+b⃗)=15(3b⃗−(a⃗+b⃗))+(a⃗+b⃗)\vec{E} = \frac{1}{5} \vec{BC} + \vec{B} = \frac{1}{5} (\vec{OC} - \vec{OB}) + (\vec{a} + \vec{b}) = \frac{1}{5} (3\vec{b} - (\vec{a} + \vec{b})) + (\vec{a} + \vec{b})

Step 3

Calculate \( \vec{DE} \)

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Answer

The vector ( \vec{DE} ) can be calculated as:

DE⃗=E⃗−D⃗\vec{DE} = \vec{E} - \vec{D} Substituting our previously found expressions for ( \vec{D} ) and ( \vec{E} ) gives us:

ight) - \left( \frac{2}{5} (\vec{a} + \vec{b}) \right) $$ This can be simplified to get \( \vec{DE} \) in its simplest form.

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