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OABC is a trapezium - Edexcel - GCSE Maths - Question 19 - 2021 - Paper 3
Question 19
OABC is a trapezium. \(ar{OA} = a\) \(AB = b\) \(OC = 3b\) D is the point on OB such that OD:DB = 2:3 E is the point on BC such that BE:EC = 1:4 Work out the vec... show full transcript
Worked Solution & Example Answer:OABC is a trapezium - Edexcel - GCSE Maths - Question 19 - 2021 - Paper 3
Step 1
D is the point on OB such that OD:DB = 2:3
Answer
Given the ratio (OD:DB = 2:3), we can express the vectors in terms of a single variable.
Let (OD = 2k) and (DB = 3k) where (k) is a scalar. Therefore, we have:
[ OB = OD + DB = 2k + 3k = 5k ]
Thus, the vector (\overrightarrow{OD} = \frac{2}{5}\overrightarrow{OB} = \frac{2}{5}(a + b)] and (\overrightarrow{D} = \frac{2}{5}(a + b)).
Step 2
E is the point on BC such that BE:EC = 1:4
Answer
Similarly, for E, we have the ratio (BE:EC = 1:4). Let (BE = m) and (EC = 4m), which gives:
[ BC = BE + EC = m + 4m = 5m ]
The vector (\overrightarrow{BE} = \frac{1}{5}\overrightarrow{BC} = \frac{1}{5}(b + 3b) = \frac{1}{5}(4b) = \frac{4}{5}b ]
So, we can express (\overrightarrow{E} = \overrightarrow{B} + \overrightarrow{BE} = (b + 3b) + \frac{1}{5}(4b) = 4b + \frac{4}{5}b = \frac{24}{5}b ].
Step 3
Work out the vector DE in terms of a and b
Answer
[\overrightarrow{DE} = \overrightarrow{E} - \overrightarrow{D} = \left(\frac{24}{5}b\right) - \left(\frac{2}{5}(a + b)\right) = \left(\frac{24}{5}b - \frac{2}{5}a - \frac{2}{5}b\right) = \left(-\frac{2}{5}a + \frac{22}{5}b\right)]
Thus, the final answer for (\overrightarrow{DE}) in its simplest form is:
[\overrightarrow{DE} = -\frac{2}{5}a + \frac{22}{5}b].