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A, B and C are three points such that \( \overrightarrow{AB} = 3a + 4b \) and \( \overrightarrow{AC} = 15a + 20b \) (a) Prove that A, B and C lie on a straight line - Edexcel - GCSE Maths - Question 17 - 2022 - Paper 1
Question 17
A, B and C are three points such that \( \overrightarrow{AB} = 3a + 4b \) and \( \overrightarrow{AC} = 15a + 20b \) (a) Prove that A, B and C lie on a straight... show full transcript
Worked Solution & Example Answer:A, B and C are three points such that \( \overrightarrow{AB} = 3a + 4b \) and \( \overrightarrow{AC} = 15a + 20b \) (a) Prove that A, B and C lie on a straight line - Edexcel - GCSE Maths - Question 17 - 2022 - Paper 1
Step 1
(a) Prove that A, B and C lie on a straight line.
Answer
To prove that points A, B, and C lie on a straight line, we can show that the vectors ( \overrightarrow{AB} ) and ( \overrightarrow{AC} ) are scalar multiples of each other.
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First, we find the vector ( \overrightarrow{BC} ):
[ \overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB} = (15a + 20b) - (3a + 4b) = 12a + 16b ]
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Next, we can express ( \overrightarrow{AB} ) in terms of ( \overrightarrow{AC} ):
[ \overrightarrow{AC} = k \cdot \overrightarrow{AB} ] where ( k ) is a scalar.
Dividing the two expressions, we have:
[ \frac{\overrightarrow{AC}}{\overrightarrow{AB}} = \frac{15a + 20b}{3a + 4b} = 5 ]
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Thus, we can see that ( \overrightarrow{AC} = 5 \cdot \overrightarrow{AB} ), demonstrating that A, B, and C are collinear.
Step 2
(b) Find the ratio length of DF : length of DE
Answer
We need to find the relationship between the lengths of ( \overrightarrow{DF} ) and ( \overrightarrow{DE} ).
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First, let's express these lengths using the given vectors:
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From ( \overrightarrow{DE} = 3e + 6f ), we can find its length: [ |\overrightarrow{DE}| = |3e + 6f| = \sqrt{(3^2 + 6^2)} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} ]
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From ( \overrightarrow{EF} = -10.5e - 21f ): [ |\overrightarrow{DF}| = |\overrightarrow{DE} + \overrightarrow{EF}| = |(3e + 6f) + (-10.5e - 21f)| ]
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Simplifying gives: [ |\overrightarrow{DF}| = |(-7.5e - 15f)| = 7.5\sqrt{1^2 + 2^2} = 7.5\sqrt{5} ]
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Now we can find the ratio:
[ \text{Ratio} = \frac{|\overrightarrow{DF}|}{|\overrightarrow{DE}|} = \frac{7.5\sqrt{5}}{3\sqrt{5}} = \frac{7.5}{3} = \frac{5}{2} ]
Thus, the ratio of the lengths is ( 5 : 2 ).