Points A, B and C have position vectors a, b and c, with 2 a + 3 b - 5 c = 0. How does C divide AB?

Question 56

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Points $A$ , $B$ and $C$ have position vectors $\vec{a}$ , $\vec{b}$ and $\vec{c}$ , with $2\vec{a} + 3\vec{b} - 5\vec{c} = \vec{0}$ . How does $C$ divide $AB$ ?

Internally, in the ratio 2 : 3Correct answerYour answerExternally, in the ratio 3 : 2Correct answerYour answerInternally, in the ratio 3 : 2Correct answerYour answerExternally, in the ratio 2 : 3Correct answerYour answer

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CorrectNot quiteAnswer C

Correct answer: C

$\vec{c} = \frac{2\vec{a} + 3\vec{b}}{5} = \frac{3\vec{b} + 2\vec{a}}{3 + 2}$ .
This is $\frac{m\vec{b} + n\vec{a}}{m + n}$ with $m = 3$ , $n = 2$ : $C$ divides $AB$ internally with $AC : CB = 3 : 2$ .

Worth noting

2 : 3 pairs each weight with its own point; in $\frac{m\vec{b} + n\vec{a}}{m + n}$ the weight on $\vec{b}$ is the $AC$ part of the ratio.

Concept: Reading the ratio of a division
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Points AAA, BBB and CCC have position vectors a⃗\vec{a}a, b⃗\vec{b}b and c⃗\vec{c}c, with 2a⃗+3b⃗−5c⃗=0⃗2\vec{a} + 3\vec{b} - 5\vec{c} = \vec{0}2a+3b−5c=0. How does CCC divide ABABAB?

Explanation

Points $A$ , $B$ and $C$ have position vectors $\vec{a}$ , $\vec{b}$ and $\vec{c}$ , with $2\vec{a} + 3\vec{b} - 5\vec{c} = \vec{0}$ . How does $C$ divide $AB$ ?