7 a-c- divides the join of points given by the position vectors a-2 b-3 c- and -2 a-3 b-5 c- in the ratio

Let A⃗=a⃗+2b⃗+3c⃗ B⃗= 2 a⃗+3 b⃗+5 c⃗ C⃗=7 a⃗ 2 c⃗ Let us assume that C⃗ divides the line segment A⃗B⃗ in the ratio x:1, then the position vector of C wi ...

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Standard XII

Mathematics

Section Formula

7 a⃗-c⃗ divid...

Question

$7 \to a - \to c$ divides the join of points given by the position vectors $\to a + 2 \to b + 3 \to c$ and $- 2 \to a + 3 \to b + 5 \to c$ in the ratio

2:3, internally

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3:1, externally

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3:2, externally

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Does not divide at all

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Solution

The correct option is C 3:2, externally
Let $\to A = \to a + 2 \to b + 3 \to c$
$\to B = - 2 \to a + 3 \to b + 5 \to c \to C = 7 \to a - 2 \to c$
Let us assume that $\to C$ divides the line segment $\to A B$ in the ratio x:1, then the position vector of C will be
$\to r = \frac{x \to B + \to A}{x + 1} 7 \to a - \to c = \frac{x (- 2 \to a + 3 \to b + 5 \to c) + (\to a + 2 \to b + 3 \to c)}{x + 1} o r 7 (x + 1) \to a - (x + 1) \to c = (- 2 x + 1) \to a + (3 x + 2) \to b + (5 x + 3) \to c o r (9 x + 6) \to a - (3 x + 2) \to b - (6 x + 4) \to c = \to 0 \Rightarrow 9 x + 6 = 0 \dots \dots (1) 3 x + 2 = 0 \dots \dots (2) 6 x + 4 = 0 \dots \dots (3)$
From (1), (2) and (3) we get a unique value of x which is $\frac{- 2}{3}$ .
Minus sign indicates external division. Also it can be written as $\frac{3}{2}$ external division if instead of $\frac{A C}{B C}$ we take $\frac{B C}{A C}$ .

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7→a−→c divides the join of points given by the position vectors →a+2→b+3→c and −2→a+3→b+5→c in the ratio

$7 \to a - \to c$ divides the join of points given by the position vectors $\to a + 2 \to b + 3 \to c$ and $- 2 \to a + 3 \to b + 5 \to c$ in the ratio