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Section Formula

Three distric...

Question

Three district points A, B and C with p.v.s. and $\to a, \to b$ and $\to c$ respectively are collinear if there exist non-zero scalars x, y, z such that

A

x \to a + y \to b + z \to c = 0

and

x + y + z = 0

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B

x \to a + y \to b + z \to c

and

x + y + z \neq 0

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C

x \to a + y \to b + z \to c \neq 0

and

x + y + z = 0

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D

x \to a + y \to b + z \to c = 3

and

x + y + z \neq 0

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Solution

The correct option is A $x \to a + y \to b + z \to c = 0$ and $x + y + z = 0$
If three points are collinear, we can see that as one point dividing the line joining the other two points either internally or externally in some ratio.

So, $x \to a + y \to b = (x + y) \to c$ when $\to c$ divides the line joining $¯ ¯ ¯ a$ and $¯ ¯ b$ in the ratio $y : x$

This can also be written as $x \to a + y \to b - (x + y) \to c = 0$

If $z = - (x + y)$ , we get $x + y + z = 0$ and $x \to a + y \to b + z \to c = 0$

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