111.
Show that the three points A (1, –2, –8) , B (5. 0. –2) and C (11, 3. 7) are collinear and find the ratio in which B divides AC.
Given points are A (1, –2, –8), B (5, 0, –2), C (11, 3, 7).
Let O be origin.
![OA with rightwards arrow on top space equals space straight i with hat on top space minus space 2 space straight j with hat on top space minus space 8 space straight k with hat on top comma space space OB with rightwards arrow on top space equals space 5 space straight i with hat on top space minus space 2 space straight k with hat on top comma space space OC with rightwards arrow on top space equals space 11 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 7 space straight k with hat on top
AB with rightwards arrow on top space equals space OB with rightwards arrow on top space minus space OA with rightwards arrow on top space equals space left parenthesis 5 space straight i with hat on top space minus space 2 space straight k with hat on top right parenthesis space minus space left parenthesis straight i with hat on top space minus space 2 space straight j with hat on top space minus space 8 space straight k with hat on top right parenthesis space equals space 4 space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 6 space straight k with hat on top
AC with rightwards arrow on top space equals space OC with rightwards arrow on top space minus space OA with rightwards arrow on top space equals space left parenthesis 11 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 7 space straight k with hat on top right parenthesis space minus space left parenthesis straight i with hat on top space minus space 2 space straight j with hat on top space minus space 8 space straight k with hat on top right parenthesis
space space space space space space space equals space 10 space straight i with hat on top space plus space 5 space straight j with hat on top space plus space 15 space straight k with hat on top](/application/zrc/images/qvar/MAEN12066707.png)
![therefore space space space AC with rightwards arrow on top space equals space 5 over 2 left parenthesis 4 space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis space equals space 5 over 2 AB with rightwards arrow on top
therefore space space space AC with rightwards arrow on top space and space AB with rightwards arrow on top space are space parallel space vectors](/application/zrc/images/qvar/MAEN12066707-1.png)
But A is their common point.
∴ points A. B. C are collinear and B divides AC in the ratio 2 : 3.
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