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Question

# Show that the 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.

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Solution

## Given points $A\left(1,-2,-8\right),B\left(5,0,-2\right),C\left(11,3,7\right)$. Therefore, $\stackrel{\to }{AB}=5\stackrel{^}{i}+0\stackrel{^}{j}-2\stackrel{^}{k}-\stackrel{^}{i}+2\stackrel{^}{j}+8\stackrel{^}{k}=4\stackrel{^}{i}+2\stackrel{^}{j}+6\stackrel{^}{k}$ $\stackrel{\to }{BC}=11\stackrel{^}{i}+3\stackrel{^}{j}+7\stackrel{^}{k}-5\stackrel{^}{i}+2\stackrel{^}{k}=6\stackrel{^}{i}+3\stackrel{^}{j}+9\stackrel{^}{k}$ and, $\stackrel{\to }{AC}=11\stackrel{^}{i}+3\stackrel{^}{j}+7\stackrel{^}{k}-\stackrel{^}{i}+2\stackrel{^}{j}+8\stackrel{^}{k}=10\stackrel{^}{i}+5\stackrel{^}{j}+15\stackrel{^}{k}$ Clearly, $\stackrel{\to }{AB}+\stackrel{\to }{BC}=\stackrel{\to }{AC}$ Hence $A,B,C$ are collinear. Suppose $B$ divides in the ratio AC in the ratio $\lambda :1$. Then the position vector $B$ is $\left(\frac{11\lambda +1}{\lambda +1}\right)\stackrel{^}{i}+\left(\frac{3\lambda -2}{\lambda +1}\right)\stackrel{^}{j}+\left(\frac{7\lambda -8}{\lambda +1}\right)\stackrel{^}{k}$ But the position vector of $B$ is $5\stackrel{^}{i}+0\stackrel{^}{j}-2\stackrel{^}{k}.$ $\frac{11\lambda +1}{\lambda +1}=5,\frac{3\lambda -2}{\lambda +1}=0,\frac{7\lambda -8}{\lambda +1}=-2\phantom{\rule{0ex}{0ex}}⇒11\lambda +1=5\lambda +5,3\lambda -2=0,7\lambda -8=-2\lambda -2\phantom{\rule{0ex}{0ex}}⇒6\lambda =4,3\lambda =2,9\lambda =6\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{2}{3},\lambda =\frac{2}{3},\lambda =\frac{2}{3}$

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