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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.
Solution
Given points \[A\left( 1, - 2, - 8 \right), B\left( 5, 0, - 2 \right), C\left( 11, 3, 7 \right)\]
Therefore,
\[\overrightarrow{AB} = 5 \hat{i} + 0 \hat{j} - 2 \hat{k} - \hat{i} + 2 \hat{j} + 8 \hat{k} = 4 \hat{i} + 2 \hat{j} + 6 \hat{k}\]
\[\overrightarrow{BC} = 11 \hat{i} + 3 \hat{j} + 7 \hat{k} - 5 \hat{i} + 2 \hat{k} = 6 \hat{i} + 3 \hat{j} + 9 \hat{k}\]
and, \[\overrightarrow{AC} = 11 \hat{i} + 3 \hat{j} + 7 \hat{k} - \hat{i} + 2 \hat{j} + 8 \hat{k} = 10 \hat{i} + 5 \hat{j} + 15 \hat{k}\]
Clearly, \[\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\]
Hence A, B, C are collinear.
Suppose B divides in the ratio AC in the ratio \[\lambda: 1\]. Then the position vector B is \[\left( \frac{11\lambda + 1}{\lambda + 1} \right) \hat{i} + \left( \frac{3\lambda - 2}{\lambda + 1} \right) \hat{j} + \left( \frac{7\lambda - 8}{\lambda + 1} \right) \hat{k}\]
But the position vector of B is \[5 \hat{i} + 0 \hat{j} - 2 \hat{k} .\]
\[\frac{11\lambda + 1}{\lambda + 1} = 5, \frac{3\lambda - 2}{\lambda + 1} = 0 , \frac{7\lambda - 8}{\lambda + 1} = - 2\]
\[ \Rightarrow 11\lambda + 1 = 5\lambda + 5, 3\lambda - 2 = 0, 7\lambda - 8 = - 2\lambda - 2\]
\[ \Rightarrow 6\lambda = 4, 3\lambda = 2, 9\lambda = 6\]
\[ \Rightarrow \lambda = \frac{2}{3}, \lambda = \frac{2}{3}, \lambda = \frac{2}{3}\]
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