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Question
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the yz - plane .
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Solution
\[ \text{ The equation of the line through the points (5, 1, 6) and (3, 4, 1) is } \]
\[\frac{x - 5}{3 - 5} = \frac{y - 1}{4 - 1} = \frac{z - 6}{1 - 6}\]
\[ \Rightarrow \frac{x - 5}{- 2} = \frac{y - 1}{3} = \frac{z - 6}{- 5}\]
\[\text{ The coordinates of any point on this line are of the form} \]
\[\frac{x - 5}{- 2} = \frac{y - 1}{3} = \frac{z - 6}{- 5} = \lambda\]
\[ \Rightarrow x = - 2\lambda + 5; y = 3\lambda + 1; z = - 5\lambda + 6\]
\[\text{ So, the coordinates of the point on the given line } are\left( - 2\lambda + 5, 3\lambda + 1, - 5\lambda + 6 \right).\]
\[\text{ Since this point lies on the YZ- plane,} \]
\[x = 0\]
\[ \Rightarrow - 2\lambda + 5 = 0\]
\[ \Rightarrow \lambda = \frac{5}{2}\]
\[\text{ So, the coordinates of the point are} \]
\[\left( - 2\lambda + 5, 3\lambda + 1, - 5\lambda + 6 \right)\]
\[ = \left( - 2 \left( \frac{5}{2} \right) + 5, 3 \left( \frac{5}{2} \right) + 1, - 5 \left( \frac{5}{2} \right) + 6 \right)\]
\[ = \left( 0, \frac{17}{2}, \frac{13}{2} \right)\]
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