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Find the Coordinates of the Point Where the Line Through (5, 1, 6) and (3, 4, 1) Crosses The Zx - Plane . - Mathematics

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प्रश्न

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the  zx - plane .

योग

उत्तर

\[ \text{ The equation of the line through the points (5, 1, 6) and (3, 4, 1) is } \]

\[\frac{x - 5}{3 - 5} = \frac{y - 1}{4 - 1} = \frac{z - 6}{1 - 6}\]

\[ \Rightarrow \frac{x - 5}{- 2} = \frac{y - 1}{3} = \frac{z - 6}{- 5}\]

\[\text{ The coordinates of any point on this line are of the form } \]

\[\frac{x - 5}{- 2} = \frac{y - 1}{3} = \frac{z - 6}{- 5} = \lambda\]

\[ \Rightarrow x = - 2\lambda + 5; y = 3\lambda + 1; z = - 5\lambda + 6\]

\[\text{ So, the coordinates of the point on the given line are } \left( - 2\lambda + 5, 3\lambda + 1, - 5\lambda + 6 \right).\]

\[\text{ Since this point lies on the ZX- plane,} \]

\[y = 0\]

\[ \Rightarrow 3\lambda + 1 = 0\]

\[ \Rightarrow \lambda = \frac{- 1}{3}\]

\[\text{ So, the coordinates of the point are} \]

\[\left( - 2\lambda + 5, 3\lambda + 1, - 5\lambda + 6 \right)\]

\[ = \left( - 2 \left( \frac{- 1}{3} \right) + 5, 3 \left( \frac{- 1}{3} \right) + 1, - 5 \left( \frac{- 1}{3} \right) + 6 \right)\]

\[ = \left( \frac{17}{3}, 0, \frac{23}{3} \right)\]

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अध्याय 29: The Plane - Exercise 29.12 [पृष्ठ ६५]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
Exercise 29.12 | Q 1.2 | पृष्ठ ६५

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