The distance of the point of intersection of lines 1 x − 4 = − 4 y + 3 = 7 z + 1 and 2 x − 1 = − 3 y + 1 = 8 z + 10 from (7,-4, 7) is

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Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Mathematics3D-Coordinate GeometryLine ( Equation etc.)Medium2 minQB

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The distance of the point of intersection of lines $\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7}$ and $\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}$ from (7,-4, 7) is

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Solution:

The point of intersection is (5, -7,6) (parameter 1 on the first line and 2 on the second). Its distance from (7, -4, 7) is $4 + 9 + 1 = 14$ .

Stream:JEESubject:MathematicsTopic:3D-Coordinate GeometrySubtopic:Line ( Equation etc.)

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