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Find the vector equation of the line which is parallel to the vector ijk3i^-2j^+6k^ and which passes through the point (1, –2, 3). - Mathematics

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प्रश्न

Find the vector equation of the line which is parallel to the vector `3hat"i" - 2hat"j" + 6hat"k"` and which passes through the point (1, –2, 3).

योग

उत्तर

We know that the equation of line is `vec"r" = vec"a" + vec"b"lambda`

Here, `vec"a" = hat"i" - 2hat"j" + 3hat"k"` and `vec"b" = 3hat"i" - 2hat"j" + 6hat"k"`

∴ Equation of line is `vec"r" = (hat"i" - 2hat"j" + 3hat"k") + lambda(3hat"i" - 2hat"j" + 6hat"k")`

⇒ `(xhat"i" + yhat"j" + zhat"k") = (hat"i" - 2hat"j" + 3hat"k") + lambda(3hat"i" - 2hat"j" + 6hat"k")`

⇒ `(xhat"i" + yhat"j" + zhat"k") - (hat"i" - 2hat"j" + 3hat"k") = lambda(3hat"i" - 2hat"j" + 6hat"k")`

⇒ `(x - 1)hat"i" + (y + 2)hat"j" + (z - 3)hat"k" = lambda(3hat"i" - 2hat"j" + 6hat"k")`

Hence, the required equation is

`(x - 1)hat"i" + (y + 2)hat"j" + (z - 3)hat"k" = lambda(3hat"i" - 2hat"j" + 6hat"k")`

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अध्याय 11: Three Dimensional Geometry - Exercise [पृष्ठ २३५]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 11 Three Dimensional Geometry
Exercise | Q 2 | पृष्ठ २३५

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