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Find the Vector Equation of the Plane Passing Through (1, 2, 3) and Perpendicular to the Plane Vecr.(Hati + 2hatj -5hatk) + 9 = 0 - Mathematics

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प्रश्न

Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the plane `vecr.(hati + 2hatj -5hatk) + 9 = 0`

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उत्तर

The position vector of the point (1, 2, 3) is `vecr_1 = hati +2hatj + 3hatk`

The direction ratios of the normal to the plane, , `vecr.(hati+2hatj -5hatk)+9 = 0`are 1, 2, and −5 and the normal vector is `vecN = hati + 2hatj - 5hatk`

The equation of a line passing through a point and perpendicular to the given plane is given by, `vecl = vecr + lambdavecN` ,`lambda in R`

`=> hatl = (hati + 2hatj + 3hatk) + lambda(hati +  2hatj -5hatk)`

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अध्याय 11: Three Dimensional Geometry - Exercise 11.4 [पृष्ठ ४९८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 11 Three Dimensional Geometry
Exercise 11.4 | Q 7 | पृष्ठ ४९८

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