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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 - CBSE (Science) Class 12 - Mathematics

ConceptDirection Cosines and Direction Ratios of a Line

#### Question

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

#### Solution

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 = 0are 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)

Is there an error in this question or solution?

#### APPEARS IN

NCERT Solution for Mathematics Textbook for Class 12 (2018 to Current)
Chapter 11: Three Dimensional Geometry
Q: 7 | Page no. 498

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Solution Find the Vector Equation of the Plane Passing Through (1, 2, 3) and Perpendicular to the Plane Vecr.(Hati + 2hatj -5hatk) + 9 = 0 Concept: Direction Cosines and Direction Ratios of a Line.
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