#### 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 = 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)`

Is there an error in this question or solution?

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.