Question

Find the vector and Cartesian equation of the planes that passes through the point (1, 0, −2) and the normal to the plane is `hati + hatj - hatk`

Answer 1

The position vector of point (1, 0, −2) is `veca = hati - 2hatk`

The normal vector `vecN` perpendicular to the plane is `vecN = hati + hatj - hatk`

The vector equation of the plane is given by, `(vecr-veca).vecN = 0`

`=>[vecrr - (hati -2hatk)].(hati + hatj - hatk) = 0`

`vecr `is the position vector of any point P (x, y, z) in the plane.

`:. vecr = xhati + yhatj + zhatk`

Therefore, equation (1) becomes

This is the Cartesian equation of the required plane.