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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`
Solution
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.
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