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Find the equation of plane in Cartesian ...

Find the equation of plane in Cartesian form which is at a distance `3/sqrt(41)` from origin and its
normal vector from origin is `3hati-4hatj-4hatk`

Text Solution

Verified by Experts

We know that equation of plane in Cartesian from is
`lx+my+nz=d`
here `(l, m, n) -= (3/sqrt(41),(-4)/sqrt(41),(-4)/sqrt(41))`
`d=3/sqrt(41)`
so equation of plane in Cartesian form is given by
`3x- 4y- 4z = 3`
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