Question

Find the vector and cartesian equations of the planes that passes through (1, 0, – 2) and the normal to the plane is `hati + hatj - hatk`

पर्याय

• `vecr * (hati + hatj - hatk) - 3` = 0

• `hati + hatj - hatk` = – 1

• `vecr * (hati - 2hatj + hatk) + 1` = 0

• `(hati - 2hatj + hatk - 1)` = 0

Answer 1

`vecr * (hati + hatj - hatk) - 3` = 0

Explanation:

Equation of the plane passing through `veca` and normal to `vecn` is

`(vecr - veca) * vecn` = 0

`veca = (1, 0, -2) = hati - 2hatk`

`vecn = hati + hatj - hatk`

∴ Equation of the plane required is `[vecr - (hati - 2hatk)] * (hati + hatj - hatk)` = 0

`vecr * (hati + hatj - hatk) - (hati + 2hatk) * (hati + hatj - hatk)` = 0

or `vecr * (hati + hatj - hatk) - (1 + 2)` = 0

or `vecr * (hati + hatj - hatk) 0 - 3` = 0