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Find the vector and cartesian equation of plane which passes through the point `(1,3,-2)` and normal to the vector `(2hati+hatj-3hatk)` .

A

`2x+y+3z=11`

B

`2x+y-3z=11`

C

`2x-y-3z=11`

D

`2x-y+3z=11`

Text Solution

Verified by Experts

The correct Answer is:
B
Here `veca=hati+3hatj-2hatk` and `vecn=2hati+hatj-3hatk`
Vector equation of plane
` (vecr-veca).vecn=0`
`rArr vecr.vecn-veca.vecn=0`
`rArr vecr.(2hati+hatj-3hatk)=(hati+3hatj+2hatk) .(2hati+hatj-3hatk)`
`= 2+3+6`
`rArr vecr.(2hati+hatj-3hatk) = 11`.
For cartesian equation `vecr = xhati+yhatj+zhatk`
`:. (xhati+yhatj+zhatk).(2hati+hatj-3hatk)=11`
`rArr 2x+y-3z=11`, which is the cartesian equation of the plane.
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