Given ` vecalpha=3 hat i+ hat j+2 hat ka n d vecbeta= hat i-2 hat j-4 hat k` are the position vectors of the points `Aa n dBdot` Then the distance of the point ` hat i+ hat j+ hat k` from the plane passing through `B` and perpendicular to `A B` is a. `5` b. `10` c. `15` d. `20`

Options

A. 5
B. 10
C. 15
D. 20

Answer

a

Explanation

`vec(AB)=vecbeta-vecalpha= -2hati-3hatj-6hatk`
Equation of the plane passing through B and perpendicular to AB is
`" "(verr-vec(OB))*vec(AB)=0`
`vecr*(2hati+3hatj+6hatk)+ 28=0`
Hence, the required distance from
`" "vecr=-hati+hatj+hatk`
`" "=|((-hati+hatj+hatk)*(2hati+3hatj+6hatk)+28)/(|2hati+3hatj+ 6hatk|)|`
`" "= |(-2+3+6+ 28)/(7)|`
`" "=5` units