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Given vecalpha=3 hat i+ hat j+2 hat ka ...

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

Text Solution

Verified by Experts

The correct Answer is:
a

`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
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