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The distance from the point -hati+2hatj+...

The distance from the point `-hati+2hatj+6hatk` to the straight line through the point (2,3,-4) and parallel to the vector `6hati+3hatj-4hatk`, is

A

6

B

7

C

8

D

9

Text Solution

Verified by Experts

The correct Answer is:
B

From the figure,
`vec(AP) =-3hati-hatj+10hatk`
`|vec(AP)|=sqrt(100)`
`vec(PN)` is perpendicular to the line.
`vec(AN)` = The projection of `vec(AP)`, on the line, i.e., on `6hati+3hatj-4hatk`
`=|(vec(AP).(6hati+3hatj)-4hatk)/(6hati+3hatj-4hatk)|`
`=|(-18-3-40)/sqrt(60)|=sqrt(61)`
`therefore PN^(2)=AP^(2)-AN^(2)=110-61=49`
`therefore PN=7`
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