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Question
A line passing through the point A with position vector `veca=4hati+2hatj+2hatk` is parallel to the vector `vecb=2hati+3hatj+6hatk` . Find the length of the perpendicular drawn on this line from a point P with vector `vecr_1=hati+2hatj+3hatk`
Solution
Let the equation of the line be `vecr=veca+lambda vecb`
Here ` veca=4hati+2hatj+2hatk and vecb=2hati+3hatj+6hatk`
equation of the line `=4hati+2hatj+2hatk+lambda (2hati+3hatj+6hatk)`
Let L be the foot of the perpendicular and P be the required point from which we have to find the length of the perpendicular
`P(vecalpha)=hati+2hatj+3hatk`
vec(PL)=position vector of L -position vector of P
`=4hati+2hatj+2hatk+lambda (2hati+3hatj+6hatk)-(hati+2hatj+3hatk)`
`=3hati-hatk+lambda(2hati+3hatj+6hatk).................(i)`
Now, `vec(PL).vecb=0[Since vec(PL)" is perpendicular to "vecb]`
`=3hati-hatk+lambda(2hati+3hatj+6hatk).(2hati+3hatj+6hatk)=0`
`=>(3+2lambda)2+(3lambda)3+(-1+6lambda)6=0`
`=>6+4lambda+9lambda-6+36lambda=0`
`=>49lambda=0`
`therefore lambda=0`
`vec(PL)=3hati-hatk ["from "(ii)]`
`|vec(PL)|=|sqrt(3^2+(-1)^2)|`
`therefore |vec(PL)|=sqrt(10)`
Length of the perpendicular drawn on the line from `P=sqrt(10)`
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