#### 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)`