Question

Find the vector equation of the plane passing through three points with position vectors ` hati+hatj-2hatk , 2hati-hatj+hatk and hati+2hatj+hatk` . Also find the coordinates of the point of intersection of this plane and the line `vecr=3hati-hatj-hatk lambda +(2hati-2hatj+hatk)`

 

Answer 1

Let the position vectors of the three points be `veca= hati+hatj-2hatk , vecb=2hati-hatj+hatk and vecc=hati+2hatj+hatk`

So, the equation of the plane passing through the points `vec a,vecb and vecc` is

`(vecr-veca).[(vecb-vecc)xx(vecc-veca)]=0`

`[vecr-(hati+hatj-2hatk)][(hati-3hatj)xx(hatj+3hatk)]=0`

`[vecr-(hati+hatj-2hatk)](hatk-3hatj-9hati)=0`

`vecr(-9hati-3hatj+hatk)-(-9-3-2)=0`

`vecr(-9hati-3hatj+hatk)+14=0`

`vecr(9hati+3hatj-hatk)=14  ............(1)`

So, the vector equation of the plane is `vecr(9hati+3hatj-hatk)=14`

The equation of the given line is `vecr=3hati-hatj-hatk lambda +(2hati-2hatj+hatk)`

Position vector of any point on the give line is

`vecr=(3+2lambda)hati+(-1-2lambda)hatj+(-1+lambda)hatk ...................(2)`

the point(2) lies on plane (1) if ,

`[(3+2lambda)hati+(-1-2lambda)hatj+(-1+lambda)hatk](9hati+3hatj-hatk)=14`

`9(3+2lambda)+3(-1-2lambda)-(-1+lambda)=14`

`11 lambda+25=14`

`lambda=-1`

Putting `lambda=-1` in (2) we have

`vecr=(3-2)hati+(-1+2)hatj+(-1-1)hatk`

`=hati+hatj-2hatk`

Thus, the point of intersection of the given line and the plane (1) is `hati+hatj-2hatk`