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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)`
Solution
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`
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