# Find the vector equation of the plane passing through three points with position vectors  i+j-2k , 2i-j+k and i+2j+k . Also find the coordinates of the point of intersection of this plane and the line r=3i-j-k λ+(2i-2j+k) - Mathematics

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

Concept: Vector and Cartesian Equation of a Plane
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