Find the Cartesian from the equation of the plane `vecr=(s-2t)hati+(3-

1.	Find the Cartesian from the equation of the plane `vecr=(s-2t)hati+(3-t)hatj+(2s+t)hatk`.
Answer» We have, `vecr=(s-2t)hati+(3-t)hatj+(2s+t)hatk` `ltimplies (xhati+yhatj+zhatk)=(s-2t)hati+(3-t)hatj+(2s+t)hatk` `ltimplies x-2y+6=1/2(y+z-3)` [equating the values of x] `ltimplies 2x-4y+12=y+z-3 rArr 2x-5y-z+15=0`. This is the required Cartesian form of the equation of the given plane.

Discussion

Find the equation of a line passing through the point `(2hati-3hatj-5hatk)` and perpendicular to the plane`vecr.(6hati-3hatj+5hatk) +2=0`. Also find the point of intersection of this line and the plane.
If the points `(1,1,p)` and `(-3,0,1)` be equidistant from the plane `vecr.(3hati+4hatj-12hatk)+13`, find the values of p.
Find the distance of the point (1,2,5) from the plane `vecr.(hati+hatj+hatk)+17=0`
Find the co-ordinates of the foot of perpendicular and the length of perpendicular drawn from the point `(2,3,7)` to the plane `3x-y-z=7`.
Find the length of the foot of the perpendicular from the point (1,1,2) to the plane `vecr.(2hati-2hatj+4hatk)+5=0`
Find the coordinates of the image of the point P`(1, 3, 4)` in the plane `2x-y+z+3=0`.
Find the equationof the plane passing through each group of points. i) A(2,2,-1), B(3,4,2) and C(7,0,6) ii) A(0,-1,-1), B(4,5,1) and C(3,9,4) iii) A(-2,6,-6), B(-3,10,-9) and C(-5,0,-6).
Find the coordinates of the foot of the perpendicular and the perpendicular distance from the point P(3,2,1) to the plane `2x-y+z+1=0`.
Find the distance of the point `(-1,-5,-10)` from the point of the intersection of the line `vecr = 2hati-2hatk+lambda(3hati+4hatj+2hatk)` and the plane `vecr.(hati-hatj+hatk) = 5`.
Find the distance of the point `(2hati-hatj-4hatk)` from the plane `vecr.(3hati-4hatj+12hatk)=9`.