The equation of the plane passing through the point (-1, 2, 1) and perpendicular to the line joining the points (-3, 1, 2) and (2, 3, 4) is · · · · · ·

Options

A. `rdot(5hati+2hatj+2hatk)=1`
B. `rdot(5hati+2hatj+2hatk)=-1`
C. `rdot(5hati-2hatj+2hatk)=-5`
D. `rdot(5hati-2hatj-2hatk)=1`

Answer

A

Explanation

Direction ratios of line joining points (-3,1,2) and (2,3,4) are (2+3,3-1,4-2) i.e.,(5,2,2)
Since plane is perpendicular to the line joining the given points therefore direction ratio of normal to the plane are (5, 2, 2).
`therefore` Required equation of plane passing through the point (-1, 2, 1) and having normal with direction ratio (5, 2, 2) is
`a(X-x_(1)+b(y-y_(1))+c(z-z_(1))=0`
`!5(x+1)+2(y-2)+2(z-1)=0`
`!5x+5+2y-4+2z-2=0`
`!5x+2y+2z-1=0`
`!rdot(5ati+2hatj+2hatk)-1=0`