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Question
Find the distance of a point (2, 5, −3) from the plane `vec r.(6hati-3hatj+2 hatk)=4`
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Solution
Consider the vector equation of the plane.
`vec r.(6hati-3hatj+2 hatk)=4`
`=>(xhati+yhatj+zhatk).(6hati-3hatj+2hatk)=4`
6x-3y+2z=4
6x-3y+2z-4=0
Thus the Cartesian equation of the plane is
6x-3y+2z-4= 0
Let d be the distance between the point (2,5,-3) to the plane
Thus, `d=|(ax_1+by_1+cz_1+d)/(sqrt(a^2+b^2+c^2))|`
`=>d=|(6xx2-3xx5+2xx(-3)-4)/(sqrt(6^2+(-3)^2+2^2))|`
`=>d=|(12-15-6-4)/sqrt(36+9+4)|`
`=>d=|(-13)/sqrt49|`
`=>d=13/7 units`
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