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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`

Concept: Distance of a Point from a Plane

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