# Find the distance of a point (2, 5, −3) from the plane - Mathematics

Find the distance of a point (2, 5, −3) from the plane vec r.(6hati-3hatj+2 hatk)=4

#### 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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