Find the Cartesian Equation of the plane passing through P(8, -2,0) and perpendicular to a- 5i+3j-k What is the distance of this plane to the point 0(2,2, 2)? (a) (b)

Answer:equation of plane, 5x+3y-z-36=0Distance of point (2,2,2) from plane = 4.05 unitsStep-by-step explanation:Given,Plane passing through the point = (8, -2, 0)Let's say, [tex]x_1\ =\ 8[/tex]                [tex]y_1\ =\ -2[/tex]                 [tex]z_1\ =\ 0[/tex]Plane perpendicular to the vector, a= 5i + 3j- kSince, the vector is perpendicular to the plane, hence the equation of plane can be given by[tex](5i + 3j- k).((x-x_1)i+(y-y_1)j+(z- z_1)k)=\ 0[/tex][tex]=>(5i + 3j- k).((x-8)i+(y+2)j+(z-0)k)=\ 0[/tex][tex]=>\ 5(x-8)+3(y+2)-z=0[/tex][tex]=>\ 5x\ -\ 40\ +\ 3y\ +\ 6\ -\ z\ =\ 0[/tex][tex]=>\ 5x\ +\ 3y\ -\ z\ -\ 36\ =\ 0[/tex]Hence, the equation of plane can be given by, 5x+3y-z-36=0Now, we have to calculate the distance of the point O(2,2,2) from the plane 5x+3y-z-36=0Let's say,a= 5, b= 3, c= -1, d=-36[tex]x_0=2,\ y_0=2,\ z_0=2[/tex]So, distance of a point from the plane can be given by,[tex]d=\dfrac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}[/tex]  [tex]=\dfrac{\left |5\times 2+3\times 2+(-1)\times 2-36\right |}{\sqrt{5^2+3^2+(-1)^2}}[/tex]  [tex]=\dfrac{24}{\sqrt{35}}[/tex]   = 4.05 unitsSo, the distance of the point O(2,2,2) from the given plane will be 4.05 units.