Could somebody solve this vector problem please?

the equation of the plane is x+3y-2z+d=0 and as it passes through (6,1,-2)

6+3+4+d=0 so d=-13 and the plane is x+3y-2z-13=0

Let P be the point P ( 6, 1, -2) and O be the origin O (0, 0, 0)

Vector ( OP ) = 6 i + j - 2 k , where i, j, and k are the unit vectors in the directions of the

x, y and z axes respectively

Similarly, if A is a point A (1, 3, - 2) , then Vector ( OA ) = i + 3j - 2 k

The length of the Vector OA = √(1² + 3² + 2²) = √14

Unit Vector in the direction OA = (1 / √14)(i + 3 j - 2 k)

Since OA is perpendicular to the the plane, then the unit vector normal to the plane is

( n ) = (1 / √14)(i + 3 j - 2 k)

Now the projection of OP on OA = ( OP ) ∙ ( n ) = (6 i + j - 2 k) ∙ {(1 / √14)(i + 3 j - 2 k)}

= (1 / √14) ( 6 + 3 + 4) = (13 / √14) = Length of the perpendicular from the origin on the plane

Hence the vector equation of the plane is ( r ) ∙ ( n) = (13 / √14)

For the cartesian equation : Put ( r ) = x i + y j + z k and ( n ) = (1 / √14)(i + 3 j - 2 k)

and you get x + 3y - 2z = 13

A plane is a locus .- All vectors in the space that are normal to a given direction .-

If R (x,y,z( is a position vector of the plane ( It is a point on the plane ) , ie

R= x i+yj+zk and passes by a other position vector like P = ai+bj+zk now , A VECTOR on the plane is V= R-P .- This vector must be normal to a given direction , N , so

V dot N =0

(R-P) dot N=0 This is the vectorial equation of a plane .-

This is ,

((x-a) i+(y-b)j+(z-c) k) dot( Nxi+Nyj+Nzk) =0

Nx(x-a)+Ny(y-b)+Nz(z-c)=0

Nx x +Ny y +Nz z - (Nxa+Nyb+Nzc) =0

- (Nxa+Nyb+Nzc)= a cosnstant = D so finally

Nx x+Ny y+Nz z +D =0

is the cartesian eq of a plane .-

Is very important that you note that given a cartesian equation of the plane , the factors of x, y, z are the Normal components of the plane , ie its NORMAL vector .-

In your case , passes by (6,1,-2) , so a vector on the plane is R-(6,1,-2) and is Normal to N(1,3,-2) .- So all is given .-

The plane is (x-6) i +(y-1) j + (z+2) k dot (1,3,-2)=0

/x-6) +3(y-1)-2( z+2)=0

x+3y-2z -13=0