Make a straight line through point C, L perpendicular to AD, with point C as the origin.

L is the x-axis, CD is the y-axis, and CF is the z-axis. The coordinates of points b, d, e and f are: B(0, -a, 0), D (0, a, 0), E (a, -a, 0) and F (0, 0, 2a). Vector EB=(-a, 0, 0), vector FD = (.

So EB vertical FD

Question 2: Let the normal vector of the fed surface be n(x, y, z).

Vector EF=(-a.a, 2a), vector DF=(0, -a, 2a)

The coordinates of the normal vector, the coordinates of the vector EF and the coordinates of the vector DF are combined into two equations.

Solve the coordinates of the normal vector

I figured out that the coordinates of the n vector are n(4a, 2a, a).

The modulus length of n vector is (root number 21) a.

Take point B as a vector, not as a vector of the given plane.

The vector I take is the vector BE(a, 0, 0).

Then the distance d from point b to the plane = vector BE times the modulus length of vector n/n.

The final calculation result is: (0 under 265438+4) a/21.