Distance of top of pole to wall

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An architectural feature of a building is a flat wall that is angled out over the footpath. Measurements (in meters) of three vertices of the wall were recorded against the x,y and z axes:
x- axis is horizontal, the straight boundary of the footpath and the property on which the building sits.
y- axis is horizontal directed at right angles away from the footpath.
z- is vertical

The co-ordinates of the three vertices are:
A(4,1,0), B(1,-4,11), C(24,-1,9)

a) Give two vectors that lie in the plane of the wall
AB=(-3,-5,11) and BC=(23,-5,-2)
Is this correct?

b) Find a vector normal to the wall
Not sure what this means or how to do it

c) Give the equation to the plane of the plane of the wall
??

d) What angle does the wall make with the footpath
?

A 5m vertical pole stands with its base at the point (15,-3,0)
e) ?

f) What is the shortest distance from the top of the pole to the wall
?

Any help would be great thanks.
 
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hints:
1) draw a diagram
2) normal means perpendicular to
3) review the properties of cross-product and dot-product
 


a)AB(-3,-5,11),BC(23,3,-2)
b)n1=ABxBC=(65,247,130)
c)E1:AB.r=Ab.n(hope you can do further dot product)
d)equation of plane of footpath is E2:z=o.So,n2=(0,0,1).So, angle=cos^-1(n1.n2/m(n1)m(n2))...(m means magnitude)
f)if f(x,y,z)=0 is the of wall then, distance=f(15,-3,0)/m(15,-3,0)...
i am not sure about the distance.....
hope this helps you...
 


I think he directly saw the quetions...never saw the rules and formulas...
 
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