Let E to be the origin ,

and EF be x-axis , EH be y-axis , EA be z-axis .

Then we have the following coordinates:

A(0,0,8) , F(8,0,0) , G(8,8,0) , H(0,8,0) , M(0,4,4)



(a)

FM

= ∣(-8,4,4)∣

= √ [ (-8)^2+4^2+4^2 ]

= 4 √ [ 2^2+1^2+1^2 ]

= 4√6 ...Ans



(b)

N

= A + s( vector AG ) , for some s

= (0,0,8) + s(8,8,-8)

= (0,0,8) + t(1,1,-1) , let t = 8s

= (t,t,8-t)

So, NM = (-t,4-t,-4+t)



NM．AH

= (-t,4-t,-4+t)．(0,8,-8)

= 8(4-t) - 8(-4+t)

= 64 -16t

= 0

Hence, t = 4 , N = (4,4,4)



FN

=∣(-4,4,4)∣

= √ [ (-4)^2+4^2+4^2 ]

= 4 √ [ 1^2+1^2+1^2 ]

= 4√3 ...Ans



(c)

FA = (-8,0,8) , FH = (-8,8,0)

FA×FH = (-64,-64,-64) , where × is "cross product"

And (-64,-64,-64) is the vector perpendicular to the plane AFH

GA = (-8,-8,8) , GH = (-8,0,0)

GA×GH = (0,-64,-64) , the vector perpendicular to the plane AGH

Suppose the angle between the plane AFH and AGH is θ

(-64,-64,-64)．(0,-64,-64) = ∣(-64,-64,-64)∣ ∣(0,-64,-64)∣ cosθ

2 * 64^2 = 64√3 * 64√2 * cosθ

cosθ = 2 / √6 = √6 / 3



θ

= arccos (√6 / 3 )

≒ 0.615479708670387 rad

= ( 0.615479708670387 * 180 / π )°

≒ 35.2643896827547 ° ...Ans

a)

FM² + AM² = AF²

FM² + 8²/2 = 8² + 8²

FM² = 96

FM = 4√6



b)

NM ⊥ AH and GH ⊥ AH so NM // GH then N is the mid point of AG.

(2FN)² = DH² + FH²

4FN² = 8² + (8² + 8²)

FN² = √48

FN = 4√3



c)

The required angle = ∠NMF.

cos∠NMF = (NM² + FM² - FN²) / (2NM × FM)

cos∠NMF = (4² + 96 - 48) / (2 × 4 × 4√6) = 2/√6

∠NMF = 35.264... = 35.3° (3 sig. figures)