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Halbrecht
6 months ago

This is the 🙁

.

.

B is A + u

.

xb – 0 = 1

yb – 2 = 3

zb – 3 = -2

ergo

B( 1 / 5 / 1 )

.

For if one had A and B and should u find (so normal vector determination ) one would have to

1 – 0

5 – 2

1 – 3

( 1 / 3 / -2 )

Hamburger02
6 months ago

B = A + u
D = A + v
E = A + E

C = B + v
G = C + w

F = E + u

H = E + v

(b)
Diagonal:

d1 = A – G
d2 = D – F
d3 = H – B
d4 = E – C

Hamburger02
6 months ago
Reply to  Hausbaum32

Of course you have to calculate it. I’ll do it once for point B:

B = A + u = (0/2/3) + (1/3/-2) = (1/5/1)
B(1/5/1)

Halbrecht
6 months ago
Reply to  Hausbaum32

of course: the coordinates