maths vertices of a triangle

這裏可以幫到你
http://yayantw.test.hansawell.net/yahoo.com.hk/hk/auction/1118037410

cos@ = 〈1/sqrt3 (i + j - k) ‧1/sqrt2 (i + j)〉/│1/sqrt3 (i + j - k) │‧│1/sqrt2 (i + j)│ = 2/sqrt(6), right?
normal vector of xy-plane = k
required angle = angle between the 2 normals.

a. Vector AB = (4,1,3) - (1,3,2) = (3,-2,1)

Vector BC = (0,1,-1) - (4,1,3) = (-4,0,-4)

Vector CA = (1,3,2) - (0,1,-1) = (1,2,3)

Perimeter of the triangle ABC

= │AB│+│BC│+│CA│

= sqrt[(3)^2 + (-2)^2 + (1)^2] + sqrt[(-4)^2 + (-4)^2] + sqrt[(1)^2 + (2)^2 + (3)^2]

= 2sqrt14 + 4sqrt2

= 13.14 units (2 d.p.)


b. Area of the triangle ABC

= 1/2 │AB X BC│ (Cross product of vectors)

│i j k│
= 1/2 │3 -2 1│
│-4 0 -4│

= 1/2 │8i + 8j - 8k│

= 1/2 sqrt(3 X 8^2)

= 4sqrt3 sq. units


c. The normal vector to the plane = 1/sqrt3 (i + j - k) (From part b)

The required angle is 90* - angle between the normal vector and the xy plane.

The unit vector in the xy plane = 1/sqrt2 (i + j)

cos@ = 〈1/sqrt3 (i + j - k) ‧1/sqrt2 (i + j)〉/│1/sqrt3 (i + j - k) │‧│1/sqrt2 (i + j)│

= 1/sqrt3

@ = 54.74*

So, the required angle = 90* - 54.74* = 35.26* (2 d.p.)